the tetration of (1+i) and the (a+bi)^(c+di) formula
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- Опубліковано 21 вер 2024
- We will find a formula for a complex number raised to a complex power, namely (a+bi)^(c+di)=?
Thanks to Leon M best blader for his email from almost a year ago, • Complex number to a co...
i^i, • i^i
Euler's Formula, • Euler's Formula (but i...
Math for fun, sin(z)=2, • Math for fun, sin(z)=2
☀️support this channel: / blackpenredpen
blackpenredpen
This must be the wrong channel. This looks like blackpenredpenbluepen
Yup, bprp's copycat! : )
@@blackpenredpen imposter ;)
Pen(BlackRedBlue)
5:06
redpenblackpen
Extending the list of keyphrases (for drinking games & such) in bprp, from two entries to four:
#1: "isn't it?"
#2: "for you guys"
#3: "Let's do some math for fun!"
#4: "And as always . . . that's it!"
Fred
Nice!!!!
May I also add "here is the deal"?
And "I don't like to be in the bottom". It's a famous quote of his. XD
@@blackpenredpen Be my guest!
@Doge It would be neat to include that, but he only said it once. This was meant to be a list of his recurring phrases.
Maybe it could be first on a new list; a list of "famous phrases."
If any of these get repeated enough times, they can jump to the recurrent list.
Fred
is ur name Fred?
@@GDPlainA Yup.
I love the way he smiles and teaches. He is so happy when he teaches. He is the first person i have seen who loves his job moreover teaching. He is Steve Chow. Who is in love with Black and red colours. If some one insults him, he ignores that guy. I really appreciate his behavior. Hats off to you ! I hope you success in future or and become a mathematician, sir !
he is already a mathematician, wtf
i think it is a human trait called enthusiasm
All the above, and it reminds one of other enthusiastic and exceptional teachers. Richard Feynman comes to mind.
Fred
Who's insulting him?
1:28 in my "theoretical solid state physics" class i once had to integrate a function with regard to a variable that was called d. moreover the function was a linear function in d, so the integral was int ddd
so, d(d)=d?
f(d) = d
🤯
∫(3dd²-7dd)ddd=dd³-((1/2)*7dd²))+C
So when my gf said we had an argument yet to be solved she was referring to a+bi, that's good to know
Actually, referring to the angle that a+bi makes with the +real axis in the complex plane ;-)
Fred
@@ffggddss Cool, it really sounded complex for me to win this battle anyway- Pretty sure you are not my girlfriend, though
@@Wecoc1 Nor are you mine!
Fred
lol
That's Arctan(b/a) = Arcsin(b/√(a^2+b^2)) = Arccos(a/√(a^2+b^2)).
So please tell your gf that you understand this is the argument you need to solve together. With a big grin on your face. It has many solutions, so pick one. 😉
My juniors are just now using complex roots to find the standard form of a polynomial. Should I show this to them next?! XD
Some might be able to follow along and find interested in it. If you every do, let me know how it goes!
I consider that you analyze parts of maths that anybody think, you should write a book of math for fun, i will be your first fan, thanks for your brightness and for make the people think! Kudos to you blackpenredpen
JORGE ALEJANDRO GUERRA ALFARO thank you very much for your support! I will send you a book for free if I do write one in the future!
I found a formula which is more compact than this, and which also accounts for the non-principal values of (a + bi)^(c + di), and I found it without using polar coordinates. I simply wrote (a + bi)^c*(a + bi)^di, and when I simplified (a + bi)^di, my final result was (a + bi)^(c + di) = (a + bi)^c*e^(- d atan2(b, a))*(cos(d Ln((a^2 + b^2)^(1/2))) + i sin(d Ln((a^2 + b^2)^(1/2))))*e^(- 2πn), where n is an integer. The benefit of this compact expression is that it involves no addition inside the trigonometric functions whatsoever, and it reduces the exponentiation of 2 complex numbers to multiplication of 2 complex numbers, which is a nice property to have.
