# Inverse tangent function in real and complex domain

1. May 12, 2012

### bluecode

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

2. Relevant equations

3. The attempt at a solution
I've only been able to do part (a) of this question.
I ended up with:
$$tanz= i ({\frac{1-e^{(2iz)}}{1+e^{(2iz)}}})$$
I'm not sure how to approach the next two parts. If anyone could give me any pointers, I'd be very grateful! Thanks!

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2. May 12, 2012

### jackmell

Why not just for starters approach it formally, get the answer, then justify what you did. I'll use the alternate formula:

$$\arctan(z)=i/2\log \frac{i+z}{i-z}$$
and you have:
$$tan(w)=i\left(\frac{1-e^{2iw}}{1+e^{2iw}}\right)$$
ok then, how about if I let:
$$w=i/2\log\frac{i+z}{i-z}$$
then can you not just muscle-through:
$$\tan(w)=\tan\left(i/2\log\frac{i+z}{i-z}\right)$$

Last edited: May 12, 2012