Q16 (d)

Criteria	Marks
Provides correct solution	3
Shows $z_1z_2+z_1z_3+z_2z_3=1$	2
Treats $z_1$, $z_2$ and $z_3$ as zeros of a polynomial OR shows $	z_1

Sample answer:
As $z_1z_2z_3=1$,
$|z_1||z_2||z_3|=1$
But $|z_1|=|z_2|=|z_3|$, so
$|z_1|^3=1,\quad |z_1|=1$
Hence $|z_1|=|z_2|=|z_3|=1$.

Let $z_1=e^{i\theta}$ for some $\theta$.
$\frac1{z_1}=e^{-i\theta}=\overline{z_1}$
Similarly,
$\frac1{z_2}=\overline{z_2},\qquad\frac1{z_3}=\overline{z_3}$
Since $z_1z_2z_3=1$,
$z_1z_2=\frac1{z_3}=\overline{z_3}$
Similarly,
$z_1z_3=\overline{z_2},\qquad z_2z_3=\overline{z_1}$
Hence

$$z_1z_2+z_1z_3+z_2z_3 =\overline{z_3}+\overline{z_2}+\overline{z_1} =z_1+z_2+z_3=1$$

Therefore $z_1,z_2,z_3$ are the zeros of the polynomial
$z^3-z^2+z-1=0$

$$(z-1)(z^2+1)=0$$

$z=1,\ i,\ -i$
In some order $z_1,z_2,z_3$ are $1$, $i$, $-i$.