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.
The easy method to do this kind of exercise is to notice that $Re(z)=0$ if and only if $z=-\overline z$. Notice that if $|z|=1$ then $\overline z=\frac{1}{z}$.
In your case $$ \overline{\left(\frac{1-z}{1+z}\right)}=\frac{1-\overline z}{1+\overline z}=\frac{1-\frac{1}{z}}{1+\frac{1}{z}}=\frac{z-1}{1+z}=-\frac{1-z}{1+z}$$
Usually, the replacement of the algebraic form $z=x+iy$ makes things much more complicated in problems like this. First you should try applying the standard results from the general case, such as $|z|^2=z\overline z$, or $z \in \Bbb{R}$ if and only if $z=\overline z$, etc.
Let $z=a+bi$, where $a^2+b^2=1$ and $z\neq -1$. Then $$\begin{eqnarray} \Re\left(\frac{1-z}{1+z}\right) &=&\Re\left(\frac{(1-z)\overline{(1+z)}}{|1+z|^2}\right)\\ &=&\Re\left(\frac{(1-a-bi)(1+a-bi)}{(a+1)^2+b^2}\right)\\ &=&\Re\left(\frac{1-(a^2+b^2)-2bi}{(a+1)^2+b^2}\right)\\ &=&\Re\left(\frac{-2bi}{(a+1)^2+b^2}\right)=0\end{eqnarray}$$
Using polar form of a complex number, $z=e^{i\theta},\quad\theta\in\mathbb{R}$.
Hence $\frac{1-z}{1+z}=\dfrac{1-e^{i\theta}}{1+e^{i\theta}}=\dfrac{e^{-\frac{i\theta}{2}}-e^{\frac{i\theta}{2}}}{e^{-\frac{i\theta}{2}}+e^{\frac{i\theta}{2}}}=\dfrac{-2i\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}}=-i\tan\frac{\theta}{2}$
Since we are adding multiple proofs, here are two more.
1) Geometric proof.
$\displaystyle z$ corresponds to point $\displaystyle B$. $\displaystyle 1+z$ corresponds to $\displaystyle H$ and $\displaystyle 1-z$ to $\displaystyle G$.
$\displaystyle ACGD$ and $\displaystyle ADHB$ are rhombii, with $CD$ parallel to $\displaystyle AH$. In rhombii the diagonals intersect at right angles, and so $\displaystyle AG$ is perpendicular to $\displaystyle AH$.
Thus $\displaystyle \frac{1-z}{1+z} = ci$ for some $\displaystyle c$.
2) Using vectors.
We refer to the above figure.
$\displaystyle H = (1+\cos \theta, \sin \theta)$.
$\displaystyle G = (1- \cos \theta, -\sin \theta)$.
Their dot product = $\displaystyle (1 + \cos \theta)(1- \cos \theta) - \sin^2 \theta = 1 - \cos^2 \theta - \sin^2 \theta = 0$
So $\displaystyle \vec{AH}$ and $\displaystyle \vec{AG}$ are perpendicular.
Incidentally, the converse is also true:
If $\displaystyle \text{Re}\left(\frac{1-z}{1+z}\right) = 0$, then $\displaystyle |z| = 1$.
Hint $\: $ Realizing the denominator produces a purely imaginary numerator, i.e.
$$\displaystyle\ z\bar z\: =\: 1\ \:\Rightarrow\:\ \frac{(1-z)\:(1+\bar z)}{(1+z)\:(1+\bar z)}\: =\: \frac{\bar z-z}{|1+z|^2}\: =\: \frac{r\:i}{s},\ \ r,s\in \mathbb R$$
Note $\ $ This is an instance of the powerful method of rationalizing denominators to simplify arithmetic of algebraic numbers or functions. Generally, $\rm F$-rationalize means coerce to an element of the base ring $\rm F$ of an algebraic extension, which classically is $\mathbb Q = $ rationals. Above we have the base ring $\rm F =\mathbb R\subset \mathbb R[{\it i}] = \mathbb C,\:$ so rationalize means realize.
I want to point out that the problem in your approach is that, generally,
$$\mathrm{Re}\left(\frac{1-z}{1+z}\right) \ne \frac{1-\mathrm{Re}(z)}{1+\mathrm{Re}(z)}.$$
In other words, just because there's a Re on the left of the equation doesn't mean you can replace a complex variable $z$ with its real part $x$; that's tantamount to switching the order of evaluation of two functions composed (as in $f\circ g$ vs. $g\circ f$, here with $f=\mathrm{Re}$ and $g(z)=\frac{1-z}{1+z}$).
The general approach you want to take is replacing $z$ with $x+yi$, then trying to get the whole thing inside the real part in the form $\text{blah}_1+\text{blah}_2i$, where both $\text{blah}$s are entirely real, so you can just chop off $\text{blah}_2$ to finish evaluating. Frequently this process involves slapping the conjugate $x-yi$ in there somewhere to make things real.
Yet another approach, is to note that $Re(\overline z_1z_2)=z_1 \cdot z_2$, where $\cdot$ denotes the euclidean dot product in $R^2$.
Let $z=x+iy$ for $x,y$ in $R$, then we have $Re(\frac{1-z}{1+z})=\overline{1-z}{\frac{1}{1+z}}=(1-x,y)\cdot (\frac{1+x}{1+2x+x^2+y^2},\frac{-y}{1+2x+x^2+y^2})=\frac{1-(x^2+y^2)}{2+2x}=0$
Now, $$T(1)= 0$$ $$T(-1)= \infty$$ $$T(i)=\frac{1-i}{1+i}= \frac{(1-i)\overline{(1+i)}}{|1+i|^2}=\frac{(1-i)^2}{2} = \frac{-2i}{2}=-i$$ That is the image of the unit circle is the imaginary axis and in particular $$\Re\frac{1-z}{1+z}=0, \qquad \text{when $|z|=1$}$$
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 \ \ . \ ] $
