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?

  • $\begingroup$ 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. $\endgroup$ – 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 $R=\left|\dfrac{a^2-1}{2a}\right|$


enter image description here

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.

enter image description here

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

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.