If $z=x-i y$ and $z^{\frac{1}{3}}=p+i q$, then $\frac{\left(\frac{x}{p}+\frac{y}{q}\right)}{\left(p^2+q^2\right)}$ is equal to

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Aicharya · Accepted Answer

$z=(p+i q)^3=p\left(p^2-3 q^2\right)-i q\left(q^2-3 p^2\right)$ $\Rightarrow \frac{x}{p}=p^2-3 q^2 \& \frac{y}{q}=q^2-3 p^2 \Rightarrow \frac{\frac{x}{p}+\frac{y}{q}}{\left(p^2+q^2\right)}=-2$

If $z=x-i y$ and $z^{\frac{1}{3}}=p+i q$, then…

Solution