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

Can you help me figure out what is the image of line segments $${z =x+iy: -π/2 under $$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?

• math.stackexchange.com/questions/1312756/… Your question asks specifically for the shape when $y=cst$, while this linked post asks for the image of the complex plane when $y$ describe all $\mathbb R$, so this is more general. I addressed it in the second part of my post. – zwim Mar 17 at 22:05

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 $$\tan(\Omega)=\mathbb C\setminus\{iy\mid |y|\ge 1\}$$.