<s>[INST] if $|\tan z |=1$, then what $\Re(z)$ is equal to? 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.
 [/INST] $\tan (x+iy) = \frac{\sin (x+iy)}{\cos (x+iy)}$
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
$|\tan (x+iy)| = \frac {|\sin 2x + i\sinh 2y|}{\cos 2x + \cosh 2y}\\
\frac {\sqrt {\sin^2 2x + \sinh^2 2y}}{\cos 2x + \cosh 2y}\\
\frac {\sqrt {1-\cos^2 2x + \cosh^2 2y-1}}{\cos 2x + \cosh 2y}\\
\frac {\sqrt {(\cosh 2y + \cos 2x)(\cosh 2y - \cos 2x)}}{\cos 2x + \cosh 2y}\\
|\tan (x+iy)| = \sqrt {\frac {\cosh 2y - \cos 2x}{\cos 2x + \cosh 2y}}=1$
$\cosh 2y - \cos 2x = \cos 2x + \cosh 2y\\
\cos 2x = 0$
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