Find all $z$ such that $\left|\tan z\right| = 1$

Question

Find all z such that $$\left|\tan z\right| = 1$$

The first thing that came to my mind was to write tangent in terms of $e^z$ and take its modulus, but I couldn't solve it in this way.

Answer 1

Here is one way outlined:

1) Verify that (for $x\in\mathbf R$ and $y\in\mathbf R$) $$ \overline{\tan(x+iy)}=\tan(x-iy) $$

2) Make the following calculation $$ |\tan(x+iy)|^2=\tan(x+iy)\tan(x-iy)=\cdots=\frac{-\cos 2x+\cosh 2y}{\cos2x+\cosh2y}. $$

3) Conclude from the right-hand side (I leave that to you).

Answer 2

$$\tan z= \frac{\sin z}{\cos z}=\frac{e^{iz}-e^{-iz}}{i(e^{iz}+e^{-iz})}$$

So $$|\tan z|= \left|\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}\right|=1$$

$$\Rightarrow |e^{iz}+e^{-iz}|=|e^{iz}-e^{-iz}|$$

$$\Rightarrow |e^{2iz}+1|^2=|e^{2iz}-1|^2$$

$$\Rightarrow (e^{2iz}+1)\overline{(e^{2iz}+1)}=(e^{2iz}-1)\overline{(e^{2iz}-1)}$$

$$\Rightarrow (e^{2iz}+1){(e^{-2iz}+1)}=(e^{2iz}-1){(e^{-2iz}-1)}$$

$$\Rightarrow 1+e^{2iz}+e^{-2iz}+1=1-e^{2iz}-e^{-2iz}+1$$

$$\Rightarrow e^{2iz}+e^{-2iz}=0$$

$$\Rightarrow \cos(2z)+i\sin (2z)+\cos(2z)-i\sin (2z)=0$$

$$\Rightarrow \cos(2z)=0$$

So$$2z=n\pi+\frac{\pi}{2}$$

Answer 3

WLOG let $\tan z=\tan(x+iy)=\cos A+i\sin A$ where $A,x,y$ are real

$\tan(x+iy+x-iy)=\cdots=\dfrac{2\cos A}{1-1}=0\implies2x=n\pi+\dfrac\pi2\ \ \ \ (1)$

$\tan\{x+iy-(x-iy)\}=\cdots=\dfrac{2i\sin A}{1+1}$ $\iff i\sin A=\tan(2iy)=i\tanh(2y)\iff\dfrac{e^{2y}-e^{-2y}}{e^{2y}+e^{-2y}}=\sin A$

Applying Componendo and Dividendo, $$e^{4y}=\dfrac{1+\sin A}{1-\sin A}$$

$$\implies4y=\ln\dfrac{1+\sin A}{1-\sin A}\ \ \ \ (2)$$