For complex $z$, $|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right) = 0$

Solution 1:

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.

Solution 2:

Let $z=a+bi$, where $a^2+b^2=1$ and $z\neq -1$. Then $$\begin{eqnarray} \Re\left(\frac{1-z}{1+z}\right) &=&\Re\left(\frac{(1-z)\overline{(1+z)}}{|1+z|^2}\right)\\ &=&\Re\left(\frac{(1-a-bi)(1+a-bi)}{(a+1)^2+b^2}\right)\\ &=&\Re\left(\frac{1-(a^2+b^2)-2bi}{(a+1)^2+b^2}\right)\\ &=&\Re\left(\frac{-2bi}{(a+1)^2+b^2}\right)=0\end{eqnarray}$$

Solution 3:

Using polar form of a complex number, $z=e^{i\theta},\quad\theta\in\mathbb{R}$.

Hence $\frac{1-z}{1+z}=\dfrac{1-e^{i\theta}}{1+e^{i\theta}}=\dfrac{e^{-\frac{i\theta}{2}}-e^{\frac{i\theta}{2}}}{e^{-\frac{i\theta}{2}}+e^{\frac{i\theta}{2}}}=\dfrac{-2i\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}}=-i\tan\frac{\theta}{2}$

Solution 4:

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 $\displaystyle |z| = 1$.