$\frac{z_1+z_2}{1+z_1z_2} \in \mathbb{R}$. Is my solution correct? Why?

Question

I have to show that if numbers:

• $z_1$ and $z_2 \in \mathbb{C} $
• $ |z_1| = |z_2| = 1 $
• $z_1*z_2 \neq -1$

then: $$\frac{z_1+z_2}{1+z_1z_2} \in \mathbb{R}$$

So I thought I can transform that expression like that: \begin{align*} \frac{z_1+z_2}{1+z_1z_2} &= \frac{(z_1+z_2)(1-z_1z_2)}{(1+z_1z_2)(1-z_1z_2)} = \frac{z_1-z_1*z_1z_2+z_2-z_2*z_1z_2}{|1+z_1z_2|^2} \\ &= \frac{z_1-|z_1|^2*z_2+z_2-|z_2|^2*z_1}{|1+z_1z_2|^2} \\ &= \frac{z_1-z_2+z_2-z_1}{|1+z_1z_2|^2} = \frac{0}{|1+z_1z_2|^2} \\ &= 0 \end{align*}

Is that correct? Could you tell me exactly why $(1+z_1z_2)(1-z_1z_2) = |1+z_1z_2|^2$ ?

I know that $z * \bar z = |z|^2$. But that equality gets me confused (if it is even right).

Answer 1

Since @Aqua has resolved your query here ia an alternate proof$$\frac{z_1+z_2}{1+z_1z_2}=\frac{\frac{1}{\bar {z_1}}+\frac{1}{\bar {z_2}}}{1+\frac{1}{\bar {z_1}\bar{z_2}}}=\frac{\bar z_1+\bar z_2}{1+\bar z_1\bar z_2}$$

As $X=\bar X \Rightarrow X$ is real we complete the proof

Answer 2

This is not correct: $$(1+z_1z_2)(1-z_1z_2)=|1+z_1z_2|^2$$

it should be $$(1+z_1z_2)(\overline {1+z_1z_2})=|1+z_1z_2|^2$$

as $|w|^2 =w\cdot \overline{w}$. (Notice further that $\overline {1+z_1z_2} = \overline{1}+\overline {z_1z_2} =1 + \overline {z_1}\cdot \overline{z_2})$

Also, you could easyl check it is not correct if you take $z_1=z_2=1$, you don't get $0$ at the end.

Answer 3

I would like to add a geometric interpretation explaining this property:

if $z_1 \ne z_2$, $z=z_3=\frac{z_1+z_2}{1+z_1z_2}$ is the intersection of line $[z_1,z_2]$ with the real axis.

Indeed, line $[z_1,z_2]$ has equation:

$$\begin{vmatrix}z_1&\overline{z_1}&1\\ z_2&\overline{z_2}&1\\ z&\overline{z}&1 \end{vmatrix}=0 \tag{1}$$

(see here).

As we are looking for a real $z$, we have $\overline{z}=z$.

Taking into account the fact that $\overline{z_1}=\tfrac{1}{z_1}$ and $\overline{z_2}=\tfrac{1}{z_2}$, it is not difficult, by multiplying by $z_1z_2$ to obtain by Laplace expansion of (1) followed by a factorization by $(z_2-z_1$):

$$(z_2-z_1)(z(1+z_1z_2)-(z_1+z_2))=0$$

$$\iff z_1=z_2 \ \ \text{or} \ \ z=\frac{z_1+z_2}{1+z_1z_2}$$

as desired.

Remark: if $z_1=z_2$, by continuity, point $z$ is the intersection point of the tangent to the unit circle with the real axis.