The thing is, for a fair comparison, even with radius 1 you need to allow x and y to be complex. a^2-b^2+c^2-d^2+2(ab+cd)i=1 This basically makes the projections the same family of slices as with a radius of i, just rotated/from a different perspective. What this is beginning to hint at is the idea that in the complex projective plane, *all* conics are basically the same shape: ellipses, parabolas, circles of radius 1 or i, etc.
There are still three cases, once you include degeneracies: the conic could degenerate to the union of two crossing lines (e.g. x^2 = y^2) and even worse the two lines could fall atop one another (e.g. x^2 = 0). But the ones defined by quadratic equations _that don't factor_ are all related by projective coordinate transformations. Note that @diribigal is using "complex plane" here (quite correctly) to mean a space with two complex coordinates, not one. What people usually call the complex plane, i.e. the real plane, would be better called the complex line. @quantranhong1092 On a complex manifold we can always multiply our thumb by i and point our finger that direction to figure out what's right-handed, so no we won't meet nonorientable surfaces.
Kinda hearkens back to the notions of ancient Greek philosophers and "Ideal" versions of which all tangible versions are just lesser replicas. All "chairs" are degrades copies of the idealized spiritual form of "Chair" which can only exist in a non-physical fashion. There's only *one* of this particular curved shape that exists in a higher complex space. And with it, we can intersect it into lower spaces in different ways, taking "slices" out of it to represent all the conic "sections" (circles, ellipses, parabolas, hyperbolas, etc.) But, at its core, it's *still* the one singular thing that continues to exist in its perfect ideal form even when it isn't really intersecting the real plane in a way that's meaningful to *us.* It's not much different from realizing that the "root" of a polynomial can be an imaginary/complex value because it "intersects" the X-Axis sort of "floating above and below" the plane of the graph that we had been used to working with.
Can you make 4 different views, where each view locks respectively "a", "bi", "c" & "di". Then you display each view simultaneously in the video, sync up each of them & increases the "locked" number, so each view changes "in synchronization". I have played a game called: "4D miner", and have a somewhat rough idea on how to "understand" 4D spaces, and this visualization might help. Link in next reply (in case UA-cam decides to be stupid & auto-delete that comment, just because it contains a link, even to UA-cam itself.)
Complex Analysis is an amazing branch of mathematics that opens up vistas into structures like: *Conformal Mappings *Riemann Surfaces *Fractals *Chaotic Attractors *Hilbert Spaces The thing about these topics too, is that there's a beautiful 'visual' aspect to them, despite the advanced, underlying mathematics governing their structures. I'd suggest to any serious undergrad pursuing Math to go the distance with C.A. it's WORTH it!! 🤔
@@rickwilliams967 I've seen some really creative applications of conformal mappings in figuring out electric capacitances of conductors with complicated shapes And fractals like the Sierpinski triangle are sometimes used for wideband antenna design, but I'm not sure how much detailed analysis is involved in that case.
@@rickwilliams967 To an extent, u are right, as a lot of high-end Math eg. PhD, MPhil level is more of an abstract, non-physical applicability in terms of real-world processes BUT....that's the sheer beauty of the discipline as it advances into uncharted mathematical frontiers, whether practical or not. From this example alone, just look at the whole new world of discoveries created by the introduction of the imaginary unit 'i' & the extension of the number system to have crazy ideas like 0=0+0i, 1=1+0i, etc. So with all due respect, I wouldn't necessarily say it's a 'complete' waste. That's why C.A. is my personal fave in all advanced Math Subject classifications! & I even forgot to mention 'Quaternions,' a brilliant application leaning into another crazy topic of 4D Space...🤣🤣
@@rickwilliams967 you can not be sure what is practical and what is not. any mathematical subject can be suitable for use in any unexpected scientific case. also mathematicians do not care if something they work on is practical or not
This reminds me of one triangle I "invented". Side lengths are: -1, i, and 0, all of which are impossible. Yet, it is a right angled triangle (by converse of pythagorean theorem).
Unfortunately it doesn't, the Pythagoras theorem doesn't generalize this easily to the comples place - the c^2 = a^2 + b^2 version only works in euclidian real space. The general formula would be ||v+w||^2 = ||v||^2 + ||w||^2, where ||x|| is the "norm" or colloquially length of the vector. (note that v and w are vectors such that v = (a, 0) and w = (0, b), a triangle has to be in a plane, not a line, so it's actually the R^2 case, it just simplifies because we assume the triangle starts at the origin - the other coordinates are 0) Mathematically, the length of a vector (abstractly scalar is just a 1x1 vector) is defined as the square root of the inner product of the vector with itself (also called dot or scalar product in euclidian space). In the R^2 case above the inner product is : = v_1*w_1 + v_2*w_2, or in our specific simplified case : ||a|| = = a*a + 0*0 = a^2. The same process goes for ||b||. If we then plug them into the general formula we get the familiar ||v+w||^2 = a^2 + b^2 To maintain some integral properties of the inner product, angles and norms (most notably in this case that it always has to be positive, since it's the unsigned distance from origin geometrically), the complex case of inner product is instead = v_1(w_1)'+ v_2(w_2)', with the (w)' being the complex conjugate of w - which means the imaginary part has the opposite sign. If we then apply this to your example of (i, 0) we then get : i*(-i) + 0*0 = 1, which then results in the general formula giving sqrt(1^2 + 0^2) = 1, not -1
Thank you. You may not have intended this, but you made my brain click, and now I have intuition into why you need sinh and cosh for transformations in Special Relativity.
In the olden days, they used to formulate special relativity by using Pythagorean formulae but with an imaginary time coordinate "ict". That formalism is out of fashion now (too much opportunity for confusion), but it does make it clear that the geometry of special relativity can be thought of as continuing Euclidean geometry into a complex world. There was a sort of mathematical joke going around a while back showing a right triangle with the legs labeled "1" and "i" and the hypotenuse "0". In Minkowski's formulation of special relativity, that's just a light-like interval.
@@MattMcIrvin I still do that when teaching SR to mathematically inclined but otherwise lay audiences. I don't use it as a formalism, however. Not because it's confusing--pretty much all ways of dealing with SR are confusing, especially that relativistic mass stuff--but because I prefer quaternions. Essentially, space rather than time is imaginary. It works out the same if you remember where to put the minus signs. I can remember that a lot easier than the right hand rule, and when I look like I'm counting on my fingers, people look at me funny. This also seems to fit better with the Minkowski metric of [1, -1, -1, -1] used as the diagonal in GR. Using the Cayley-Dickson numbers also provides an interesting possibility of evolution of the universe. First there was time alone: real. Then came Everett spacetime with one dimension of space. Then quaternions for GR space-time. Then sedenions can be used to embed all the dimensions needed for string formulations (like Susskind, I don't consider strings a theory but rather a fairly nice bunch of mathematics in which theories can be represented). Maybe octonions are for some simplified version of string mathematics, but I don't know if such has been formulated. It would be nice to find a use for trigintaduonions, because it's such a great name. But I digress. Back in the 1980s, when I was on the faculty of FSU with the Supercomputer Computations Research Institute, we had an unusual view of scientific visualization. Everywhere else, such as the NSF centers, assembled _renaissance teams_ of people using user-hostile programs such as AVS and APE, who would take almost a year to make a movie. Our attitude was to do the visualizations first. So I wrote an influential but now forgotten package called SciAn that threw away all that icky visual dataflow stuff and was inspired by MacDraw for the UI, sort of. I had a mini AI that could figure out how to do some visualization just from the data. Whether this turned out to be the best one or not didn't matter; the point was to get something up on the screen quickly to excite the user, who could then change and refine it. I could spend 30 minutes showing a scientist how to use it, and they'd go off and do something that surprised me. Anyway, it worked, and ideas from it got into the genome of visualization programs prehistorically. I still see QCD visualizations in videos that look as if they were done in SciAn, whether they were or not. The visualization here made my brain click exactly the same way we wanted to do, which is why it impressed me so much.
