It is given that complex numbers $z=x+iy$ and $w=u+iv$ are on the unit circle such that $z^2+w^2=1 ...\left(\mathrm{i}\right)$

So, $\overline{\left(z^2\right)}+\overline{\left(w^2\right)}=1$

$\Rightarrow {\left(\overline{z}\right)}^2+{\left(\overline{w}\right)}^2=1$

$\Rightarrow {\left(\frac{\overline{z}z}{z}\right)}^2+{\left(\frac{\overline{w}w}{w}\right)}^2=1$

$\Rightarrow \frac{1}{z^2}+\frac{1}{w^2}=1\Rightarrow z^2+w^2=z^2{w}^2$

 $\Rightarrow z^2{w}^2=1 ...\left(\mathrm{ii}\right)$

$\because$ The number of solution of equations $x^2+y^2=1$ and $x^2{y}^2=1$ is eight 

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