image of $\{x+iy:\frac\pi 4<|y|<\frac {3\pi}4\}$ under $\coth(z)$

Question

Find the image of the set $\{x+iy:\frac\pi 4<|y|<\frac {3\pi}4\}$ under the map $\coth(z)$.

I tried to write $w=\frac{e^{2z}+1}{e^{2z}-1}$ and then $e^{2z}=\frac{w+1}{w-1}$.

I took $w=u+iv$ and tried to find conditions on $u$ and $v$, but we cannot extract logarithm on both sides so it doesn't help.

Following a comment:

Let $f(z)=2z$, $g(z)=e^z$ and $h(z)=\frac{z+1}{z-1}$, then $\coth(z)=h\Big(g\big(f(z)\big)\Big)$.

Now, if $D=\{x+iy:\frac\pi 4<|y|<\frac {3\pi}4\}$, then

$$f(D)=\{x+iy:\frac\pi 2<|y|<\frac {3\pi}2\}$$ $$g\big(f(D)\big)=\{re^{i\theta}:r>0,\frac{\pi}{2}<\theta<\frac{3\pi}{2}\}=\{z:Re(z)<0\}$$ $$h\Big(g\big(f(D)\big)\Big)=\{z:Re(z)>0\}$$

Is it so?

Answer 1

Your solution is 2/3 correct, $f(D)$ and $g(f(D))$ are determined correctly.

But $h(z)=\frac{z+1}{z-1}$ maps the left half-plane to the unit disk: One (of many possible) argument is that the left half-plane is the location of all points which are closer to $z=-1$ than to $z=+1$, i.e. the location of all points for which $|z+1| < |z-1|$.