Complex Analysis - Transcendental Solutions Help

[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!

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

I tried turning the right hand side into -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.

 

[Office_Shredder]

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?

 

[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.

 

[Office_Shredder]

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

 

[gbu]

Right. Which leaves me with

<br /> e^{-2b} = 1<br />

Which we can log both sides...

<br /> -2b = \ln 1 = 0<br />

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 into 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. :(

 

[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.

 

[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.

 

[Office_Shredder]

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

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