Question

If $(x+iy{)}^{\frac{1}{3}}$ = a + ib, then $\frac{x}{a}$ + $\frac{y}{b}$ =

• A. 0
• B. -1
• C. 1
• D.

Answer 1

The correct option is D

x + iy = $(a+ib{)}^3$

= $a^3$ + $i^3{b}^3$ + 3a$i^2{b}^2$ + 3i$a^2$b

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

$\Rightarrow$ x = $a^3-3a{b}^2$ , y = $3a^{2}b-b^3$

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

= $4(a^2-b^2)$