The complex solutions to x^2+y^2=-1 (or actually, to any quadratic that doesn't have a double root) really form a sphere in terms of topology. If you rewrite it as (x+iy)(x-iy)=-1 and make a substitution u=x+iy,v=x-iy, then we're now looking at solutions to uv=-1. For each nonzero value of u, we get exactly one value of v. That means we can essentially ignore v and just think about the values of u as determining points on the curve. But u can take any complex value, so that just gives a new copy of the complex plane. It's just like how with real numbers, the graph of y=x^2 gives one y value for each x value, so we get a "bent" copy of the real line, in the form of a parabola. So the solutions form a plane with a point removed to avoid u=0. But in some sense, that's just like a sphere. One little detail: if we're working in projective space, it's okay to let u=0, and we say that it gives a "point at infinity" on the curve, coming from division by 0. Without the point for u=0 and the point at infinity, a plane is the same as a sphere with two holes in it: take the points out of a sphere, making two holes, then flatten what's left to get a plane with the hole at u=0. But when we include those points, we patch the holes, and we get... at long last, a sphere.
I don't agree. Topologicaly, the solutions to a quadratic equation in the complex affine plane form a sphere with 2 holes, that is a cylinder. As you said, the solutions in the complex projective space are homeomorphic to a a sphere, but there two points of intersection with the line at infinity, so the intersection with the affine plane consists in a sphere minus two points.
My first thought upon looking at this is to wonder what the area of this circle would be. For r=i and A= πr^2 this means A=-π. Also, the circumference would be π2i. So, the area is a real number and the circumference is imaginary.
I think the first thing to do, would have been to look at x^2 + y^2 = t^2 while allowing x and y to be complex, with r real and non-zero, before making r imaginary. Doing this, I believe you would find that you get essentially the same shape either way. Indeed, take the set {(x,y) in C^2 | x^2 + y^2 = 1^2 } For and (x,y) in this set, and any complex number r, (r x)^2 + (r y)^2 = r^2 . So, for different (non-zero) values of r, the shapes you get are the same, except for scaling and rotating (of phase for both x and y coordinates) . So, In this sense, The shape of a circle with radius i, is the same as the shape of a circle of any radius.
The 4d Minecraft project really helps with imagining actual 4th dimensional structures. Actually seeing a 3 dimensional slice turning around in 4th dimensional space really helped me understand 4th dimensional structures. As well as a 2 dimensional slice turning throughout 3 dimensional space as a comparison. Also you were so close to a 3 minute 14 second video!
well that would be a boring video as i by definition is just 1. Just rotated by 90 degrees. so it is just 1 in the y direction. And that is just a circle with radius 1.
Another way to think about purely real part of this object is how it acts on real points. We know that circles act on points via inversion (OX*OY=R^2). This object acts like a composition of inversions in 3 mutually orthogonal circles, or, alternatively as a composition of inversion in a circle and then point-reflection in the circle's center. One notable property is that action of purely imaginary circle has no real fixed points. It also appears as a second bisector between 2 non-intersecting circles.
I think it is well known that in the complex plane ^2, all the conic sections are essentially the same formula a(x-b)^2+(y-c)^2=0 already in ellipse form. transform y to yi and you get the hyperbola. with circles and parabolas being limit/special cases even normally. To make this formal you have to introduce points at infinity and the projective complex plane. Then all conic sections well be sections of the same shape.
After some reflections, I have understood that a circle in imaginary plane is just a specal case, not the general one. If (a+b*i)^2+(c+d*i)^2=i^2, a and c can be any real numbers, as long as a^2 < b^2, c^2 < d^2 and 2*a*b*i+2*c*d*i=0 If a=b=0, it's just a circle on imaginary plane This means that the complex shape x^2+y^2=i^2 projects on the real plane as the whole plane
Dear Ron & Math, there is also a nice way to visualize a surface in 4D space. You take two squares which will represent two variables. Left square will be x = a+bi, where a is horizontal axe and b is vertical axe. Right square will be y = c+di. Now you start taking some points on left square and paint them. Take for example 0+0i and paint it red. Now we find all solutions of x^2+y^2=-1 where x=0+0i, it is y= 0+1i and 0-1i. So we paint this points red on right square. Then we take another point on left square, say x=0+1i, paint it green. Find a solution, it is y=0+0i, paint it green on right. And so on and on.
Cool, the projection is a 3D Penrose diagram! I like to think that the resulting 4D shape is a good description of the shape of space-time near a black hole.
The circle becomes a twisted line. Like when you take a 3d piece of paper put a twist in it and glue the ends in become a 2d piece of paper. All in the Z plane, it comes down the y wraps around the origin and continues up the x in that arc trajectory.
when you pulled up x^2 + y^2 = -1 my immediate thought was that its the same as a circle with r=1, except the regions considered inside and outside are swapped
I was just thinking about this topic in context of constructorial spaces where time can be described as measurement potential. Measurement as indicated by complex numbers and potential indicated as imaginary numbers. Another way to describe time as being structural information visible against infinity like the event horizon of a black hole. Might also look at this kind of algebra as a model for the relationship between calculus (point space) and geometry (cube) as some sort of harmonic oscillator where by the corners of such geometries cannot exist at the same time as the connective lines between them. Could say something about angular momentum in particle physics and also the nature of the universe. All considerations for experimental design. Thanks so much Ron. Skippy is just back from her walk on the beach so she’s just going to be so excited about that as the girly woofs is also part of my science team.
Having a negative radius doesnt change the shape of a circle in a practical sense. It just changes the angles observed. Many people might just take the absolute value of the radius as an argument. The complex continuation of absolute value is modulus and modulus of i is just 1. The angles would be weird if you leaned into that aspect of it but yeah
I'm not an English-speaker, so excuse my explanation. If you have a line, draw a 3 cm segment, then add 4*i cm, so how many cm would you get and where they would be? The best answer is that on the line there would be 3 cm, but on the plane -- 5 cm so that the projection of 5 cm segment on the line is the same 3 cm. Furthermore, if you draw a line and try to mark on it i cm segment, on the plane there will be 1 cm segment which wll be projected on the line as a point A circle is a plain shape, its projectons on x and y axes are line segments with length of 2*r. If r = i, the projections will be just 2 points, so it would be exactly 1 point on the real plane. But on imaginary plane there will be just a simple circle with |r| = 1
"Why do we consider that there is an imaginary axis orthogonal to each real axis? I believe there is only one imaginary axis orthogonal to all real dimensions."
