## Task A-4.5.

Let $z_{1}$, $z_{2}$, and $z_{3}$ be complex numbers such that $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1$ and $4 z_{3}=$ $3\left(z_{1}+z_{2}\right)$. What is $\left|z_{1}-z_{2}\right|$?
## First Solution.

By applying the absolute value to the last equality, we get $4=4\left|z_{3}\right|=3\left|z_{1}+z_{2}\right|$.

Squaring the obtained equality $\left|z_{1}+z_{2}\right|=\frac{4}{3}$, we get

$$
\begin{aligned}
\frac{16}{9} & =\left|z_{1}\right|^{2}+z_{1} \overline{z_{2}}+z_{2} \overline{z_{1}}+\left|z_{2}\right|^{2} & & 1 \text { point } \\
& =2+z_{1} \overline{z_{2}}+z_{2} \overline{z_{1}}, & & 1 \text { point }
\end{aligned}
$$

which means $z_{1} \overline{z_{2}}+z_{2} \overline{z_{1}}=\frac{-2}{9}$.

Now we calculate

$$
\left|z_{1}-z_{2}\right|^{2}=\left|z_{1}\right|^{2}-z_{1} \overline{z_{2}}-z_{2} \overline{z_{1}}+\left|z_{2}\right|^{2}=2+\frac{2}{9}=\frac{20}{9}
$$

We conclude that $\left|z_{1}-z_{2}\right|=\frac{2}{3} \sqrt{5}$.