As for dealing with the (a + bi)^c factor, I found a general formula which is valid for all natural numbers, and which can easily be extended to integers. It is not as trivial to extend it to rational numbers. The ultimate goal is to extend the formula such that c can be any real number. The only problem is that c would need to become the upper boundary of the summation, and this requires defining summation such that its boundaries of summation can be real numbers as opposed to only integers. Alternatively, one can simply use the polar form of the numbers to then create an identity such that the summation can be extended to complex numbers such that it satisfies the equality, which is the reverse direction of the development I suggested. If you want to see the formula I derived, then I can post it in the comments.
Another nice development is that in order to get other values from the principal value of the exponentiation, all one has to do is multiply by e^(-2πn).
Applying it to (1 + i)^(1 + i), we obtain (1 + i)*e^(-π/4)*(cos(Ln(2)/2) + i sin(Ln(2)/2))*e^(-2πn), which can easily be expanded further by using the half-angle formulas. I assume this is equivalent to the expression shown by Wolfram Alpha by virtue of the angle-sum identities.
Shortest math paragraph be like :
I derived this a couple of months ago, but your way of doing it is way nicer and cleaner. Good job!
Around 2m 50s & on, about finding θ - There's a function in computer languages that's useful for that, atan2(x,y). It starts out as
{ tan⁻¹(y/x), x≥0, y≠0
{ π + tan⁻¹(y/x), x
A nice step by step teacher for math problems. This brings less complexity in understanding to complex numbers Euler Complex Circle mathematics.
4:30 Curb your calculus. :-)
this is so good
Before I saw this video, or any other video regarding this topic for that matter, I tried to accurately calculate (a+bi)^(c+di) and I managed to get the real component to compute correctly but not the complex component. Keep in mind, I did this without any external help, including without any tutorial on the internet, or from a peer. It turns out, I had tried to distribute the " e^(-dArg(a+bi)) " component into both the complex and real components, but had forgotten to do so to the complex component. Until I watched this video to see if I had gotten anything remotely correct, I had no clue that it wasn't a computational error that I made, but simply forgetting to distribute properly. Because of this, I feel I have now greatly expanded my knowledge on mathematics, and definitely have learned to pay attention to subtle errors like the one I had made so I don' t lose motivation to give up on a personal mathematical endeavor. Thank you for creating this video, because I believe that I have learned not only more in the realm of mathematics, but also about myself personally. I love your stuff, and thank you for teaching the world mathematics!
You are the best maths teacher on utube
My preferred way to get the argument is piecewise. calculate r as normal, then use { if b>=0 then θ = arccos(a/r) } { if b
This channel is very underrated with views.
I very much enjoy watching your videos. Your enthusiasm is infectious.
NOBODY CAN TEACH MATHEMATICS LIKE YOU - PERIOD.
Why do I love maths so much.... _i_ am so complex
Best comment and I can relate!!!!
*i* am not complex, *i* am just different from other real things
2005 VincentChui that’s pretty complex
@@mathematicalhuman6858 you're imagining things
Interesting problem along these lines: Find a curve in the complex plane describing all complex numbers z such that z^z is a positive real number.
Paco Libre
Wow, that is interesting and I will to think about it a bit.
Copied for doing it (if @blackpenredpen doesn't do it) in free time.
EDIT: that weird value I talked about is actually at the number 1/e, so that the one line actually touches the real number line. Also (1/e)^(1/e) = e^(-1/e), NOT ln(2). Those two values are so close to each other that I mistakenly said it's ln(2)
Another observation. If you zoom out enough, that is, so you are at the level of -10*10^50 to 10*10^50, and even more, then the lines seem to be approaching the number line (line b=0). So what really happens is that those "lines" are not lines but almost linear curves which curve towards the number line extremely extremely slowly
I think I'm gonna be seeing this in my electrical engineering classes next semester!!
I like that he thought ahead enough to break out [re^i(theta)]^(c+di) into a product early on. I didn’t do that until later and it added more work for me. Of course, he may have spent time off-camera working out his approach!
Sir,I'm from Bangladesh!
It's really very helpful.
Thanks a lot
That feeling when I wrote this down last week... Thanks man
*Checks his paper while watching video*
It is interesting to observe what condition is required for the end result to be real, namely cθ + d.ln(r)= 0, or integer multiples of π. In the special case when c = 0, r = sqrt(a²+ b²) must = 1 (principle value). The expression i^i is one to satisfy this condition.
