Question

Let $z=x+iy$ and $w=u+iv$ be complex numbers on the unit circle such that $z^2+w^2=1$. Then the number of ordered pairs $(z,w)$ is

• A. 0
• B. 4
• C. 8
• D. infinite

Answer 1

Answer: (C)

It is given that complex numbers $z=x+iy$ and $w=u+iv$ are on the unit
circle such that $z^2+w^2=1\dots (i)$
So, $\overline{\left(z^2\right)}+\overline{\left(w^2\right)}=1$
$\Rightarrow {\overline{\left(z\right)}}^2+{\overline{\left(w\right)}}^2=1$
$\Rightarrow {\left(\frac{\bar{z}z}{z}\right)}^2+{\left(\frac{\bar{w}w}{w}\right)}^2=1$
$\Rightarrow \frac{1}{z^2}+\frac{1}{w^2}=1$
$\Rightarrow z^2+w^2=z^2{w}^2$
$\Rightarrow z^2{w}^2=1\dots \left(ii\right)$
$\because$ The number of solution of equations
$x^2+y^2=1$ and $x^2{y}^2=1$ is eight

$\therefore$ Number of ordered pair $(z,w)$ is also eight