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)

$-2$

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

$\overline{z}=a^3+3a^2\left(ib\right)+3a(ib{)}^2+(ib{)}^3$

$\overline{z}=(a^3-3a{b}^2)+(3a^{2}b-b^3)i$

Given, $z=x+iy$ and $\overline{z}=x-iy$

Therefore, comparing the real and imaginary coefficients, we get

$x=a^3-3a{b}^2$

$y=-3a^{2}b+b^3$

Therefore, $\frac{x}{a}+\frac{y}{b}$

$=\frac{a^3-3a{b}^2}{a}+\frac{-3a^{2}b+b^3}{b}$

$=(a^2-3b^2)+(-3a^2+b^2)$

$=-2(a^2+b^2)$

Therefore,

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