I think that, for me to be confident calling something a circle with radius=i, I would want some function to be called a "distance function" and take a locus of points with "distance"=i from the centre. But I don't know if it's possible to adapt the requirements for a distance function to having a complex output in a sensible way. Maybe for points in a higher dimensioned space, idk.
A question. When visualizing 4D objects can’t you just choose one of the dimensions to be time, and the other three are space. Then the 4D shape would just be an animated 3D object that we can see?
Once upon a time I wondered: with y²+x²=r², what are the y values beyond |x| > r∈ℝ+ ? And saw that it was a hyperbola connected to and extending out from the sides of the graphed circle, but "perpendicular" to the plane, as if the graphed x and y were each real with their own complex planes. Of course there's another hyperbola above and below, when looking at it with respect to x. y²+x²=-1 simply exists completely outside the strictly real part of this "complex 2D" graph.
It's probably easiest to solve for y as a function of x and then plot Abs(sqrt(-1-x^2)) over the complex plane and then color the output according to Arg(sqrt(-1-x^2)) this way you visualize almost the entire function at once.
It seems almost fibrative.. like a bundle of strings twisted and the ends go off in every direction to form circles that are circumferential to 3d spheres. The state of all of them being the 4d hypersphere, while simultaneously the larger the imaginary component the more space between these bundles moving more into imaginary space, while at the same time the smaller the imaginary component the tighter these fibers and twists get until the inflection point where b->0 and then the imaginary component collapses, the twists get so tight they cinch, and then the radius briefly becomes zero then undefined
This is so cool!! The equations for circles/spheres/n-speres define only the locus of points on the "outer" "shell" of the points they enclose, which are really disks/balls/n-balls. So for these n-spere loci, there will always be a geodesic path between any two points in the set such that the geodesic is coincident with a great circle, which is always a closed loop. In this case of an imaginary radius, the hyperbola extends to infinity, but mathematically, would it still obey this principle that for any pair of points, there exists a great circle containing the geodesic between those points? The 4-hyperboloid shape extends to two opposite infinities, but perhaps under a different projection, like the real projective plane (or complex projective "plane" if that's a thing??) this principle could be shown? I'm way out of my depth here but it would be great if you (or some other commenter) could pitch in!
Thank you very much @ayane_m but geometry is really not my strength. I made this video up to a point that I am still 100% certain what I am talking about. I will sit with you and wait here as well.
Yes, the complex projective plane is a thing, and under the real projective plane, hyperbolas are fully connected across points at infinity, so I image the same would apply to hyperboloids in the complex projective plane.
@@ron-math I'd say you really stretched the definition of a circle in this one, but still, when you set the real parts to 0 you had your pure imaginary circle of radius 1 in imaginary space alone. Because everyone and their mother knows that i is 1 just shifted 90 degrees.
I did some work on this and I think I found a very satisfying way to visualize the full complex circle in 3d, but only under some interpretation. Change of variables from x,y to r, θ, with both being complex, using the normal polar coordinates formulas. This works because sin and cos are analytic. Now you very simply have r^2 = R^2, which just means the two parts of r (real and imaginary) are fixed and uninteresting. So you can scale by R and just use the unit circle and focus on the real and imaginary parts of θ and how they affect x and y. Now, if you let r=R=1, you can get (from the polar coordinates transform equations) equations for the real and imaginary parts of x and y in terms of the real and imaginary parts of θ. This is a two dimensional parametrized surface in four dimensions. You will see that the real parts of x and y form a circle parametrized by the real part of theta, with radius equal to cosh of the imaginary part of theta. Similarly, the imaginary parts of x and y form a circle with radius equal to sinh of the imaginary part of theta. So when the angle θ between (x, y) and the x axis is real, you get a real unit circle. As you change the imaginary part of θ you start to get a larger real circle, plus a pure imaginary circle which starts at zero size and gets bigger as θ gets more complex. These both get one turn as the real part of θ goes from 0 to 2π. The best way i can think of to visualize this is a torus with primary radius cosh(Im(θ)) and secondary radius sinh(Im(θ)). The primary radius is the real parts of x and y, the local coordinates relative to the secondary radius give the imaginary parts of x and y. You trace out a spiral around both radii of the torus simultaneously as the real part of θ changes. The real part of the solution is where you are at the secondary center of the torus, which forms a flat disk extending out from the real circle. The imaginary part is like a Moebius strip, but with a full twist instead of a half, unfortunately 😅. I have some analysis and some graphs I can share by other means that show the result. It's a very nice concept: the real unit circle is the part of the complex unit circle with no imaginary components in either coordinate. All the other stuff grows from that in two sheets outwards: one is a flat disk and the other is a cool Moebius strip-like structure.
I think it would be pretty easy to take the parametric equations for the surface with R=1 and scale and flip them for different values of R. A purely imaginary circle will swap the primary and secondary radius of the torus (think a torus fatter than it is around) and a complex number will mix the contributions in a way that might yield some interesting visualizations under this interpretation.
Taking into account trigonometrical form of i and how it is defined - then circle of r = i, is same as circle of r = 1. Overall circle is infinite root (sum of all roots) of i
Very cool problem and video! Though it IS actually possible to plot curves or surfaces like this in 4D, projected onto a 2D screen, if you know how. I solved the this parametrically, as the solution is a parametric surface with two parameters, embedded in 4D. I won't go through the solution here but you can verify it solves both constraints: a = sinh(t)cos(theta) c = sinh(t)sin(theta) b = cosh(t)sin(theta) d = -cosh(t)cos(theta) Think of (a,c) - the two real parts - as a point lying on the "real plane" and (b,d) as lying on the "imaginary plane". Separately, both points lie on circles in their respective planes (theta varying from 0 to 2*pi), but for any value of theta, the two points will be 90 degrees apart in their respective planes. The two circles have different radii of sinh(t) and cosh(t). In the 4D plot, for any value of t you will have a circle lying in 4D space. The special solution a = c = 0 corresponds to when t = 0. This is a pure unit circle in the imaginary plane. But as you increase t, the two radii both expand exponentially, and are essentially equal in the limit. The circle rotates from the imaginary plane to about 45 degrees between the real and imaginary planes, and it's radius grows exponentially. So in the limit, the growing circle generates a plane. Picture the solution like this: a flat film of soap, bounded by a large circular ring and a small circular ring in the middle, which is empty. Now rotate the inner ring 45 degrees to the plane, and the soap film will adjust accordingly. Now make the larger ring have infinite radius, and that's your surface. If you're interested in seeing a picture or animation, or how to plot in 4D in general, let me know. Subscribe to my channel as I will be putting a lot of 4D visualization content soon. I'd love to do a response video to this problem.
Distance is always taken to it's absolute value. You can't have a circle with radius negative 1 because negative length doesn't exist. Similarly, a radius of i also can't exist. In either case, you would use the absolute value. And the absolute value of i is 1 Negative distance isn't a thing, and neither is negative area. Distances are absolute, and i has a distance of 1
It appears that the "negativeness" of i² has transformed circle in hyperbola. Hyperbola is curve DIFFERENCE of squares, so the next question is: What happens if we take equation for hyperbola & introduce imaginary into it? Will it create a circle?