I used to hate complex, but you gave me a whole new view of it with all your calculus
Thanks for the video!
how can this comment be 4 days old if the video was uploaded 20minutes ago
Vitor Sasaki
It wasn’t uploaded 20 mins ago , this was uploaded before 4 days and it was unlisted because there wasn’t a thumbnail to it , until 20 mins!
@@omarifady and how the f*** people found the video before
Vitor Sasaki
bprp put the link of this video before, in the description of another videos😅
man, your videos still get me. love them!
daniel thank you!!!!
That's cool! More and more interesting problems. Thanks!
Imaginary power converts absolute value into argument and argument into absolute value.
(R * exp(i*theta))^i = exp(-theta) * exp(i*lnR)
Awesome formula!!! Nice work! I 🧡it. You can also write Arg(a+bi) as tan⁻¹(b/a), can't you?
could you do integral of 1/(x^2+1) with partial fractions? I know the answer is just arctan x + C but if you factor it with the complex roots you get a very interesting result
That is how i integrate 1/(x^2 +1) .
Same result ?
And in the end you just get a complex definition of arctan! You can also get it by solving the equation (e^(ia)-e^(-ia))/(e^(ia)+e^(-ia)) = x
he literally blew my mind when he said a =e^(ln(a)) i had to do a research before continuing the video... thanks a lot.
Thank you ❤️❤️. I've been trying to understand Zeta function and this has helped me a lot.
Just wanted to say i started watching your videos a few months ago. Im almost 18 but your explanation are so good that i keeped looking at them until i anderstood what you were saying. Thank you for your amazing videos. Btw sorry about my english
Brilliant, always enjoy your work.
I tried this in the other form and did (a+bi)^(c+di)=e^((c+di)ln(a+bi))=e^((c+di)(ln((a^2+b^2)^(1/2))+arctan(b/a)i), where arctan(b/a) is subjective to the quadrant. From here it was e^((cln((a^2+b^2)^(1/2))-darctan(a/b))+(carctan(b/a)+dln((a^2+b^2)^(1/2)))i)=e^(z+xi) where z is being used as a variable for the real part and x is being used for the imaginary part. From here I realised that (z+xi) is just ln(the answer)=ln(L+Mi). The working follows e^(2z)=L^2+M^2 and x=arctan(M/L), via substitution (e^(2z)-L^2)/((Ltan(x))^2)=M^2/M^2=1, L=(+-)((1)/((tan(x)^2)+1))e^(z) and mi=Ltan(x)i. My issue is I haven’t been able to figure out the different cases of when the real and imaginary are negative or positive. Do you know how I can see or test which quadrant the answer should be in, or have you already done a video explaining which quadrant the answer should lie in without doing both methods?
If you finish the calculations to get the final answer in polar coordinates, it's interesting to see that there are real solutions to the problem when d ln (a^2 + b^2) + c arctan (b/a) = pi n, n = 0, 1, 2, ...
Can I just say I LOVE this channel so much. it's helped me so much in school (I'm from the UK and the curriculum over here is a little simplistic). The other day our teacher set an extra qs to differentiate an integral (which is wrt x) wrt a to obtain the result for another integral and the only reason I knew to take the derivative sign inside as a partial derivative was cus of BPRP :)
If you make a+bi into a linear function that goes through the origin like. y=mx, then slope is Δy/Δx,
Δy/Δx=b/a, to find the angle from the slope use tan^-1(b/a)
Are you kidding me right now !!😱😵
How can I remember that bigggggggggg formula but it is pretty impressive 😁😀
We want black board and "whitechalkredchalk" again in later videos
Riemann stated the Riemann hypothesis in a paper in 1859. He had to input complex numbers into the Riemann-Zeta function and calculate lots of points such that he could see enough of the Riemann-Zeta landscape to think of his hypothesis. He would have had to do the above calculation over and over. He had no computers. It always boggled my mind as to how he did that. Maybe he had lots of minions doing the calculations for him.
Impressive - though I have not investigated this enough to know exactly how he arrived at the hypothesis. Did he write it down that he used such a method?
