Question

If $z=x+iy$ is a complex number such that ${\overline{z}}^{\frac{1}{3}}=a+ib$, then the value of $\frac{1}{a^2+b^2}(\frac{x}{a}+\frac{y}{b})=$

• A. $-1$
• B. $-2$
• C. $0$
• D. $2$

Answer 1

Answer: (B)

$-2$

Given $z=x+iy$ , ${\overline{z}}^{\frac{1}{3}}=a+ib$

$\overline{z}=x-iy={(a+ib)}^3$

$\Rightarrow x-iy=a^3-3b^{2}a+i(3a^{2}b-b^3)$

By comparing real and imaginary parts , we get $x=a(a^2-3b^2)$ and $y=-b(3a^2-b^2)$

Therefore we get $\frac{x}{a}=a^2-3b^2$ and $\frac{y}{b}=-3a^2+b^2$

Therefore $\frac{1}{a^2+b^2}(\frac{x}{a}+\frac{y}{b})=\frac{1}{a^2+b^2}(a^2-3b^2-3a^2+b^2)=\frac{1}{a^2+b^2}(-2)(a^2+b^2)=-2$