@@rickwilliams967 Well I suppose that if mathematicians tried to work on those things they’d fail horribly as they’re not trained nor have the interest or even the skill for that. Did you know that much of mathematics used in both of those applications was invented because people were doing just that- discovering mathematics? Let us be, what we do is probably gonna be useful in the next 200 years or so lmao
Since both x and y are complex, both would be points on the imaginary plane, so that makes me wonder what you would get for each point x that you select on that plane as points (0 or more?) y on the same plane that would fulfill the circle equation for radius i. From the various 3D projections of a 4D shape that I still cannot wrap my mind around, I have no idea what that mapping from x to y points that fulfill the equation would be. Will there always be 2 y points for each x? Which points x have valid y values and which do not? I feel this video could go way deeper into all this.
The first thing I thought was that for a circle of radius, r = i area would be pi*r² , That is, Area = -pi. (And an object with a negative area was enough to get me to doubt my own existence). And If we add the area of a circle with r = 1 to the area of a circle with r = i then The total area would be 0. And if we did that with three dimensional objects, then they cancel out each other's volume or existence, And maybe antimatter is just matter with dimensions that are measured in units of i. And that's why matter / antimatter pairs annihilate each other out of existence. 🤔.
imaginary numbers having much to do with 4D objects and being very applicable to our 3D space makes me wonder if we truly live in a 3D space or are simply the 4D equivalent of living on a screen Somehow, a whole bunch of people who are way better at physics and math than me took interest in my comment. Thanks for being here, I guess?
Correct me if im wrong, but iirc by calculating the energy loss of gravitational waves through space, it was calculated that we should pretty much only be in a 3+1D space. unless the extra dimensions are extremely "small", whatever that means (like string theory tries to suggest)
Lots of physics has a "complex number"-like structure! you see it in relativity where you can get negative norms, and in quantum (not completely unrelated if you're talking about the dirac algebra). When it's connected to relativity it's a manifestation of the constancy of the speed of light. Although mostly what we see is three real-like dimensions and one imaginary-like one (or the reverse) as opposed to this two-real two-imaginary structure, about the only thing I can think of are spinors which to my knowledge are sort of connected to quaternions but I know almost nothing about them. I'd suggest looking into geometric algebras if you want your mind to be really blown! Almost all of the complex numbers that show up in physics can be thought of as stemming from some sort of geometric algebra afaik. They're used in modern theoretical treatments of physics.
Simple question: how is this possible? From what I know, |z - zₒ| = r is equation for a circle... which IMMEDIATELY leads to CONTRADICTION, with r=i, as on the left side of equation is complex number's modulus (i.e. real number), while on the right side we have imaginary number. There can't possibly be solution for this. What am I missing??
When I imagine a circle, it's still a circle... It's not like imaginary numbers where the number I imagine is always different to everyone elses.. A circle is a circle, end of story.
While you Aren't wrong, The way One visualizes A concepts Can change, When exposed To different Dimensions. A circle Maybe a Circle, but What would That actually Look like, Based on The math? The method You use To visualize, And articulate the math You want To communicate, Can result On different Results.
@@dr.blockcraft6633 when I was about 10. I actually discovered how to use sin and cos to draw circles pixel by pixel... I inadvertently had to find the value of pi*2.. In my adult life I wanted faster circles, So I copied some code of Wikipedia that draw circles really fast without sin and cos... You actually get different circles... There are probably other ways to draw circles, and they probably give slightly different pixel results.. Is this what you mean ?
Your title said, "Shape Of An Imaginary Circle" so I naturally assumed in the plane only. Sooo...I guessed it would be a circle of radius i in the complex plane. But you went to 4 dimensions without a motivation or justification.
i is Real in a particular 4D algebra that models 2D geometry. In this particular modeling, i becomes the Real origin, rather than an orthogonal direction, and it's actually the product of two orthogonal lines through the origin. I will have to mention though that this particular 4D algebra is not the quaternions. (It might be the split quaternions though, but I'm not certain since it's constructed with a completely different technique.)
The thing is, for a fair comparison, even with radius 1 you need to allow x and y to be complex. a^2-b^2+c^2-d^2+2(ab+cd)i=1 This basically makes the projections the same family of slices as with a radius of i, just rotated/from a different perspective. What this is beginning to hint at is the idea that in the complex projective plane, *all* conics are basically the same shape: ellipses, parabolas, circles of radius 1 or i, etc.
maybe it can be a 4d mobius strip kind of surface instead of a normal circle
@@quantranhong1092the surface you get should still be orientable I think
@@quantranhong1092there is no difference
There are still three cases, once you include degeneracies: the conic could degenerate to the union of two crossing lines (e.g. x^2 = y^2) and even worse the two lines could fall atop one another (e.g. x^2 = 0). But the ones defined by quadratic equations _that don't factor_ are all related by projective coordinate transformations.
Note that @diribigal is using "complex plane" here (quite correctly) to mean a space with two complex coordinates, not one. What people usually call the complex plane, i.e. the real plane, would be better called the complex line.
@quantranhong1092 On a complex manifold we can always multiply our thumb by i and point our finger that direction to figure out what's right-handed, so no we won't meet nonorientable surfaces.
Kinda hearkens back to the notions of ancient Greek philosophers and "Ideal" versions of which all tangible versions are just lesser replicas. All "chairs" are degrades copies of the idealized spiritual form of "Chair" which can only exist in a non-physical fashion.
There's only *one* of this particular curved shape that exists in a higher complex space. And with it, we can intersect it into lower spaces in different ways, taking "slices" out of it to represent all the conic "sections" (circles, ellipses, parabolas, hyperbolas, etc.) But, at its core, it's *still* the one singular thing that continues to exist in its perfect ideal form even when it isn't really intersecting the real plane in a way that's meaningful to *us.* It's not much different from realizing that the "root" of a polynomial can be an imaginary/complex value because it "intersects" the X-Axis sort of "floating above and below" the plane of the graph that we had been used to working with.
its a circle obviously
That's not how I imagined it
@@JonWilsonPhysicsbe realistic, don't make it complex.
@@onradioactivewavesBut it being complex can also be real :)
Well if a 2D projection of 3D is real, then my waifu is real.
@@gdmathguy well I would make that statement retrospectively, but that would be convoluted.
The circle became more complex than i imagined!
than i*
i see what you did there!
😂
Ask Marc to add it to 4D Toys, then you can visualise it much more easily
I will pin the most accessible, intuitive explanation when it appears in the comment section.
Someone got an idea of i-metric music with quaternion frequency
Now this seems to be more possible than I thought
Can you make 4 different views, where each view locks respectively "a", "bi", "c" & "di".
Then you display each view simultaneously in the video, sync up each of them & increases the "locked" number, so each view changes "in synchronization".
I have played a game called: "4D miner", and have a somewhat rough idea on how to "understand" 4D spaces, and this visualization might help.
Link in next reply (in case UA-cam decides to be stupid & auto-delete that comment, just because it contains a link, even to UA-cam itself.)
4D miner, how to build a 4D room: ua-cam.com/video/PtKBwrIIddI/v-deo.html
i already am top comment, its not necessary
Did you make a parametrzation for a generic solution?
Complex Analysis is an amazing branch of mathematics that opens up vistas into structures like:
*Conformal Mappings
*Riemann Surfaces
*Fractals
*Chaotic Attractors
*Hilbert Spaces
The thing about these topics too, is that there's a beautiful 'visual' aspect to them, despite the advanced, underlying mathematics governing their structures.
