<s>[INST] For complex $z$, $|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right)  = 0$ 
If $|z|=1$, show that: $$\mathrm{Re}\left(\frac{1 - z}{1 + z}\right) = 0$$

I reasoned that for $z = x + iy$, $\sqrt{x^2 + y^2} = 1\implies x^2 + y ^2 = 1$ and figured the real part would be:
$$\frac{1 - x}{1 + x}$$
I tried a number of manipulations of the equation but couldn't seem to arrive at any point where I could link the two to show that the real part was = 0.
 [/INST] Not as elegant as many other approaches, but one I thought I'd try out:
Taking the ratio $ \ \frac{1 - z}{1 + z } \ = \ w \ = \ \zeta + i·\eta  \ $ to be equal to some complex number, we may arrange the equation into $ \ [ \ 1 - (\cos \theta + i·\sin \theta) \ ] \ = \ [ \ 1 + (\cos \theta + i·\sin \theta) \ ] · (\zeta + i·\eta) \ \ , \ \ z \neq -1 \ \ , $ writing $ \ z \ $ in "polar form" for $ \ |z| = 1 \ \ . $ Multiplying out the right side and equating real and imaginary parts produces the system of equations
$$  1   -   \cos \theta  \ \ = \ \ \zeta·\cos \theta \ + \ \zeta \ - \ \eta·\sin \theta \ \ \ , \ \ \ -\sin \theta \ \ = \ \ \eta·\cos \theta \ + \ \eta \ + \ \zeta·\sin \theta  $$
$$ \Rightarrow \ \ (1 + \zeta)·\cos \theta \ - \ \eta·\sin \theta \ \ = \ \ 1  -  \zeta \ \ \ , \ \ \ \eta·\cos \theta \ + \ (1 + \zeta)·\sin \theta \ \ = \ \ -\eta \ \ .   $$
We solve these to obtain
$$ \cos \theta \ \ = \ \ \frac{ (1 + \zeta)·(1 - \zeta) \ - \ \eta^2}{(1 + \zeta)^2 \ + \ \eta^2} \ \ = \ \ \frac{  1 \ - \ \zeta^2 \ - \ \eta^2}{(1 + \zeta)^2 \ + \ \eta^2}  \ \ ,      $$
$$ \sin \theta \ \ = \ \ \frac{  -\eta·(1 + \zeta) \ - \ \eta·(1 - \zeta)}{(1 + \zeta)^2 \ + \ \eta^2} \ \ = \ \ \frac{  -2·\eta }{(1 + \zeta)^2 \ + \ \eta^2}  \ \ .      $$
The sum of the squares of these expressions is
$$ \cos^2 \theta \ + \ \sin^2 \theta \ \ = \ \ \frac{ (1 \ - \ \zeta^2 \ - \ \eta^2)^2 \ + \ ( -2·\eta)^2 }{[ \ (1 + \zeta)^2 \ + \ \eta^2 \ ]^2} $$ $$ = \ \ \frac{  \zeta^4 \ + \ 2·\zeta^2·\eta^2 \ + \ \eta^4  \  - \  2·  \zeta^2 \ - \ 2·\eta^2  \   +  \ 1 }{[ \ ( \zeta + 1)^2 \ + \ \eta^2 \ ]^2}   $$
$$ = \ \ \frac{ [ \ ( \zeta + 1)^2 \ + \ \eta^2 \ ]·[ \ ( \zeta - 1)^2 \ + \ \eta^2 \ ]}{[ \ (\zeta + 1)^2 \ + \ \eta^2 \ ]^2} \ \ = \ \ \frac{  ( \zeta - 1)^2 \ + \ \eta^2 }{ (\zeta + 1)^2 \ + \ \eta^2 } \ \ , $$
which is only equal to $ \ 1 \ $ for all values of $ \ \theta \ $ when $ \ \zeta  =  0 \ \  . $  (We can verify that this becomes a constant function by calculating
$$ \frac{\partial}{\partial \eta} \ \left[ \ \frac{  ( \zeta - 1)^2 \ + \ \eta^2 }{ (\zeta + 1)^2 \ + \ \eta^2 } \ \right]_{\zeta = 0} \ \ = \ \    \frac{  8 \ · \zeta  \ · \ \eta }{[ \ (\zeta + 1)^2 \ + \ \eta^2 \ ]^2} | _{\zeta = 0}   \ \ = \ \ 0 \ \ . \ \ )  $$
Hence, $ \ w \  $ is purely imaginary.
[I believe that the converse of the proposition is shown to be true by the relation for $ \ \cos^2 \theta \ + \ \sin^2 \theta \ \ . \ ] $
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