Complex Analysis - Transcendental Solutions Help

1. Feb 23, 2012

gbu

This isn't really homework help. I'm working through a complex analysis textbook myself, and am stumped on the complex transcendentals, but I figured this was the best place for it. I would greatly appreciate any guidance here, I'm getting very frustrated!

1. The problem statement, all variables and given/known data

The problem is to find all solutions of $e^{2iz} = 1$ where $z \in \mathbb{C}$.

The correct answer is, I believe $z = n \pi$ for any integer n.

2. Relevant equations

Euler's equation: $-1 = e^{i \pi}$

3. The attempt at a solution

I tried turning the right hand side in to $-e^{i \pi}$ via Euler's equation, then taking a logarithm of both sides... gives $2iz = i \pi$... but Wolfram Alpha says the answer is $n \pi$ where $n \in Z$. Clearly not where I got to.

2. Feb 23, 2012

Office_Shredder

Staff Emeritus
If $-1=e^{i\pi}$, what is $e^{2i\pi}$? (Then consider the rest of the solutions and how they fit in)

You can find them more directly by just using Euler's formula $e^{2iz} = e^{2i(a+bi)} = e^{-2b}e^{2ia} = e^{-2b}(\cos(2a)+i\sin(2a))$

How can you solve for a and b here?

3. Feb 23, 2012

gbu

I can see intuitively the correct answer from the first line (thanks a lot for that!), it makes it obvious that $n \pi$ is an answer... I'm still not sure how to solve a and b in that equation though.

$e^{-2b}(\cos(2a)+i\sin(2a)) = 1$
$\cos(2a)+i\sin(2a) = e^{2b}$

I don't know where to go from here. Perhaps this is a simpler algebra problem than I'm making it, but it seems like there's two unknowns and only one equation?

[strike]Edit: wait, $\cos(2a)+i\sin(2a)$ is a complex number itself... I'm still not sure where to go with that. Let $w = \cos(2a) + i\sin(2a)$ and we have

$e^{-2b} w = 1$

But, still not sure how to get to n \pi.[/strike] I'm clearly very confused. I'll go back to the textbook I suppose.

Last edited: Feb 23, 2012
4. Feb 23, 2012

Office_Shredder

Staff Emeritus
What is the absolute value of $$e^{-2b}(\cos(2a)+i\sin(2a))$$? (Hint: it's $e^{-2b}$

5. Feb 23, 2012

gbu

Right. Which leaves me with

$e^{-2b} = 1$

Which we can log both sides...

$-2b = \ln 1 = 0$

Telling me b = 0.

As for a, we have $e^{-2b}(\cos(2a) + i\sin(2a)) = 1$, but we know b is 0.

[strike]So we have [itex ]\cos(2a) + i\sin(2a) = 1[/itex]

[itex ]\cos(2a) + i\sin(2a) = 1[/itex] or [itex ]e^{2ia} = 1[/itex][/itex]

Log both sides... nope?

[itex ]2ia = 0[/itex][/strike]

Still no idea what to do with a. :-( I greatly appreciate your help so far... do you have any recommended reading for this basic stuff? The textbook I'm using is a bit too "succinct" for my learning style. It was going great until we got in to something I didn't "get", and now I feel like I have no where to go with it.

I get the feeling you can't go log(e^Z) = Z where Z is a complex variable. Is this correct?

Edit:

I think when we get to

$\cos(2a) + i\sin(2a) = 1$

We can just solve by recognizing that cos and sin are periodic on $2\pi$ and thus 2a can be rewritten as $2a + n\pi$ giving

$\cos(2a + n\pi) + i\sin(2a + n\pi) = 1$

I still don't know how to get rid of all that stuff though. :(

Last edited: Feb 23, 2012
6. Feb 24, 2012

jackmell

What's wrong with just taking log of both sides?

$$z=\frac{1}{2i}\left(\ln|1|+i(0+2k\pi)\right)$$

and that's just:

$$z=k\pi,\quad k\in\mathbb{Z}$$

I think though maybe you're not recognizing that the complex log is infinitely-valued:

$$\log(z)=\ln|z|+i(\text{Arg}(z)+2k\pi)$$

and that's what I'm doing implicitly in the first equation above.

Last edited: Feb 24, 2012
7. Feb 24, 2012

gbu

Brilliant, thank you jackmell! That's exactly what I was missing. It's funny, the textbook clearly says that, but I just didn't really "see" it.

8. Feb 24, 2012

Office_Shredder

Staff Emeritus
Two complex numbers are equal to each other, so the real and complex parts are:
cos(2a)=1
sin(2a)=0

hopefully at that point it's clear that a is any integer

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