That said, though, there is a rather more verifiable example in this picture of a 3D graph of the Gamma function for complex argument, made long before computers:
enacademic.com/pictures/enwiki/74/Jahnke_gamma_function.png
Compare with a modern computer-generated graph of the same thing:
upload.wikimedia.org/wikipedia/commons/8/80/GammaAbsSmallPlot.svg
That means that yes, _someone sat down there and calculated, by hand, the values to suitably high precision, at many different points in the plane_ . So such calculations definitely have been done. Another thing I notice is that there are, actually, in many older math texts, topics that are often not covered in today's texts. While I understand that is often because many of them pertain to calculational techniques that have been "obsoleted" through the use of the computer, I tend to think that there is still a bit of a loss here, because while we may be able to use a computer to calculate far more data then we ever could by hand, knowing these techniques gives one access to more ways of _thinking_ , and in my mind you can never have enough of such ways.
Not to mention, the artistry skills it would take to draw such a thing so nicely. There really has been a lot "drained" from modern schooling, even though I think other advances are welcome.
a few months ago I was working on (a+bi)^(c+di) and I got the same answer I'm so pumped
I'm sitting next to this guy on test day.
You uploaded the video on my birthday😊
OMG! He had to use a blue pen!
YEYYYY
I looked for days to find a non-polar formula for (a+bi)^c, and now here am I, looking for a non-polar formula of this.
This is my favourite video that I’ve watched on youtube in a while. I love the formula derivation style. Please make more “Creating a formula from scratch” videos! #YAY
So proud when you say Théta (θ)
Brilliant!
But u can write cos (pi/4+x) as (cosx+sinx)/sqrt (2) and formula simplifies
The only issue is that theta is not unique: you can always add two pi to it, but that changes the result of the exponentiation.
Yea. I mentioned that in the video.
Sure. And in fact one should not be surprised that the complex power will have many values in general. After all, it is the mother of radicals, and each nth root has up to two values in the real numbers, and n values in the complex numbers. Thus the most general definition of exponentiation should contain some suitable "source" of this behavior, and that is precisely it: the ambiguity in the argument of the base.
Wonderful ❤️ complex number power explained!!
Thank u i really enjoy ur calculations
Thanks!!!
This is so interesting, but also entertaining. I had some good laughs
i don't know about math, but it's satisfies me when know the answer(s)
it is incorrect to equal argument to inverce tangent, cause inverce tangent is defined as a number from -π/2 to π/2, ang arg is sometimes defined at -π to π or 0 to 2π
I solved this a while back. But I also went on to show which combinations of a, b, c and d produce a real number result. (We know i^i is real). Then graph them! Interesting result.
Perhaps you could have a go.
3:46
(BlackRedBlue)(Pen)^3: i c theta
me: I see theta too! It is in the exponent!
regular numbers: the overworld
i : undiscovered nether world of minecraft 1.16
uno de los mejores vídeos que he visto! saludos!!
Sir you are just amazing I am so sorry for that you are not my collage teacher
Awesome. Really brilliant way of solving complex number problem. . I want to know whether it can be derived as the Cartesian form of complex number.
“I knoww, we see a d-theta, but we are not doing trig integrals, okay? So just calm down.”
u r my inspiration sir love you
Very nice explain sir
Wow I'm learning so much from UA-cam 💕.
Thanks to BLACKPEN REDPEN
YAY!!!!!
: )
What to do when we have no complex number for example: (sqrt(2))^(sqrt(3))
beautiful!
How u can manage to solve all these tough problem in some minutes.for me it took lots of days to solve a difficult problem .
U just make the maths more. Easier for me😊😊 thanks for that😊😊😊😊😊😊😊
There's a little thing called "preparation" that goes into the making of the videos; as well as into teaching a class.
A good teacher, like bprp, does that in a way that makes it easier for the viewers/students to grasp the underlying principles.
Fred
Still, this prove is rather easy for a mathemtician and can be found in 20 minutes or so.
I reached to this formula before!! How amazing is that 😍😍❤️
Me too
very nice maestro!
Your videos are so interesting!!
Writing Arg(a+bi) as arctan(b/a) would be clearer.