I'd suggest to any serious undergrad pursuing Math to go the distance with C.A.
it's WORTH it!! 🤔
Also, a complete waste of time. Practically speaking.
@@rickwilliams967 I've seen some really creative applications of conformal mappings in figuring out electric capacitances of conductors with complicated shapes
And fractals like the Sierpinski triangle are sometimes used for wideband antenna design, but I'm not sure how much detailed analysis is involved in that case.
@@rickwilliams967"Practicality" is myopic though :)
@@rickwilliams967 To an extent, u are right, as a lot of high-end Math eg. PhD, MPhil level is more of an abstract, non-physical applicability in terms of real-world processes BUT....that's the sheer beauty of the discipline as it advances into uncharted mathematical frontiers, whether practical or not.
From this example alone, just look at the whole new world of discoveries created by the introduction of the imaginary unit 'i' & the extension of the number system to have crazy ideas like 0=0+0i, 1=1+0i, etc. So with all due respect, I wouldn't necessarily say it's a 'complete' waste.
That's why C.A. is my personal fave in all advanced Math Subject classifications!
& I even forgot to mention 'Quaternions,' a brilliant application leaning into another crazy topic of 4D Space...🤣🤣
@@rickwilliams967 you can not be sure what is practical and what is not. any mathematical subject can be suitable for use in any unexpected scientific case. also mathematicians do not care if something they work on is practical or not
In order to see the shape of an imaginary-sized circle, you must first see that a circle, itself, is more than you think it to be.
This reminds me of one triangle I "invented". Side lengths are: -1, i, and 0, all of which are impossible. Yet, it is a right angled triangle (by converse of pythagorean theorem).
Unfortunately it doesn't, the Pythagoras theorem doesn't generalize this easily to the comples place - the c^2 = a^2 + b^2 version only works in euclidian real space. The general formula would be ||v+w||^2 = ||v||^2 + ||w||^2, where ||x|| is the "norm" or colloquially length of the vector. (note that v and w are vectors such that v = (a, 0) and w = (0, b), a triangle has to be in a plane, not a line, so it's actually the R^2 case, it just simplifies because we assume the triangle starts at the origin - the other coordinates are 0)
Mathematically, the length of a vector (abstractly scalar is just a 1x1 vector) is defined as the square root of the inner product of the vector with itself (also called dot or scalar product in euclidian space). In the R^2 case above the inner product is : = v_1*w_1 + v_2*w_2, or in our specific simplified case : ||a|| = = a*a + 0*0 = a^2. The same process goes for ||b||. If we then plug them into the general formula we get the familiar ||v+w||^2 = a^2 + b^2
To maintain some integral properties of the inner product, angles and norms (most notably in this case that it always has to be positive, since it's the unsigned distance from origin geometrically), the complex case of inner product is instead = v_1(w_1)'+ v_2(w_2)', with the (w)' being the complex conjugate of w - which means the imaginary part has the opposite sign.
If we then apply this to your example of (i, 0) we then get : i*(-i) + 0*0 = 1, which then results in the general formula giving sqrt(1^2 + 0^2) = 1, not -1
The converse of a true statement is not necessarily true. They are not logically equivalent.
@@hannahupperman3972the converse of P.T. is, however, true
It would have been iconic if this video would have been of 3 minute 14 seconds.
Thats what i thought too😂
Even better if it had been sightly more...
i-conic
Thank you. You may not have intended this, but you made my brain click, and now I have intuition into why you need sinh and cosh for transformations in Special Relativity.
In the olden days, they used to formulate special relativity by using Pythagorean formulae but with an imaginary time coordinate "ict". That formalism is out of fashion now (too much opportunity for confusion), but it does make it clear that the geometry of special relativity can be thought of as continuing Euclidean geometry into a complex world.
There was a sort of mathematical joke going around a while back showing a right triangle with the legs labeled "1" and "i" and the hypotenuse "0". In Minkowski's formulation of special relativity, that's just a light-like interval.
@@MattMcIrvin I still do that when teaching SR to mathematically inclined but otherwise lay audiences. I don't use it as a formalism, however. Not because it's confusing--pretty much all ways of dealing with SR are confusing, especially that relativistic mass stuff--but because I prefer quaternions. Essentially, space rather than time is imaginary. It works out the same if you remember where to put the minus signs. I can remember that a lot easier than the right hand rule, and when I look like I'm counting on my fingers, people look at me funny. This also seems to fit better with the Minkowski metric of [1, -1, -1, -1] used as the diagonal in GR.
Using the Cayley-Dickson numbers also provides an interesting possibility of evolution of the universe. First there was time alone: real. Then came Everett spacetime with one dimension of space. Then quaternions for GR space-time. Then sedenions can be used to embed all the dimensions needed for string formulations (like Susskind, I don't consider strings a theory but rather a fairly nice bunch of mathematics in which theories can be represented). Maybe octonions are for some simplified version of string mathematics, but I don't know if such has been formulated. It would be nice to find a use for trigintaduonions, because it's such a great name.
But I digress. Back in the 1980s, when I was on the faculty of FSU with the Supercomputer Computations Research Institute, we had an unusual view of scientific visualization. Everywhere else, such as the NSF centers, assembled _renaissance teams_ of people using user-hostile programs such as AVS and APE, who would take almost a year to make a movie. Our attitude was to do the visualizations first. So I wrote an influential but now forgotten package called SciAn that threw away all that icky visual dataflow stuff and was inspired by MacDraw for the UI, sort of. I had a mini AI that could figure out how to do some visualization just from the data. Whether this turned out to be the best one or not didn't matter; the point was to get something up on the screen quickly to excite the user, who could then change and refine it. I could spend 30 minutes showing a scientist how to use it, and they'd go off and do something that surprised me. Anyway, it worked, and ideas from it got into the genome of visualization programs prehistorically. I still see QCD visualizations in videos that look as if they were done in SciAn, whether they were or not.
The visualization here made my brain click exactly the same way we wanted to do, which is why it impressed me so much.
This feels like asking the question "of this scalar thing was a vector, what would happen?"
The complex solutions to x^2+y^2=-1 (or actually, to any quadratic that doesn't have a double root) really form a sphere in terms of topology. If you rewrite it as (x+iy)(x-iy)=-1 and make a substitution u=x+iy,v=x-iy, then we're now looking at solutions to uv=-1. For each nonzero value of u, we get exactly one value of v. That means we can essentially ignore v and just think about the values of u as determining points on the curve. But u can take any complex value, so that just gives a new copy of the complex plane. It's just like how with real numbers, the graph of y=x^2 gives one y value for each x value, so we get a "bent" copy of the real line, in the form of a parabola.
So the solutions form a plane with a point removed to avoid u=0. But in some sense, that's just like a sphere. One little detail: if we're working in projective space, it's okay to let u=0, and we say that it gives a "point at infinity" on the curve, coming from division by 0. Without the point for u=0 and the point at infinity, a plane is the same as a sphere with two holes in it: take the points out of a sphere, making two holes, then flatten what's left to get a plane with the hole at u=0. But when we include those points, we patch the holes, and we get... at long last, a sphere.
