exp(tanz) = 1, complex analysis

[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?

[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.

[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 ?..

[malawi_glenn]

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

[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

Answer in textbook:

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!! =(

[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).

[malawi_glenn]

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


I get t^2=(1-2*pi*n)/(1+2*pi*n). So I get a number that is always negative.

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

[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.

[Dick]

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?

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

[malawi_glenn]

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

[Dick]

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

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

[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.

[Dick]

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

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.