if $|\tan z |=1$, then what $\Re(z)$ is equal to?

Question

if $|\tan z|=1$, then what is $\Re(z)$=?

I tried both the ways i.e. $\tan z=\tan(x+iy)$ and the exponential method but I didn't get an answer that I want. Any help is appreciated.

Answer 1

Hint:

WLOG $\tan(x+iy)=\cos t+i\sin t$ where $t$ is real

$\implies\tan(x-iy)=\cos t-i\sin t$

$\tan(x+iy)\tan(x-iy)=1\iff\cos(x+iy)\cos(x-iy)=\sin(x+iy)\sin(x-iy)$

$$0=\cos(x+iy+x-iy)=?$$

Answer 2

$\tan (x+iy) = \frac{\sin (x+iy)}{\cos (x+iy)}$

Expand using trig idenities

$\frac{\sin x\cosh y + i\cos x\sinh y}{\cos x\cosh y - i\sin x\sinh y}$

Multiply top and bottom by the conjugate of the denominator to make it rea.

$\frac{\sin x\cos x(\cosh^2 y - \sinh^2 y)+i(\cos^2x + \sin^2x)\cosh y\sinh y}{\cos^2 x\cosh^2 y + \sin^2 x\sinh^2 y}$

Simplify the numerator using

$\cos^2 x + \sin^2 x = 1\\ \cosh^2 x - \sinh^2 x = 1$

$\frac{\sin x\cos x+i\cosh y\sinh y}{\cos^2 x\cosh^2 y + \sin^2 x\sinh^2 y}\\ \frac{\frac 12 (\sin 2x + i\sinh 2y)}{\cos^2 x\cosh^2 y + \sin^2 x\sinh^2 y}$

With the numerator as a function of $2x$, it would be nice to get the denominator as a function of $2x$

$\frac{\sin x\cos x+i\cosh y\sinh y}{\cos^2 x\cosh^2 y + \sin^2 x\sinh^2 y}\\ \frac{\frac 12 (\sin 2x + i\sinh 2y)}{\frac 14 (1+\cos 2x)(1+\cosh^2 2y) + \frac 14(1-\cos 2x)(\cosh 2y - 1)}\\ \tan (x+iy) = \frac {\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y}$

That is the hard part

$|\tan (x+iy)| = \frac {|\sin 2x + i\sinh 2y|}{\cos 2x + \cosh 2y}\\ \frac {\sqrt {\sin^2 2x + \sinh^2 2y}}{\cos 2x + \cosh 2y}\\ \frac {\sqrt {1-\cos^2 2x + \cosh^2 2y-1}}{\cos 2x + \cosh 2y}\\ \frac {\sqrt {(\cosh 2y + \cos 2x)(\cosh 2y - \cos 2x)}}{\cos 2x + \cosh 2y}\\ |\tan (x+iy)| = \sqrt {\frac {\cosh 2y - \cos 2x}{\cos 2x + \cosh 2y}}=1$

$\cosh 2y - \cos 2x = \cos 2x + \cosh 2y\\ \cos 2x = 0$