Let $|z_1| = |z_2| = 1$. Prove that $\frac{z_1+z_2}{1-z_1z_2}$ is an imaginary number

Question

Could someone help me:

1. prove that $\dfrac{z_1+z_2}{1-z_1z_2}$ is an imaginary number
2. prove that $\dfrac{z_1+z_2}{1+z_1z_2}$ is a real number.

Answer 1

With $2{\bf Re}w=w+\bar{w}$ we have $$2{\bf Re}\dfrac{z_1+z_2}{1-z_1z_2} = \dfrac{z_1+z_2}{1-z_1z_2}+\overline{\dfrac{z_1+z_2}{1-z_1z_2}} = \dfrac{z_1+z_2}{1-z_1z_2} + \dfrac{\overline{z_1}+\overline{z_2}}{1-\overline{z_1}\overline{z_2}} = \dfrac{0}{|1-z_1z_2|^2}=0.$$ Then $\dfrac{z_1+z_2}{1-z_1z_2}$ is imaginary. (Note that $z_1\bar{z_1}=z_2\bar{z_2}=1$).

The second is similar.

Answer 2

Note that $$\frac{z_1+z_2}{1-z_1z_2}= \frac{(z_1+z_2)(1-\bar z_1\bar z_2)}{(1-z_1z_2)(1-\bar z_1\bar z_2)}= \frac{(z_1-\bar z_1)+(z_2-\bar z_2)}{(1-z_1z_2)(1-\bar z_1\bar z_2)},$$ which is pure imaginary.

Similarly, $$\frac{z_1+z_2}{1+z_1z_2}= \frac{(z_1+z_2)(1+\bar z_1\bar z_2)}{(1+z_1z_2)(1+\bar z_1\bar z_2)}= \frac{(z_1+\bar z_1)+(z_2+\bar z_2)}{(1+z_1z_2)(1+\bar z_1\bar z_2)},$$

which is real.

Answer 3

A number is real if and only if it equals its conjugate: the conjugate of the second number is $$ \frac{\overline{z_1+z_2}}{\;\vphantom{\Big|}\overline{1+z_1z_2}\;}= \frac{\overline{z_1}+\overline{z_1}}{1+\overline{z_1}\overline{z_2}}= \frac{\dfrac{1}{z_1}+\dfrac{1}{z_2}}{1+\dfrac{1}{z_1}\dfrac{1}{z_2}} $$ because $|z|=1$ if and only if $\bar{z}=z^{-1}$.

Can you do similarly for the first number? A number $z$ is purely imaginary if and only if $\bar{z}=-z$.

Answer 4

Calling $z_1 = e^{i\phi}, z_2 = e^{i\psi}$

$$ \frac{ e^{i\phi}+e^{i\psi}}{1- e^{i\phi} e^{i\psi}} = i \cos \left(\frac{\phi-\psi}{2}\right) \csc \left(\frac{\phi+\psi}{2}\right) $$