Thank you for reporting, we will resolve it shortly
Q.
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
KVPYKVPY 2019
Solution:
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 \bar{\left(z\right)}^{2}+\bar{\left(w\right)}^{2}=1 $
$\Rightarrow \left(\frac{\bar{z} z}{z}\right)^{2}+ \left(\frac{\bar{w} w}{w}\right)^{2}=1 $
$\Rightarrow \frac{1}{z^{2}}+\frac{1}{w^{2}}=1$
$\Rightarrow z^{2}+w^{2}=z^{2}w^{2}$
$\Rightarrow z^{2}w^{2}=1 \ldots\left(ii\right)$
$\because$ 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
