Suppose $z \neq -1$ is a complex number of norm 1. Prove that $(\frac{1+z}{|1+z|})^2 =z $

Question

Suppose $z\neq -1$ is a complex number of norm 1m $(|z| =1)$. I know that $z= n + mi$, what is the most straightforward way of solving this problem? I was also given the following sketch for a possible approach.

Answer 1

Just do direct calculations $$ \frac {(z+1)^2}{|z+1|^2} - z = \frac {(z+1)(z+1)}{(z+1)(\overline z+1)} -z = \frac {z+1}{\overline z + 1} - z = \frac {z+1 - z \overline z - z}{\overline z + 1} = 0 $$

PS

I used the fact that $|z|^2 = z \overline z = 1$

Answer 2

Hint: show the LHS has module 1 (clearly, the RHS too by assumption). Writing $z=e^{i\theta}$, show that $1+z = r e^{i\theta/2}$ for some $r \in \mathbb{R}0$.

Geometrically, this is showing that the point $(1+z)/\lvert 1+z\rvert$ is (1) on the unit circle, and (2) that the angle is half the angle of $z$ (mod $\pi$). Thus, squaring it will give you $z$.

PS (edit): one thing that may be useful: recall that $\frac{e^{i\frac{\theta}{2}} + e^{-i\frac{\theta}{2}}}{2}=\cos\frac{\theta}{2}$.

Answer 3

Even shorter: $z+1 = z + z\overline{z} = z(1+\overline{z}) = z (\overline{1 + z})$.

Therefore $(z+1)^2 = z(\overline{1+z})(1+z) = z\lvert 1 + z\rvert^2$. Then isolate $z$.

Okay, okay, it’s the same calculation as Kaster’s, but since we’re at it …