Complex number condition on the modulus

Question

The set of all complex numbers $(z_1,z_2)$ which satisfy

$$\frac{|z_1 -z_2|}{|1-\overline{z_1}z_2|} \lt 1 $$ is?

(Here $\overline{z_1}$ is $z_1$'s cojugate.)

I attempted to write $z_1$ a as $x_1 + iy_1$ and $z_2$ as $x_2 + iy_2 $ and tried to simplify. But then it didn't seem to work. How do I go about this?

Answer 1

The key to answering this question is knowing that for $a \in \mathbb{D} = \{z \in \mathbb{C} \mid |z| < 1\}$, $$T_a(z) = \frac{z - a}{1 - \overline{a}{z}}$$ is an automorphism of $\mathbb{D}$. In addition, if $|z| = 1$, then $|T_a(z)| = 1$, and if $|z| > 1, z \neq \frac{1}{\overline{a}}$, then $|T_a(z)| > 1$; note, we can make sense of $T_a(z)$ at $z = \frac{1}{\overline{a}}$ if we consider $T_a$ as an automorphism of the Riemann sphere but that isn't really necessary here.

First note that

$$\frac{|z_1 - z_2|}{|1-\overline{z_1}{z_2}|} = \frac{|z_2 - z_1|}{|1-\overline{z_1}{z_2}|} = \left|\frac{z_2 - z_1}{1-\overline{z_1}{z_2}}\right| = |\,T_{z_1}(z_2)|.$$

So, if $\underline{|z_1| < 1}$, $(z_1, z_2)$ satisfies the desired inequality if and only if $|z_2| < 1$.

Now note that

$$|\,T_{z_1}(z_2)| = \left|\frac{z_2 - z_1}{1-\overline{z_1}{z_2}}\right| = \left|\frac{\frac{z_2}{\overline{z_1}z_2} - \frac{z_1}{\overline{z_1}z_2}}{\frac{1}{\overline{z_1}z_2}-1}\right| = \left|\frac{\frac{z_1}{\overline{z_1}z_2} - \frac{1}{\overline{z_1}}}{1-\frac{1}{\overline{z_1}z_2}}\right| = \left|\frac{\frac{z_1}{\overline{z_1}z_2} - \frac{1}{\overline{z_1}}}{1-\frac{1}{z_1}\frac{z_1}{\overline{z_1}z_2}}\right| = \left|\,T_{\frac{1}{\overline{z_1}}}\left(\frac{z_1}{\overline{z_1}z_2}\right)\right|.$$

So, if $\underline{|z_1| > 1}$ (so that $\left|\frac{1}{\overline{z_1}}\right| < 1$), $(z_1, z_2)$ satisfies the desired inequality if and only if $\left|\frac{z_1}{\overline{z_1}z_2}\right| < 1$. As

$$\left|\frac{z_1}{\overline{z_1}z_2}\right| = \frac{|z_1|}{|\overline{z_1}||z_2|} = \frac{|z_1|}{|z_1||z_2|} = \frac{1}{|z_2|}$$

we see that if $|z_1| > 1$, $(z_1, z_2)$ satisfies the desired inequality if and only if $\frac{1}{|z_2|} < 1$, that is, $|z_2| > 1$.

Finally, if $\underline{|z_1| = 1}$, then

\begin{align*} \left|\frac{z_1 - z_2}{1-\overline{z_1}z_2}\right|^2 &= \frac{z_1 - z_2}{1-\overline{z_1}z_2}\overline{\left(\frac{z_1 - z_2}{1-\overline{z_1}z_2}\right)}\\ &\\ &= \frac{z_1 - z_2}{1-\overline{z_1}z_2}\frac{\overline{z_1} - \overline{z_2}}{1-z_1\overline{z_2}}\\ &\\ &= \frac{|z_1|^2 - z_1\overline{z_2} - \overline{z_1}z_2 + |z_2|^2}{1 - z_1\overline{z_2} - \overline{z_1}z_2 + |z_1|^2|z_2|^2}\\ &\\ &= \frac{1 - z_1\overline{z_2} - \overline{z_1}z_2 + |z_2|^2}{1 - z_1\overline{z_2} - \overline{z_1}z_2 + |z_2|^2}\\ &\\ &= 1. \end{align*}

So, if $|z_1| = 1$, there is no $z_2$ such that $(z_1, z_2)$ satisfies the desired inequality.

Therefore, if

$$S = \left\{(z_1, z_2) \in \mathbb{C}^2 : \frac{|z_1 - z_2|}{|1-\overline{z_1}z_2|} < 1\right\}$$

then $S = \{(z_1, z_2) \in \mathbb{C}^2 : |z_1| < 1, |z_2| < 1\} \cup \{(z_1, z_2) \in \mathbb{C}^2 : |z_1| > 1, |z_2| > 1\}$.

Answer 2

Here is a solution that falls out of my comment to Michael Albanese's solution, and which seems to be what was hinted at by Philipp Schäfer in his comment. Since $\lvert z\rvert^2 = z\overline z$ for all $z \in \mathbb C$, we have

\begin{align} \frac{\lvert z_1-z_2 \rvert^2}{\lvert 1-\overline z_1 z_2 \rvert^2} &= \frac{(z_1-z_2)\bigl(\overline z_1 - \overline z_2\bigr)}{(1-\overline z_1 z_2)(1-z_1\overline z_2)} \\ &= \frac{\lvert z_1\rvert^2 - z_1\overline z_2 - \overline z_1 z_2 + \lvert z_2\rvert^2}{1-z_1\overline z_2 - \overline z_1 z_2 + \lvert z_1\rvert^2 \lvert z_2\rvert^2} \\ &= 1 - \frac{1 - \lvert z_1 \rvert^2 - \lvert z_2 \rvert^2 + \lvert z_1\rvert^2 \lvert z_2\rvert^2}{1-z_1\overline z_2 - \overline z_1 z_2 + \lvert z_1\rvert^2 \lvert z_2\rvert^2} \\ &= 1 - \frac{(1-\lvert z_1\rvert^2)(1-\lvert z_2\rvert^2)}{\lvert 1-\overline z_1 z_2 \rvert^2}. \end{align}

Thus if $\overline z_1 z_2 \neq 1$, the inequality $\frac{\lvert z_1-z_2 \rvert}{\lvert 1-\overline z_1 z_2 \rvert} < 1$ is equivalent to $(1-\lvert z_1\rvert^2)(1-\lvert z_2\rvert^2) > 0$, which is true if and only if either

• $\lvert z_1\rvert < 1$ and $\lvert z_2\rvert < 1$, or
• $\lvert z_1\rvert > 1$ and $\lvert z_2\rvert > 1$.

Since $\overline z_1 z_2 \neq 1$ is incompatible with either of the bulleted conditions, $$\begin{align}&\left\{(z_1, z_2) \in \mathbb C^2 : \frac{\lvert z_1-z_2 \rvert}{\lvert 1-\overline z_1 z_2 \rvert} < 1\right\} \\ &\mspace{25mu}= \left\{(z_1, z_2) \in \mathbb C^2 : \text{$\lvert z_1 \rvert < 1$ and $\lvert z_2 \rvert < 1$}\right\} \cup \left\{(z_1, z_2) \in \mathbb C^2 : \text{$\lvert z_1 \rvert > 1$ and $\lvert z_2 \rvert > 1$}\right\}.\end{align}$$