It makes me think about what my professor said about research. “It’s nice having new and even naive eyes to look at a problem, because sometimes they’ll try something you didn’t think of. Or even if they do try something you think doesn’t work, maybe they find a new way to make it work.” As we get older and more experienced, it’s easy to forget the basics and how useful they are to apply first. We get used to more powerful theorems and having to use them, so we end up skipping the basics because in a classroom setting that’s usually what the problem is going to require anyway.

@IoT_

2 жыл бұрын

There is a name to this amazing phenomenon :Einstellung effect

@andrewkarsten5268

2 жыл бұрын

@@IoT_ thanks for telling me! I learned a new term, that’s a good one

@IoT_

2 жыл бұрын

@@andrewkarsten5268 You're very welcome 🍀

@yassinesafraoui

2 жыл бұрын

And that's exactly why the first thing I thought about was using binomials to distribute that (1+ib)^N

Alternate method: We have |z| = |1 - iz| = |-i - z|, so z is equidistant from 0 and -i. QED

@drpeyam

2 жыл бұрын

Very nice

@goblin5003

2 жыл бұрын

I like this approach

@elihowitt4107

2 жыл бұрын

Geometrically in one step 'z rotated pi/2 clockwise' is between 0 and 1 (Re=1/2), reverse the rotation and z lands zero and minus i

@giacomolanza1726

2 жыл бұрын

you found it earlier than me - sorry for not paying attention!

@matthew-m

2 жыл бұрын

That's quite beaitiful!

As a graduating senior who struggled through Complex Analysis this whole semester, this was so satisfying to solve and see that I got the right answer

Lovely problem, solution, and back story as well. So often solutions are presented without all the false trails that all of us run through frequently, giving a false impression of not just the mathematics, but the joy of discovery.

@drpeyam

2 жыл бұрын

Thanks so much!!!

@johnnychinstrap

2 жыл бұрын

@@drpeyam It is almost slapstick comedy as to how simple this solution is given that it could get quite intense. When you frame it in a sense that can be understood by a 12 year old you pretty much express the mastery of the topic. If you can teach calculus to kindergarten then you know your sh&^. great video for us math geeks. humbly chins

This was such a fun video. The backstory was the best part!

I feel some math problems are put there by instructors in exams just to troll (prospective) students...lol

@drpeyam

2 жыл бұрын

It was a legit problem on a legit exam 😆

I dunno if this helps but... you can start by dividing both sides by z^n since z in non-zero: 1+(1/z-i)^n=0 do some algebra: (1/z-i)^n=-1 we know that 1/z-i has the form e^(i(pi/n)K) where K is an integer, but I'm just going to call (pi/n)K T because otherwise it's a hassle do some moreish algebra: z=1/(i+e^(iT))=1/(cos T +(1+sin T )i) which means that the imaginary part of z is -(1+sin T)/(cos^2 T+(1+sin T )^2) which if you foil just becomes -(1+sin T)/(2(1+sin T))=-1/2 liketysplik

I am applied Calculus and like engineering applied, but I am new to complex analysis, but this is brilliant. I never would have thought to do it so cleanly and simply. I can not remember the quote but it was something like never under estimate mans ability to over-complicate the obvious to oblivious.

mind-blowing Dr peyam!!

I really enjoyed that! Thanks!

@drpeyam

2 жыл бұрын

Thank you 😊

I'm about to take Complex Analysis in the Fall, and I'm pretty intimidated, especially combined with Physical Chemistry. This video made things feel a little less scary. Love your enthusiasm, Dr. Peyam! This solution was beautiful and straightforward enough to follow. Lovely work.

Lots of fun, Dr. Peyam

I am teaching complex analysis this semester, so I think I am ready for the final on Monday 😀 that I’ll give, did not watch your video yet to avoid spoilers. We were repeatedly using reverse triangle inequality for improper integrals (residue calculus), I immediately applied it because of that mindset, that gave an easy solution. More precisely, working on a lot of integrals on semicircles or similar contours with ML formula type of arguments make this very easy. Let me see what you applied. Edit: Watched your solution now. Yeah, your feedback is a funny one could take “wrong” directions easily by considering harsher machinery like you are mentioning. In my case, it helped me (not the advanced tools, but the basic tool, reverse triangle inequality, appearing with the advanced tools nearby) if I saw this question while I was doing branch cuts I could have a different mindset😀. Here is my argument very close to yours: Say a= z^n, b=(1-iz)^n. Then a+b=0 implies |a+b|=0 hence | |a| -|b| | = 0 because it is a non negative thing which is less than or equal to 0, but then |a| = |b|, the rest of my argument is almost identical. It could have been a good exam problem, but I already wrote it.

