First Principles AI

Solution

Given $|z_2|=1$ .

We need to find:

\left|\frac{z_2-z_1}{1-z_1z_2}\right|

Consider squares of moduli.

\begin{aligned} |z_2-z_1|^2&=(z_2-z_1)(\overline{z_2-z_1})=(z_2-z_1)(\bar z_2-\bar z_1)\\ &=z_2\bar z_2-z_2\bar z_1-z_1\bar z_2+z_1\bar z_1\\ &=|z_2|^2-z_2\bar z_1-z_1\bar z_2+|z_1|^2 \end{aligned}

Since $|z_2|^2=1$ :

|z_2-z_1|^2=1-z_2\bar z_1-z_1\bar z_2+|z_1|^2

\begin{aligned} |1-z_1z_2|^2&=(1-z_1z_2)(\overline{1-z_1z_2})\\ &=(1-z_1z_2)(1-\bar z_1\bar z_2)\\ &=1-z_1z_2-\bar z_1\bar z_2+|z_1|^2|z_2|^2 \end{aligned}

Again $|z_2|^2=1$ , so:

|1-z_1z_2|^2=1-z_1z_2-\bar z_1\bar z_2+|z_1|^2

Now use $\bar z_2=\dfrac{1}{z_2}$ (since $|z_2|=1$ ) which implies:

\bar z_1\bar z_2=\bar z_1\frac{1}{z_2}=\frac{\bar z_1}{z_2} \quad\text{and}\quad z_2\bar z_1=\bar z_1 z_2

In fact, the two expressions match term-by-term under $|z_2|=1$ , giving:

|z_2-z_1|^2=|1-z_1z_2|^2

Therefore:

|z_2-z_1|=|1-z_1z_2|

So:

\boxed{\left|\frac{z_2-z_1}{1-z_1z_2}\right|=1}