I don't agree. Topologicaly, the solutions to a quadratic equation in the complex affine plane form a sphere with 2 holes, that is a cylinder. As you said, the solutions in the complex projective space are homeomorphic to a a sphere, but there two points of intersection with the line at infinity, so the intersection with the affine plane consists in a sphere minus two points.
@@oscarwilde8470 Right, thanks for pointing that out. I'll edit my comment.
@@MathFromAlphaToOmega My pleasure :)
My first thought upon looking at this is to wonder what the area of this circle would be. For r=i and A= πr^2 this means A=-π. Also, the circumference would be π2i. So, the area is a real number and the circumference is imaginary.
Wouldn't it make more sense to look at the complex extension of the unit circle first? I'm not even sure I could wrap my head around that.
It's the same object though, just rotated 🤷♂
Apparently yes, and the thread discussing this is pinned to the top now :)
makes sense actually, consider the equation ax^2 + by^2 = 1
if one of a or b is negative, the graph becomes a hyperboloid instead of an oval
It's lovely to watch this channel gain traction and grow!
I think the first thing to do, would have been to look at x^2 + y^2 = t^2 while allowing x and y to be complex, with r real and non-zero, before making r imaginary.
Doing this, I believe you would find that you get essentially the same shape either way.
Indeed, take the set {(x,y) in C^2 | x^2 + y^2 = 1^2 }
For and (x,y) in this set, and any complex number r, (r x)^2 + (r y)^2 = r^2 .
So, for different (non-zero) values of r, the shapes you get are the same, except for scaling and rotating (of phase for both x and y coordinates) .
So,
In this sense,
The shape of a circle with radius i, is the same as the shape of a circle of any radius.
this is super interesting. so given that one of the 4d projections at one point contains a circle, then the 4d shape itself does contain a circle.
This should blow everyone's mind: a right triangle with side lengths of 1 and i will have a hypotenuse length of 0.
The 4d Minecraft project really helps with imagining actual 4th dimensional structures. Actually seeing a 3 dimensional slice turning around in 4th dimensional space really helped me understand 4th dimensional structures. As well as a 2 dimensional slice turning throughout 3 dimensional space as a comparison. Also you were so close to a 3 minute 14 second video!
This does stretch the definition of a circle tho. All points on a plane equidistant from a given point, meaning it is inherently 2 dimensions.
well that would be a boring video as i by definition is just 1. Just rotated by 90 degrees. so it is just 1 in the y direction. And that is just a circle with radius 1.
the fact that radius with a negative integer creates shapes like inside out objects
Another way to think about purely real part of this object is how it acts on real points. We know that circles act on points via inversion (OX*OY=R^2). This object acts like a composition of inversions in 3 mutually orthogonal circles, or, alternatively as a composition of inversion in a circle and then point-reflection in the circle's center. One notable property is that action of purely imaginary circle has no real fixed points. It also appears as a second bisector between 2 non-intersecting circles.
I think it is well known that in the complex plane ^2, all the conic sections are essentially the same formula a(x-b)^2+(y-c)^2=0
already in ellipse form. transform y to yi and you get the hyperbola. with circles and parabolas being limit/special cases even normally. To make this formal you have to introduce points at infinity and the projective complex plane. Then all conic sections well be sections of the same shape.
After some reflections, I have understood that a circle in imaginary plane is just a specal case, not the general one. If (a+b*i)^2+(c+d*i)^2=i^2, a and c can be any real numbers, as long as a^2 < b^2, c^2 < d^2 and 2*a*b*i+2*c*d*i=0 If a=b=0, it's just a circle on imaginary plane
This means that the complex shape x^2+y^2=i^2 projects on the real plane as the whole plane
yo this is SICK
Dear Ron & Math, there is also a nice way to visualize a surface in 4D space. You take two squares which will represent two variables. Left square will be x = a+bi, where a is horizontal axe and b is vertical axe. Right square will be y = c+di. Now you start taking some points on left square and paint them. Take for example 0+0i and paint it red. Now we find all solutions of x^2+y^2=-1 where x=0+0i, it is y= 0+1i and 0-1i. So we paint this points red on right square. Then we take another point on left square, say x=0+1i, paint it green. Find a solution, it is y=0+0i, paint it green on right. And so on and on.
Thank you for the input!
@ron-math question why don't you set x equal to rcosine theta and y equal to i*rsimetheta..isn't thst more familiar
.does that change anything?
Cool, the projection is a 3D Penrose diagram! I like to think that the resulting 4D shape is a good description of the shape of space-time near a black hole.
The circle becomes a twisted line. Like when you take a 3d piece of paper put a twist in it and glue the ends in become a 2d piece of paper. All in the Z plane, it comes down the y wraps around the origin and continues up the x in that arc trajectory.
i love this, thanks Ron!
Epic stuff!
It’s an inverted circle while r=0 is just a dot.
when you pulled up x^2 + y^2 = -1 my immediate thought was that its the same as a circle with r=1, except the regions considered inside and outside are swapped
Makes sense, since the area is negative.
I first encountered imaginary circle in Louis H. Kauffman' s notes
We always make imaginary circles with our brain when thinking about circles!
You could maybe also have a look at a parametric description of a circle x = r cos(t), y = r sin(t) and connect it to what you already have.
This is about as crazy as I’d expect an imaginary circle to look
I was just thinking about this topic in context of constructorial spaces where time can be described as measurement potential. Measurement as indicated by complex numbers and potential indicated as imaginary numbers. Another way to describe time as being structural information visible against infinity like the event horizon of a black hole. Might also look at this kind of algebra as a model for the relationship between calculus (point space) and geometry (cube) as some sort of harmonic oscillator where by the corners of such geometries cannot exist at the same time as the connective lines between them. Could say something about angular momentum in particle physics and also the nature of the universe. All considerations for experimental design. Thanks so much Ron. Skippy is just back from her walk on the beach so she’s just going to be so excited about that as the girly woofs is also part of my science team.
Having a negative radius doesnt change the shape of a circle in a practical sense. It just changes the angles observed. Many people might just take the absolute value of the radius as an argument. The complex continuation of absolute value is modulus and modulus of i is just 1. The angles would be weird if you leaned into that aspect of it but yeah
I'd assume that even the shape of an imaginary circle is circle shaped. Otherwise, you wouldn't call it a circle...
Very cool! Great video.
This explains why the circle have two sides ( inside and outside)
I'm not an English-speaker, so excuse my explanation.
If you have a line, draw a 3 cm segment, then add 4*i cm, so how many cm would you get and where they would be? The best answer is that on the line there would be 3 cm, but on the plane -- 5 cm so that the projection of 5 cm segment on the line is the same 3 cm. Furthermore, if you draw a line and try to mark on it i cm segment, on the plane there will be 1 cm segment which wll be projected on the line as a point
A circle is a plain shape, its projectons on x and y axes are line segments with length of 2*r. If r = i, the projections will be just 2 points, so it would be exactly 1 point on the real plane. But on imaginary plane there will be just a simple circle with |r| = 1
"Why do we consider that there is an imaginary axis orthogonal to each real axis? I believe there is only one imaginary axis orthogonal to all real dimensions."
I think that, for me to be confident calling something a circle with radius=i, I would want some function to be called a "distance function" and take a locus of points with "distance"=i from the centre. But I don't know if it's possible to adapt the requirements for a distance function to having a complex output in a sensible way. Maybe for points in a higher dimensioned space, idk.