Thanks for the trick of the modulus :) After it minute 1:30 you can take a geometrical shortcut, just observing that * |z| = Euclidean distance of the point `z` (a,b) from the origin * |1-iz| = |i+z| = Euclidean distance of the same point `z` (a,b) from the point `-i` (0,-1); * the axis of the segment joining the point `-i` with the origin has equation Im z = -1/2.

I always like the back stories of the presented problems.

@drpeyam

Жыл бұрын

Agreeeeed

Polar form was the first thing that came to mind. Sin and cosine at 45 degrees +- 90 is sqrt 2 x 1/2. I didn't try this, don't know if it even works, but that was my first thought.

Hi Dr. Peyam, The conclusion of this statement is quite similar to the conclusion of the Riemann Hypothesis. Perhaps there's a 3 minute video you can make which solves that problem as well :D

This problem is easy to solve when you know how to do it 😃

Why does reduction of order work?

Also, what is the benefit of using absolute value, here?

cool solution

I had to come back because this is brilliant and reminds me of when i was teaching Laplace for control systems. There was an overwhelming line that needed to be solved and I started to use long division to solve it and my class freaked out because they could not understand it and thought it was some new high level math I was teaching. I told them a grade 5 could solve this and its just long division. Well the short of it is they took long division out of the curriculum in grade school so these kids had no idea what I was doing. I taught them some grade 5 math and then they could see how it was useful to solve this otherwise 10 page solution for the feedback loop. You really do not understand something unless you can teach it to a child. Cheers Just curious but could you not skip the step of taking the modulus because you already know the answer is going to yield a positive and negative result.

I almost solved ... because I thought that I should use some kind of "Proof by induction" on 'n'. When I saw you taking the modulus and just throwing the 'n' away, I was kind : 'Ok... I won't worry about the 'n' anyway! 😁😅'.... Sometimes, the 'simpler' kind of solution can be the most difficult, because our minds are trained to complicate things! Good video, Peyam! 🙂😁

@drpeyam

2 жыл бұрын

Exactly!

@ardademirhan2452

2 жыл бұрын

The funny thing is when there is an “n” there “is” an induction most likely. “Is” means there can be an inductive argument making that thing work. In this example, how do we see that a^n=b^n implies a=b for positive real numbers? One way to see that is to show that the function f(x)=x^n (n>0) is an increasing function for positive reals by induction (not by calculus!), base case is trivial, if x^(n-1)>y^(n-1) when x>y, then x^n= x* x^(n-1) >x* y^(n-1)>y*y^(n-1)=y^n when x>y. So your induction is served fresh for you 😀! With that said, intuitively speaking there are two types of induction problems: 1) those who amaze us by their elegance where induction is a surprising/nice idea and 2) the ones which are “boring” like I did, “boring” means that the fact we try to prove is very obvious.

Well that was fun

Very interesting

Another (slightly advanced) geometric solution is possible: the equation says that z/(1-iz) lies on the unit circle. Let Tz=z/(1-iz), its inverse is T^{-1}z=z/(iz+1). Which maps the unit circle (using the fact that Mobius transformations preserves "general circles") onto the line Im z=-1/2. (so the answer would be the same, even if the equation were: "z^r+e^{i\theta}(1-iz)^r=0" for some nonzero r and any theta (both real))

@drpeyam

2 жыл бұрын

First hour of complex analysis 😂

@dogandonmez5274

2 жыл бұрын

@@drpeyam Let's make it 20 hours 😄

Here is another way to do it. In hindsight i realize that this is pretty much equivalent to your solution by taking magnitudes, but whatever. We can divide by (1-iz)^n (unless z = -i, but then clearly the equality does not hold) to get (z/(1-iz))^n + 1 = 0, so (z/(1-iz))^n = -1. Thus, for some complex number w with |w| = 1, we have z/(1-iz) = w. Therefore, z = w - wiz z(1+wi) = w z = w/(1+iw). The imaginary part of this is (z - conj(z))/(2i) = (w/(1+iw) - (1/w)/(1 + 1/(iw)))/(2i) = (w-i)/(1+iw) * 1/(2i). 1 + iw = i(w-i), so this is 1/i * 1/(2i) = -1/2.