老师，现在的数学三维数学！你这个太前沿了
black-red, that formula is either utterly incomplete or I didn't hear the magic word "principal". Since the complex arguments will (and do) appear in the result as amplitudes, you have to consider { arg(z) } and not the Arg(z)...
Lighting needs fixing. Video becomes darker and lighter based upon the position of your head. E.g. here: 4:38. BTW: I like your channel, very good for my math skills.
Wow, I got the answer on my own! I've been watching too many of these videos...
Here is an easier form to memorize
Result r is e^(c.ln(r)-d.Arg)
Result Arg is (d.ln(r)+c.Arg)
9:34 use sum of angle formulas to expand cos and sin please ;)
2:48 No it is not an argument... This comment is an argument!
If c is an integer don't you get different answers based on what value you choose for Arg? Because the values of Arg are all 2*pi*n away from eachother it doesn't matter what value you choose inside the sin and cos functions, but it does matter for the e^(-d*Arg) in front. How can there be multiple answers?
Nice one. I'm beginning to like complex analysis even though my field of study is physics :)
Why can't I get rid of e^theta when we have e^(i*c*theta)*e^(-d*theta)? Why does the imaginary part stop us from cancelling this seemingly obsolete power?
Hey, 2penman, can you make a video about faster complex number multiplication?
next video : tetration with complex numbers! (a+bi)^^(c+di) is that even possible?
Ritu Chandra
Not sure...
Not really. There are ambiguous definition of non-integer tetration, and no one so far has found a credible, one and only definition of tetration over non-integers
Changed the thumbnail! You thought you wouldn't get caught?
Hahaha nice catch!!!
Omg you changed the video icon
Please derive the half angle formula using complex numbers.
BPRP you are a genius dude.. Respect..
Arg(a+bi) = arctan(b/a).
Real numbers are innocent. Aren't they?
Isn't it? : )
_Ghrams number has entered the chat_
Allow me to introduce myself!
@@sacrilegiousboi1271 bruh
Graham's number not just real, it's a natural number (but a really huge one)
Rayo's number be like: am I a fool to you
Arg(a+bi) is simply arctan (b/a)
So technically I = -2.06636567916 I think based on e^-pi/2
Maybe it's different in America or wherever else you teach, but in Britain I was taught that a complex numbers has real and imaginary components. And while that means you could have purely imaginary complex numbers with a zero real component, such that you could refer to some multiple of i as a complex number, then to be consistent you would have to accept all real numbers as complex numbers too. So it's best to call real numbers real, imaginary numbers imaginary, and reserve the term complex numbers for numbers that are not on either the real or imaginary axis - i.e. have non zero values for both their real and imaginary components.
The complex numbers extend the real numbers, just like the real numbers extend the rationals. So reals are complex just like rationals are real. It is a common error to refer to "complex" solutions of an equation when what is meant is non-real complex numbers.
I'm fairly sure the answer is going to be "no", but just in case, is there a closed form like this for iterated alternating addition and squaring on complex numbers?
(((N ^ 2) + M) ^ 2) + M
etc… repeating not just twice as above but P times.
There is, but it's going to be complicated. (I almost described it as ugly, but it's beautiful if you like that kind of thing.) If N = a+bi and M = c+di, the expanded form of the simple formula you have there would be a^4 - 6 a^2 b^2 + 2 a^2 c + i (4 a^3 b + 2 a^2 d - 4 a b^3 + 4 a b c - 2 b^2 d + 2 c d + d) - 4 a b d + b^4 - 2 b^2 c + c^2 + c - d^2. Add P-2 more c+di terms and it will be very complicated.
Thanks :) It seems like it still needs to be rebuilt depending on P.
Is it possible to use imaginary units for geometry? Like construct a triangle with angles expressed as a+bi?
For angles, no. at least, not in two dimensions. I'm pretty sure that complex numbered angles would be better at representing a three dimensional angle.
I haven’t seen any answers online for this, so I’m hoping you might know: what are all of the solutions to x^sqrt(2)=1
There is only one. 1
@@adevva._ What about exp(sqrt(2)nπi)?
lmao my heart rate going up every time i see d(theta), because of all the calculus conditioning that response in me