Equation of a Line in the Complex Plane
Вставка
- Опубліковано 2 чер 2024
- We derive a general equation for a line in the complex plane, and finish by interpreting one of the constants geometrically.
00:00 Cartesian equations
00:46 Complex equation
01:31 Conjugates
03:03 Rearranging
05:03 Geometric meaning of w
the idea of defining a line by its perpendicular vector and a constant reminds me of how a plane in 3d can be uniquely defined by its normal vector and any point on the plane.
i wonder if there's a generalization for those ideas that could be extended to higher dimensional geometries via hypercomplex numbers. no idea if that's the case, but it's where my mind went while watching.
thanks for sharing Dr. Barker. this was neat. I've been enjoying your regular uploads
The easiest way to extend these ideas is to consider the orthogonal space to a given non zero vector. This gives you a general hyperplane passing through the origin.
If you wanted a more "algebraic" point of view then it depends on which number systems you allow. If you want finite-dimensional R-algebras then I think you can only work in dimension that are powers of 2 and only up to a maximum.
Degree of freedom
A realm in 4D (that is to say, a flat 3D space suspended within 4D space) is _also_ specified by a vector and offset. In fact, this is true in every dimension - an (n-1)-D flat space takes a vector and offset to orient in n-D embedding space.
Where things get interesting is if you ask about orienting a flat object of _fewer_ dimensions than just 1 fewer dimension. Like a 2D plane in 4D space. In _that_ case, the orientation requires a special object known as a "bivector", a kind of object which can be understood in some sense as the "right" way to define the cross product of a pair of vectors, because in some senses the cross product "doesn't want to be" an ordinary vector. For example, the transformation rules as its inputs are transformed in a change of basis.You can think of this as arising because of the following: to orient a 2D plane in 4D, you must first orient a 3D realm in 4D, which takes one vector in 4D, then orient the plane _within that 3D realm,_ which now takes another vector _in 3D._ You thus need a 4D vector and a 3D vector. But there is some redundancy in this description, and once you "mod that out", the orientation actually only takes 6 coordinates, not 7; and this is the exact number of coordinates in a bivector in (R^4) /\ (R^4).
Man, it's very concise and really informative. I really love this format! Keep up the good work!
The ax+by=c formula even captures all possible lines at once if a = b = c = 0 ;-)
The whole explanation would be a bit cleaner if it was made explicit that a and b can't both be 0 (and thus w can't be 0 either).
Good stuff. :)
Two comments:
1) It would have been nice if you had made at least a short comment why c _has_ to be a real number and can _not_ be purely imaginary or complex. (Simply taking the complex conjugate of the equation w^* z + w z^* = c.)
2) The original equation ax + by = c describes a line in R² in "normal form", i. e. we already know that the vector (a,b) is perpendicular to the line. But ok, probably not everyone knows that already, so it's nice that you show a short proof.
oh yeah the second point is pretty interesting, these are two lines in R2 which have the same column space(If they both can generate same solution), that means any vector that is orthogonal to the first line, should be orthogonal to the second, but since the difference of such lines gives the vectors in the same plane which is in the left null space of the column space,again, orthogonal to our plane and since what were doing is just projecting those back on to the column space the imaginary component collapses and the "sine" component vanishes giving us only the real part and fair enough or "c" is a real number.
I thought it was given that 'c' was real. Did I miss something?
@@VoteScientist Yes, that was given. My point was that it would have been nice to argue why it _has_ to be real.
Nice! It's worth noting that y=mx+c successfully expresses the general "linear" (more rigorously, affine) _function_ . When we move from graphing functions to doing geometry we need the general form of the equation of a straight line ax+by=c which is taught during the study of vectors.
Very eloquent. Congratulations.
Nice video.
👍👍
Great video, thanks! The narration reminded me of soviet maths textbooks. Or is it the style of the era from about 1960s to 1980s. Could You recommend English-speaking authors with a similar style of narration?
Lol you roast him and then come in with the "wait that's a good thing" 😅
I hope you’ll make more videos on the subject, this is really interesting
I'm recently into these complex number arithmetic videos, and they are very entertaining 🤩🤩
Thanks PROF 👍
That's.so.cool! I've studied math for 2 years but I've never heard of anything like that!
I've studied math for longer. I have heard of something like it, but it's significantly more difficult... |z - w1| = |z - w2| where w1 - w2 is perpendicular to the direction you want your line in.
I sort of quickly came to the realization that the equation of a line in the complex plane is just an angle and an offset: e^it (x+ci) = y where t is real in 0,pi and c is also real as a perpendicular offset. It's two dimensional and unique so long as you exclude pi from t
Here x is a free real parameter with the position of the point given by y, neither are actually the x or y coordinate
wow!
UR videos are quiet ADVANCED.
Really quiet, i. e. not loud? Or do you mean quite? ;)
@@bjornfeuerbacher5514 quite😊
If you want something in a line, you need to queue it
👍
I bet somebody already invented a binary operation that can make the complex line equation even more spiffy
Could a line exist in the complex plane? Yes.
Suppose we choose pairs of complex numbers. We could treat every $(x_{r}+x_{i}\sqrt{-1}, y_{r}+y_{i}\sqrt{-1})$ as points in a pair of Cartesian planes, with a real plane, and an imaginary plane.
Simple as coming up with two lines in both planes, then merging the results.
I love you , I love your eyes ❤💋
Why does c need to be real?
We have ax+by=c, where a, b, x, y are all real.
The conjugate of a product is the product of the conjugates, and double conjugate gives the original number. This means that zw*+z*w=zw*+(zw*)*=2Re(zw*), which is always a real number.
We can show that for any two complex numbers w, z, the LHS expression w* z + w z* is its own conjugate, so always has to be a real number. Or we can also understand it because in "ax + by = c" we needed c to be real, and we kept the same c throughout the derivation for the complex equation.
@@DrBarkerThank you
In one of my classes, we used the characterization
Re(e^(iφ) z) = c with φ,c∊ℝ.
This has the following interpretation: If you have a line, you can rotate the complex plane by an angle φ, i.e. multiply by e^(iφ), such that you receive a line which is parallel to the imaginary axis. A line which is parallel to the imaginary axis has a constant real component c.
Nice one! That's even better than what was shown in the video, which was just Re(w* z) = c, but taking w = r e^(-iφ), you can redefine c as c/r and get your equation.
You explained that very clearly, imho