Inverse tangent function in real and complex domain

[bluecode]

Homework Statement

See attached file.

Homework Equations

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!

 

[jackmell]

Homework Statement

See attached file.

Homework Equations

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!

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:
$$\tan(w)=\tan\left(i/2\log\frac{i+z}{i-z}\right)$$