At first I thought the video is about the finding equation of circle in the complex plane, but it wasn't. Interesting.
A question. When visualizing 4D objects can’t you just choose one of the dimensions to be time, and the other three are space. Then the 4D shape would just be an animated 3D object that we can see?
Once upon a time I wondered: with y²+x²=r², what are the y values beyond |x| > r∈ℝ+ ? And saw that it was a hyperbola connected to and extending out from the sides of the graphed circle, but "perpendicular" to the plane, as if the graphed x and y were each real with their own complex planes. Of course there's another hyperbola above and below, when looking at it with respect to x. y²+x²=-1 simply exists completely outside the strictly real part of this "complex 2D" graph.
It's probably easiest to solve for y as a function of x and then plot Abs(sqrt(-1-x^2)) over the complex plane and then color the output according to Arg(sqrt(-1-x^2)) this way you visualize almost the entire function at once.
Very nice idea. Thank you!
It seems almost fibrative.. like a bundle of strings twisted and the ends go off in every direction to form circles that are circumferential to 3d spheres. The state of all of them being the 4d hypersphere, while simultaneously the larger the imaginary component the more space between these bundles moving more into imaginary space, while at the same time the smaller the imaginary component the tighter these fibers and twists get until the inflection point where b->0 and then the imaginary component collapses, the twists get so tight they cinch, and then the radius briefly becomes zero then undefined
This is so cool!!
The equations for circles/spheres/n-speres define only the locus of points on the "outer" "shell" of the points they enclose, which are really disks/balls/n-balls. So for these n-spere loci, there will always be a geodesic path between any two points in the set such that the geodesic is coincident with a great circle, which is always a closed loop.
In this case of an imaginary radius, the hyperbola extends to infinity, but mathematically, would it still obey this principle that for any pair of points, there exists a great circle containing the geodesic between those points? The 4-hyperboloid shape extends to two opposite infinities, but perhaps under a different projection, like the real projective plane (or complex projective "plane" if that's a thing??) this principle could be shown? I'm way out of my depth here but it would be great if you (or some other commenter) could pitch in!
Thank you very much @ayane_m but geometry is really not my strength. I made this video up to a point that I am still 100% certain what I am talking about. I will sit with you and wait here as well.
Yes, the complex projective plane is a thing, and under the real projective plane, hyperbolas are fully connected across points at infinity, so I image the same would apply to hyperboloids in the complex projective plane.
@@ron-math I'd say you really stretched the definition of a circle in this one, but still, when you set the real parts to 0 you had your pure imaginary circle of radius 1 in imaginary space alone. Because everyone and their mother knows that i is 1 just shifted 90 degrees.
If you take the absolute values of x,y you get a regular circle.
I did some work on this and I think I found a very satisfying way to visualize the full complex circle in 3d, but only under some interpretation.
Change of variables from x,y to r, θ, with both being complex, using the normal polar coordinates formulas. This works because sin and cos are analytic.
Now you very simply have r^2 = R^2, which just means the two parts of r (real and imaginary) are fixed and uninteresting. So you can scale by R and just use the unit circle and focus on the real and imaginary parts of θ and how they affect x and y.
Now, if you let r=R=1, you can get (from the polar coordinates transform equations) equations for the real and imaginary parts of x and y in terms of the real and imaginary parts of θ. This is a two dimensional parametrized surface in four dimensions.
You will see that the real parts of x and y form a circle parametrized by the real part of theta, with radius equal to cosh of the imaginary part of theta.
Similarly, the imaginary parts of x and y form a circle with radius equal to sinh of the imaginary part of theta.
So when the angle θ between (x, y) and the x axis is real, you get a real unit circle. As you change the imaginary part of θ you start to get a larger real circle, plus a pure imaginary circle which starts at zero size and gets bigger as θ gets more complex. These both get one turn as the real part of θ goes from 0 to 2π.
The best way i can think of to visualize this is a torus with primary radius cosh(Im(θ)) and secondary radius sinh(Im(θ)). The primary radius is the real parts of x and y, the local coordinates relative to the secondary radius give the imaginary parts of x and y. You trace out a spiral around both radii of the torus simultaneously as the real part of θ changes. The real part of the solution is where you are at the secondary center of the torus, which forms a flat disk extending out from the real circle. The imaginary part is like a Moebius strip, but with a full twist instead of a half, unfortunately 😅.
I have some analysis and some graphs I can share by other means that show the result. It's a very nice concept: the real unit circle is the part of the complex unit circle with no imaginary components in either coordinate. All the other stuff grows from that in two sheets outwards: one is a flat disk and the other is a cool Moebius strip-like structure.
I think it would be pretty easy to take the parametric equations for the surface with R=1 and scale and flip them for different values of R. A purely imaginary circle will swap the primary and secondary radius of the torus (think a torus fatter than it is around) and a complex number will mix the contributions in a way that might yield some interesting visualizations under this interpretation.
Nice explanation!
Taking into account trigonometrical form of i and how it is defined - then circle of r = i, is same as circle of r = 1. Overall circle is infinite root (sum of all roots) of i
it might be wildly unconnected but is this how we find the shape for weird orbitals like the d f g and so on?
Probably the worst part about this idea is that the area of the circle with r=i is that its area is negative pi.
really cool! never thought about it
Very cool problem and video! Though it IS actually possible to plot curves or surfaces like this in 4D, projected onto a 2D screen, if you know how. I solved the this parametrically, as the solution is a parametric surface with two parameters, embedded in 4D. I won't go through the solution here but you can verify it solves both constraints:
a = sinh(t)cos(theta)
c = sinh(t)sin(theta)
b = cosh(t)sin(theta)
d = -cosh(t)cos(theta)
Think of (a,c) - the two real parts - as a point lying on the "real plane" and (b,d) as lying on the "imaginary plane". Separately, both points lie on circles in their respective planes (theta varying from 0 to 2*pi), but for any value of theta, the two points will be 90 degrees apart in their respective planes. The two circles have different radii of sinh(t) and cosh(t). In the 4D plot, for any value of t you will have a circle lying in 4D space. The special solution a = c = 0 corresponds to when t = 0. This is a pure unit circle in the imaginary plane. But as you increase t, the two radii both expand exponentially, and are essentially equal in the limit. The circle rotates from the imaginary plane to about 45 degrees between the real and imaginary planes, and it's radius grows exponentially. So in the limit, the growing circle generates a plane.
Picture the solution like this: a flat film of soap, bounded by a large circular ring and a small circular ring in the middle, which is empty. Now rotate the inner ring 45 degrees to the plane, and the soap film will adjust accordingly. Now make the larger ring have infinite radius, and that's your surface.
If you're interested in seeing a picture or animation, or how to plot in 4D in general, let me know. Subscribe to my channel as I will be putting a lot of 4D visualization content soon. I'd love to do a response video to this problem.
Subscribed!
Would love to have your response video and please let me know if I can help in any way.
What if you try to convert polar coordinates to Cartesian coordinates using rSin(theta) and rCos(theta) using the third axis as the imaginary axis?
Very interesting video!
I gotta say, though, a map (2D) was in interesting choice of graphic to use to emphasize that we live in 3D space!
Makes the point about projecting onto a lower-dimensional space..
