Complex Analysis Problems

Question

Suppose the product $z_1z_2$ of two complex numbers is a nonzero real constant. Show that $z_2$ = $k\overline{z}_1$, where $k$ is a real number.

Hint: $z_2$=$k$ conjugate of $z_1$

Answer 1

Let $z_1 = a + bi, z_2 = c + di$ where $a,b,c,d \in \mathbb{R}$.

So $z_1z_2 = (ac-bd) + i(bc+ad)$.

If the result is purely real, then $bc = -ad$. Either $b = d = 0$ in which case $z_1$ and $z_2$ are both real and the result immediately follows or $\frac db = - \frac ca = -k$ (say) where $k$ is a real number.

So $d = -kb$ and $c = ka$, giving $z_2 = k(a - bi) = k\bar{z_1}$ as required.

Answer 2

If $0 \ne z_1z_2=c \in \mathbb R$, then $z_2= \frac{c}{z_1}=\frac{c \overline{z_1}}{z_1\overline{z_1}}=\frac{c \overline{z_1}}{|z_1|^2}=k \overline{z_1}$, where $k= \frac{c }{|z_1|^2}$.