Too many tools, and you don't know which one to use (or you need longer time to know which one to use). I think this can be common for highly educated people.

Dr. Peyam, in your opinion, is it ever too late to go to grad school? I finished my BS in Math & CS about 2 years ago and have been working as a software engineer since then, but I really really miss pure Mathematics. Looking back I think that's what I enjoyed learning the most. But, I'm afraid I've forgotten too much and wouldn't do well enough in a grad program. In your experience, have you ever seen someone in a situation similar to mine go back and succeed? Thanks :)

@drpeyam

2 жыл бұрын

Never too late, I know some people in their 60s who got their math PhD and they enjoyed it

@HasXXXInCrocs

2 жыл бұрын

@@drpeyam that's inspiring, thank you!

two years ago (age 16) i solved this on a test, it was considered to be a intermediate exc. (I live in Sweden)

Thanks it works also for riemann hypothesis 😂

I don't understand how z squared is a squared plus b squared

Who made up all these rules ?

Dear Dr.Peyam, do we not lose quite a bit of information when collapsing these complex number down to just their length? Starting from 2 we could: z^N= -(1-iz)^N Ln on both sides, and Eulers identity: Ln(z) = ln(-1)/N + ln(1-iz)= iπ/N +ln(1-iz) Ln((1-iz)/z)= ln(1/z - i)= - iπ/N e^ om both sides: z^-1 - i = e^(-iπ/N) z = 1/(i+ e^(-iπ/N)) Multiplying with the complex conjugate both in denominator and numerator we have: z = (e^(iπ/N) - i)/(2(1+sin(π/N))) And so we get something thats skewed from a circle with center -i/2, stretching toward infinity whenever π/N= 3π/2 +2πm, and shrinking into radius 1/4 as π/N =π/2+ 2πm. And i think thats neat. And maybe the differences in our answers could say something cool about what exactly taking the absolute value did to the number :) it collapses it down to it's would be center line and i dont really have time to think about it more but it's very cool.

@drpeyam

2 жыл бұрын

You should be careful when applying ln, for complex numbers we don’t have ln(ab) = ln(a) + ln(b)

@MeetTheStoneMan

2 жыл бұрын

@@drpeyam is it not so that the difference between ln(ab) and ln(a) + ln(b) in complex numbers a matter of i2πm, that is by a number whole rotations? And if so would that not mean that the final answer is functionally the same?

@XJWill1

2 жыл бұрын

@@MeetTheStoneMan What do you mean by functionally the same? It is true that the complex logarithms are multi-valued and in general they can differ by i*2*pi*k for integers k, but that does not mean you can ignore it. For example, Log(-1*i) = Log(exp(-i*pi/2)) = -i * pi/2 Log(-1) + Log(i) = i*pi + i*pi/2 = 3*pi/2 So they differ by 2*pi, but they are NOT equal. If you want to use Log(w*z) = Log(w) + Log(z) in a proof, you either need to include the i*2*pi*k term, or you need to ensure that (Arg(w) + Arg(z)) is in (-pi, pi]

@XJWill1

2 жыл бұрын

By the way, your solution for z is not correct. The correct solution (I checked for various values of n such as n=8) is: z = 1/2 * cos(pi/n) / (1 - sin(pi/n)) - i / 2 If I put that into a form similar to that written in your comment, it is: z = (exp(i*pi/n) - i) / (2 * (1 - sin(pi/n)) I guess you made an error when multiplying the denominator by its complex conjugate. So it was not applying identities loosely that led you astray, but just a math error. However, applying identities to complex numbers where they do not always hold is bound to get you sometime in the future if you keep doing it.

Easy problem! x^2 + y^2 = 1 + 2y + x^2 + y^2. So y= -0.5

Namaste🙏 sirji💐.

This problem was on my complex analysis midterm, I couldn't solve it...

How do you just cancel b = 1 + b because it doesn't make sense? I would have thought that means there was a mistake somewhere for that to show up.

