What is image of $f(z)=\tan(z)$ where $\Im(z)=cst$?

Question

Can you help me figure out what is the image of line segments ${z =x+iy: -π/2<x<π/2, y=const}$ under $f(z)=\tan(z)$.

I've got $tan(x+iy) = sin(2x)/(ch(2y)+cos(2x)) + i sh(2y)/(ch(2y)+cos(2x)) = u + iv$ and $tan(z)=w => z=-i1/2\log((wi+1)/(1-wi))$ What should I do next? Is that a good way to start?

Answer 1

You can develop $\tan(x+iy)=\dfrac{\tan(x)+i\tanh(y)}{1-i\tan(x)\tanh(y)}$

And let's call $a=\tanh(y)$ constant and $t=\tan(x)\in]-\infty,+\infty[$

We have the curve $X+iY=\dfrac{t+ia}{1-iat}\iff\begin{cases}X=\dfrac{(a^2-1)t}{1+a^2t^2}\\Y=\dfrac{a(t^2+1)}{1+a^2t^2}\end{cases}$

When $t=0$ then the curve passes by $(0,a)$ and at infinity by $(0,\frac 1a)$.

We can verify by calculation, that it is a circle tangent to the lines $Y=a$ and $Y=\frac 1a$ of origin $Y_0=\dfrac{a+\frac 1a}2=\dfrac{a^2+1}{2a}$

$X^2+(Y-Y_0)^2=R^2$ with $R=\left|\dfrac{a^2-1}{2a}\right|$

https://www.desmos.com/calculator/7l3l66kuyq

---

Now if you consider the set $\Omega=\{x+iy\mid x\in(-\frac\pi 2,\frac\pi 2),y\in\mathbb R\}\sim\{(a,t)\mid a\in(-1,1),t\in\mathbb R\}$

Then $\tan(\Omega)$ is a family of circles covering the plane.

• in fact the points $(0,\pm 1)$ cannot be reached since $|a|<1$
• also since $|t|<\infty$ the points $\{(0,y)\mid |y|>1\}$ cannot be reached either.

I enhanced the effect on the visual below by limiting $|t|<1000$ and $|a|<0.9$.

You can see the "areas" not reached.

This is of course exaggerated, in reality $\tan(\Omega)=\mathbb C\setminus\{iy\mid |y|\ge 1\}$.