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

Question

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

Professor PC Thomas Classes · Answer

(d) : $z^{1 / 3}=p+i q \Rightarrow x-i y=(p+i q)^{3}$\ $\Rightarrow \quad x-i y=p^{3}-3 p q^{2}+i\left(3 p^{2} q-q^{3}ight)$\ $\Rightarrow \quad x=p^{3}-3 p q^{2}$ and $y=-\left(3 p^{2} q-q^{3}ight)$\ $\Rightarrow \quad \frac{x}{p}=p^{2}-3 q^{2}$ and $\frac{y}{q}=-\left(3 p^{2}-q^{2}ight)$\ Adding the equations of $\left({ }^{*}ight)$ we get $\frac{x}{p}+\frac{y}{q}$\ $=-2\left(p^{2}+q^{2}ight)$\