@drpeyam

2 жыл бұрын

Well if b = b + 1 then 0 = 1. There is no mistake here, it’s like cancelling 2 in 2x = 0

@KholofeloMoyaba

2 жыл бұрын

@@drpeyam thanks for the reply. I'm still confused. Cancelling 2 with 2x = 0 makes sense because there exists an x that makes the equation valid which is x=0. But I don't see a b that can make b = b + 1 valid, as that would imply 0=1 which is not possible. I'm probably missing something really fundamental and basic here maybe... It's like with quadratic equations you can have answers that seem like they don't make sense but once you introduce complex numbers they make sense. With that b equation I don't see any frame where it would make sense. Excuse my lack of understanding 😵. Like I get we can ignore certain results because they are not relevant, but in this case its not that the solution isn't relevant, it's just not possible

@drpeyam

2 жыл бұрын

That’s the point, b = b + 1 is not valid, but since the equality is valid it must be the second case. If P => (Q or R) is true and P is true and Q is false, then R must be true

@KholofeloMoyaba

2 жыл бұрын

@@drpeyam thank you very much for that explanation. I've learnt something new. In my mind, I was seeing it as P => (Q and R). So both Q and R were to be true, but maybe only one of them relevant for the question we are asking e.g. I'm solving quadratic roots but I'm only interested in one of the solutions but both solutions are true. But with the "or" analogy it makes sense. Thanks again.

@KholofeloMoyaba

2 жыл бұрын

@@drpeyam actually I have a rebuttal. If you don't take a short cut and expand the right side it actually gives one solution, and b = b + 1 is never even an option. starting from b^2 = (1+b)^2, if you expand it out it is: b^2 = 1 + 2b + b^2 , (the squared bs cancel out and you are left with) 0 = 1 + 2b therefore 2b = -1, which leaves b = -1/2 So I think the problem was the shortcut taken - both sides can be negative or positive not just the right hand side when taking the square root. Does this make sense?

If one has the equation z^8 = (1-i * z)^8 one should note that is means z = w*(1- i *z ) ,where w is a solution of w^8 = 1 ,that is , w= Exp[ i * k*π/4) , k= 0, 1 .....7. One then has linear equations for z which are straightforward to solve.

@drpeyam

Жыл бұрын

There’s a much easier way

@renesperb

Жыл бұрын

@@drpeyam If for eample N=8 ,how do you find all 7 solutions ?

I'm gonna try to solve this on my own, by induction. So far I have proven the base case for N=1. The difficulty is in the induction step. I've never proven an implicature with induction before. So we suppose A implies B when N=k Now I have to prove A implies B when N=k+1 So, I am allowed to assume that A is true for N=k+1. Am I also allowed to assume that A is true for N=k? My question is, without giving a hint for this problem, how do you use the assumption that A implies B for N=k in proving that A implies B for N=k+1, in general??

@drpeyam

2 жыл бұрын

There’s a much easier way :)

@txikitofandango

2 жыл бұрын

@@drpeyam Hopefully not involving the binomial theorem on (a+bi)^N or polar form

@cavalieri2946

2 жыл бұрын

I actually tried it using the induction method as well. I will give you a hint: if z^n+(1-iz)^n=0, then try to change the equation z^(n+1)+(1-iz)^(n+1) to the form z*z^n+(1-iz)*(1-iz)^n. Try to arrange the expression so that you have terms on one side of degree n and terms on the other of degree 1. Add something to make the equation zero out and conclude from there.

2b or not 2b hahaha

Looking at the angle arguments ∠(a+ib) = ∠(1-b-ia) atan2(1/2,a) = atan2(-a,-1/2) ; atan2(y,x) format you find a=0 This equation only has solutions when N is even. z=±i/2 Maybe we need to invent a new type of number that gives this equation a solution when N is odd!!!

@drpeyam

2 жыл бұрын

There are other solutions though

@jessstuart7495

2 жыл бұрын

@@drpeyam, You are right. I used Maxima (CAS) to look at some of the solutions for odd N. I'll figure out where I went wrong. Thanks for the reply, and the interesting videos.

That's actually tricky, but for grad school its probably fine, and a tiny bit evil

@danielleza908

2 жыл бұрын

In my country we solve these kind of problems in high school...

