# Exp(tanz) = 1, complex analysis

1. Jun 15, 2007

### malawi_glenn

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

Find all solutions to:

$$e^{\tan z} =1, z\in \mathbb{C}$$

2. Relevant equations

z = x+yi

$$\log z = ln|z| + iargz +2\pihi, h\in \mathbb{Z}$$

$$\log e^{z} = x + iy +2\pihi, h\in \mathbb{Z}$$

$$Log e^{z} = x + iy$$

3. The attempt at a solution

I do not really know how to approach this, I tried to beging with writing tan(z) as Alots of cos(x)sinh(y) etc..

But can I do the Log(e^tan(z)) at the left side, then do the log(1) at the right side? I mean the "only" difference is that you get this $$2\pihi, h\in \mathbb{Z}$$ on both sides, so you always reduce both these terms to one: $$2\piUi, U\in \mathbb{Z}$$

What do you think?

2. Jun 15, 2007

### Dick

Ok, taking logs you get tan(z)=2*pi*i*n (yes, no need for separate 2*pi*i*n's on both sides. Why not write tan(z) in terms of complex exponentials, let t=e^(iz) and solve the resulting quadratic for t? Guess I'm not sure where you are having problems.

3. Jun 15, 2007

### malawi_glenn

so if I write

$$\tan z =\dfrac{1}{i} \dfrac{t-t^{-1}}{t+t^{-1}} = \dfrac{1}{i} \dfrac{t^{2}-1}{t^{2}+1}$$

then perform the Log(e^tanz) on left side, then log(1) on right side, is that a "legal" act? or do I have to take log - log /// Log - Log ?..

Last edited: Jun 15, 2007
4. Jun 15, 2007

### malawi_glenn

or you mean AFTER taking log's i write tanz as that?

5. Jun 15, 2007

### malawi_glenn

okay this is as far I can get:

$$e^{iz} = t$$

$$z = - i \ln |t| + argt$$

$$\tan z = \dfrac{1}{i} \dfrac{t^{2}-1}{t^{2}+1}; t^{2} \neq -1; \Rightarrow t \neq \pm i$$

$$\tan z = \log 1 = 0 +0i + 2\pi ni; n \in \mathbb{Z}$$

$$\Rightarrow \dfrac{1}{i} \dfrac{t^{2}-1}{t^{2}+1} = 2\pi ni$$

$$\Rightarrow \dfrac{t^{2}-1}{t^{2}+1} = -2\pi n$$

$$\Rightarrow t^{2}-1 = -2\pi n(t^{2}+1)$$

...

$$\Rightarrow t = \pm \sqrt{\dfrac{2\pi n-1}{2\pi n+1}} \neq -1$$

$$\Rightarrow n \neq 0$$

$$t_{1} = "+"sqrt(.. \Rightarrow arg(t_{1}) = 0 + 2\pi k, k \in \mathbb{Z}$$
$$t_{2} = "-"sqrt(.. \Rightarrow arg(t_{2}) = \pi + 2\pi k$$

it can be shown that t is always real, if n = -2, we get a negative nominator and a negative denominator, hence the number inside the squareroot is always positive.

$$z_{1} = - \dfrac{i}{2} \ln \left( \dfrac{2\pi n-1}{2\pi n+1} \right) + \pi + 2\pi k$$

$$z_{2} = - \dfrac{i}{2} \ln \left( \dfrac{2\pi n-1}{2\pi n+1} \right) + 2\pi k$$

$$z = k\pi$$

and:

$$z = \dfrac{i}{2} \ln \left( \dfrac{2\pi n-1}{2\pi n+1} \right) + \frac{\pi}{2} + 2\pi k$$

LOL I am soooo cloose!! =(

Last edited: Jun 15, 2007
6. Jun 15, 2007

### Dick

I get t^2=(1-2*pi*n)/(1+2*pi*n). So I get a number that is always negative. So I agree with the pi/2 phase in the book answer. But I get +/-pi/2+2*pi*k which is the same thing as pi/2+pi*k. But I'm also getting -i/2 in front of the log. (But I seem to get the right answer with either sign - this is troubling me).

7. Jun 15, 2007

### malawi_glenn

yes I did a relly sucky thing in getting the t^2.. I should be ashamed! :)

if n = 0 then it is positive, right? But can not be zero according to the aswer, now why is that?

Last edited: Jun 15, 2007
8. Jun 15, 2007

### Dick

Aggghhh. And get this, the sign in front of the log doesn't matter either. Take for example the cases n=1 and n=-1. The arguments of the logs are just reciprocals of each other. Must be fun to create obfuscated answers for these things.

9. Jun 15, 2007

### Dick

n=0 gives you the real solutions, the k*pi list. Not to hard so see working it as a separate case.

10. Jun 15, 2007

### malawi_glenn

I got it now, thanx again for all the help. I hope i did not made you indignant.

11. Jun 15, 2007

### Dick

I was working to hard to correct my own blunders to become indignant over yours.

12. Jun 15, 2007

### malawi_glenn

I dont understand why the answers must be so "clean".. it only cunfuses and does not get you learn more about the actual subject.

13. Jun 16, 2007

### Dick

It can be helpful to numerically evaluate a few cases of a solution for small n,k etc and then compare with yours. It's pretty clear when they don't agree - and if they do it can help to make it clearer why.