Distance is always taken to it's absolute value. You can't have a circle with radius negative 1 because negative length doesn't exist. Similarly, a radius of i also can't exist. In either case, you would use the absolute value. And the absolute value of i is 1
Negative distance isn't a thing, and neither is negative area. Distances are absolute, and i has a distance of 1
Do it in the complex numbers for x^2 + y^2 = 1.
It appears that the "negativeness" of i² has transformed circle in hyperbola. Hyperbola is curve DIFFERENCE of squares, so the next question is: What happens if we take equation for hyperbola & introduce imaginary into it? Will it create a circle?
The circle is defined by the absolute value. Absolute value aka the norm can't be anything but a positive real number (by definition).
Really cool! You’re essentially intersecting a hyperbolic paraboloid with a hyperboloid.
Good thing people are working on this instead of how to deflect asteroids or cure diseases or something.
@@rickwilliams967 Well I suppose that if mathematicians tried to work on those things they’d fail horribly as they’re not trained nor have the interest or even the skill for that.
Did you know that much of mathematics used in both of those applications was invented because people were doing just that- discovering mathematics?
Let us be, what we do is probably gonna be useful in the next 200 years or so lmao
Since both x and y are complex, both would be points on the imaginary plane, so that makes me wonder what you would get for each point x that you select on that plane as points (0 or more?) y on the same plane that would fulfill the circle equation for radius i. From the various 3D projections of a 4D shape that I still cannot wrap my mind around, I have no idea what that mapping from x to y points that fulfill the equation would be. Will there always be 2 y points for each x? Which points x have valid y values and which do not? I feel this video could go way deeper into all this.
I agree. I might make a followup video on this one sometime in the future.
My engineer brain: isn't it going to be an ordinary unit circle on an Im-Re plane???
Apparently not, since it would also mean a radius of 1... 🤦🏻♂️
The first thing I thought was that for a circle of radius, r = i
area would be pi*r² ,
That is, Area = -pi.
(And an object with a negative area was enough to get me to doubt my own existence).
And If we add the area of a circle with r = 1 to the area of a circle with r = i then The total area would be 0.
And if we did that with three dimensional objects, then they cancel out each other's volume or existence, And maybe antimatter is just matter with dimensions that are measured in units of i. And that's why matter / antimatter pairs annihilate each other out of existence. 🤔.
Not really.
The unit circle has radius 1 or i. These radii describe the same unit circle.
This reminds me of an air vortex.
When you set x^2+y^2=-1 in Desmos 3d you get a hyperbloid
I just tried. It didn't work. Can you share a link?
@@ron-mathi think it was x^2 - y^2 to make a hyperbloid
What program do you use to animate your explanations?
Manim.
imaginary numbers having much to do with 4D objects and being very applicable to our 3D space makes me wonder if we truly live in a 3D space or are simply the 4D equivalent of living on a screen
Somehow, a whole bunch of people who are way better at physics and math than me took interest in my comment. Thanks for being here, I guess?
Correct me if im wrong, but iirc by calculating the energy loss of gravitational waves through space, it was calculated that we should pretty much only be in a 3+1D space. unless the extra dimensions are extremely "small", whatever that means (like string theory tries to suggest)
@@knopfir3+1=4
Lots of physics has a "complex number"-like structure! you see it in relativity where you can get negative norms, and in quantum (not completely unrelated if you're talking about the dirac algebra). When it's connected to relativity it's a manifestation of the constancy of the speed of light. Although mostly what we see is three real-like dimensions and one imaginary-like one (or the reverse) as opposed to this two-real two-imaginary structure, about the only thing I can think of are spinors which to my knowledge are sort of connected to quaternions but I know almost nothing about them. I'd suggest looking into geometric algebras if you want your mind to be really blown! Almost all of the complex numbers that show up in physics can be thought of as stemming from some sort of geometric algebra afaik. They're used in modern theoretical treatments of physics.
we could vert well be living in 32D space, to Math it's just a few more variables
The 4D equivalent of living on a screen would be a 3D space. Now I'm imagining 4D creatures watching us on their 3D screens.
I'm confused . Isn't this a 4D object projected into 3D space which is projected on to a 2D plane - my screen ?
If it's in 4 dimensions, it's not a circle, since a circle is 2-dimensional.
The area might be -pi assuming pi r^2 holds in this case, which is pretty funny
x^2 + y^2 = 1 is not a circle if x and y are complex numbers.
Simple question: how is this possible?
From what I know, |z - zₒ| = r is equation for a circle... which IMMEDIATELY leads to CONTRADICTION, with r=i, as on the left side of equation is complex number's modulus (i.e. real number), while on the right side we have imaginary number. There can't possibly be solution for this. What am I missing??
If I use my imagination any shape can be a circle.
commented before watching the video
According to the rules of geometry you'll get a circle inside out.
So, we can't draw a circle of radius i on a plane. Instead, it exist in a 4D space using quaternions or hypercomplex algebra.
Are you talking about a circle whose area is -π ?
Asking for an imaginary friend...
When I imagine a circle, it's still a circle... It's not like imaginary numbers where the number I imagine is always different to everyone elses..
A circle is a circle, end of story.
While you Aren't wrong, The way One visualizes A concepts Can change, When exposed To different Dimensions.
A circle Maybe a Circle, but What would That actually Look like, Based on The math?
The method You use To visualize, And articulate the math You want To communicate, Can result On different Results.
@@dr.blockcraft6633 when I was about 10. I actually discovered how to use sin and cos to draw circles pixel by pixel... I inadvertently had to find the value of pi*2..
In my adult life I wanted faster circles, So I copied some code of Wikipedia that draw circles really fast without sin and cos... You actually get different circles... There are probably other ways to draw circles, and they probably give slightly different pixel results..
Is this what you mean ?
Is this how regular trig relates to hyperbolic trig?
Related to it, but I wouldn't see that this is the cause of the relation.
I wonder what an imaginary hyperbola (x² - y² = i) might look like
Thats cool and all but can someone explain what is the point of this other than just maths for maths sake
Weird!
Area of the i-circle is -π
Volume of the i-sphere is -4πi/3
Bulk of the i-hypersphere is π²/2
I am a higher being and understand 4D spaces
And it's not a circle, it's a black hole
It is not really a 4D curve... It looks like some kind of a different manifold to me...
Of course it is still a circle... wait what? it is a square?
I think it goes into the square hole
"i" isn't less then, greater than, or equal to 0 so what is it
This is used in relativity.
Is that the Desmos 3D Calculator?
Yes.
A hyper sphere.
Your title said, "Shape Of An Imaginary Circle" so I naturally assumed in the plane only. Sooo...I guessed it would be a circle of radius i in the complex plane. But you went to 4 dimensions without a motivation or justification.
Without doing any computation or watching the video, I’m guessing it’s a hyperbola.
hyperbola ❌
imaginary circle ✅
imagine if i is imaginary in 3d but a real number in 4d
i is Real in a particular 4D algebra that models 2D geometry. In this particular modeling, i becomes the Real origin, rather than an orthogonal direction, and it's actually the product of two orthogonal lines through the origin.
I will have to mention though that this particular 4D algebra is not the quaternions. (It might be the split quaternions though, but I'm not certain since it's constructed with a completely different technique.)