2:24 it makes perfect sense b = b - e^iπ

@mrocto329

2 жыл бұрын

??? what you're saying is b = b - e^iπ e^iπ = b - b e^iπ = 0 e^iπ = -1 because Euler doing his thing -1 = 0

@DeadJDona

2 жыл бұрын

@@mrocto329 I'm stupid, but I believe you could stick Euler here somewhere.

@DeadJDona

2 жыл бұрын

like e^2πi = 1, eyee suppose

@mrocto329

2 жыл бұрын

@@DeadJDona You can't really do anything here, b = b + 1 just doesn't make sense as you can subtract b from both sides and suddenly 0 = 1

@DeadJDona

2 жыл бұрын

@@mrocto329 but... e^πim for m = 0..inf

2

Will this be it? Sorry if mistakes z^N + (1 - iz)^N = 0 (1 - iz)^N = - z^N (1 - iz)^N / z^N = -1 (1/z - i)^N = -1 1/z - i is a root of -1, all of which fall on the unit circle. Put 1/z - i = exp(2 i t) t real 1/z = exp(2 i t) + i z = 1 / (exp(2 i t) + i) z = exp(-i t) / (exp(i t) + i exp(-i t)) Make the denominator real z = exp(-i t) (exp(i t) - i exp(-i t)) / (exp(2 i t) + exp(-2 i t)) z = (1 - i exp(-2 i t)) / (2 cos(2t)) z = (1 - i (cos(2t) - i sin(2t))) / (2 cos(2t)) z = (1 - i cos(2t) - sin(2t))) / (2 cos(2t)) Im(z) = - cos(2t) / (2 cos(2t)) Im(z) = -1/2

@drpeyam

2 жыл бұрын

Too complicated

@pwmiles56

2 жыл бұрын

@@drpeyam Yes your method is cleverer, mine is crude but it's fairly easy to actually do

@XJWill1

2 жыл бұрын

I do not know if the original problem specified that n is an integer. If n is a rational (or real or complex) number, then your proof fails on line 4 because (z/w)^n is not necessarily equal to z^n / w^n when n is not an integer.

@Apollorion

2 жыл бұрын

@@XJWill1 Quote: (z/w)^n is not necessarily equal to z^n / w^n when n is not an integer. Really? Even if we define a^b to be exp(b*ln(a)) ? edit: for a and b being complex & unequal to 0.

@pwmiles56

2 жыл бұрын

@@XJWill1 I believe that for non-real r in general |z|^r is not real (someone says that on stack exchange). So the modulus proof wouldn't work with non-real N either.

i never taken that and did it quickly : z^n + ( 1 - iz )^n = 0 1 - iz = z 1 = ( 1 + i )*z 1 / ( 1 + i ) = z ( 1 - i ) / ( 1 + 1 ) = z Im ( z ) = -0.5

@drpeyam

Жыл бұрын

Not true, a^n = b^n doesn’t imply a = b

@michaelempeigne3519

Жыл бұрын

@@drpeyam ok but if you remember, a person said to think like a person that hardly took any of that class; not that it true or not.

How come (a+bi)²=a²+b²?

@drpeyam

2 жыл бұрын

No, |a+bi|^2 not (a+bi)^2

@ozargaman6148

2 жыл бұрын

@@drpeyam oh yeah sorry, but still why?

@drpeyam

2 жыл бұрын

Complex absolute value is not the same as real absolute value

@drpeyam

2 жыл бұрын

|a+bi| = sqrt(a^2 + b^2) by definition

@ozargaman6148

2 жыл бұрын

@@drpeyam oh ok thanks!

you didn't think of absolute square as your first choice, so I guess you're not familiar with quantum mechanics...

im not a phd, sad story :v

@drpeyam

2 жыл бұрын

One day :3

I wanted to solve your problem in my head but your thumbnail misses =0 part so the problem is not what I thought it was. Who gave you a PhD if you are so inaccurate in your video thumbnails!!!

@drpeyam

2 жыл бұрын

K

JESUS CHRIST OF NAZARETH YHVH LOVES YOU

I let z = r.e^i.theta and 1-iz = rho.e^i.phi which gets rid of n and then sub back to r.e^i.theta = (1-i r.e^i.theta) = (1/2)(1-i) = sqrt(2)/2 e^i.(-pi/4) ...