Let z=x+iy and w=u+iv be complex numbers on t | vedclass

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(C) Given that $z$ and $w$ are complex numbers on the unit circle,we have $|z|=1$ and $|w|=1$. This implies $z\bar{z}=1 \Rightarrow \bar{z}=\frac{1}{z

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$8$

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(C) Given that $z$ and $w$ are complex numbers on the unit circle,we have $|z|=1$ and $|w|=1$. This implies $z\bar{z}=1 \Rightarrow \bar{z}=\frac{1}{z}$ and $w\bar{w}=1 \Rightarrow \bar{w}=\frac{1}{w}$. Given $z^2+w^2=1$. Taking the conjugate of both sides,we get $\bar{z}^2+\bar{w}^2=1$. Substituting $\bar{z}=\frac{1}{z}$ and $\bar{w}=\frac{1}{w}$,we get $\frac{1}{z^2}+\frac{1}{w^2}=1$,which simplifies to $\frac{z^2+w^2}{z^2w^2}=1$. Since $z^2+w^2=1$,we have $\frac{1}{z^2w^2}=1$,so $z^2w^2=1$,which means $(zw)^2=1$,so $zw=1$ or $zw=-1$. Case $1$: $zw=1$. Then $w=\frac{1}{z}$. Substituting into $z^2+w^2=1$,we get $z^2+\frac{1}{z^2}=1 \Rightarrow z^4-z^2+1=0$. This is a quadratic in $z^2$,giving $z^2 = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2} = e^{\pm i\pi/3}$. Thus $z^2 = e^{i\pi/3}$ or $z^2 = e^{-i\pi/3}$. Each gives $2$ values for $z$,total $4$ values. Case $2$: $zw=-1$. Then $w=-\frac{1}{z}$. Substituting into $z^2+w^2=1$,we get $z^2+\frac{1}{z^2}=1$,which is the same equation as Case $1$. This also gives $4$ values for $z$. Total number of ordered pairs $(z, w)$ is $4+4=8$.

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

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