Question

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

• A. 1
• B. $-1$
• C. 2
• D. $-2$

Answer 1

Answer: (D)

$z^{1/3}=+iq$ $\Rightarrow$ ${(x-iy)}^{1/3}=(p+iq)$ $(\because z=x-iy)$ $\Rightarrow$ $(x-iy)=p{(p+iq)}^3$ $\Rightarrow$ $(x-iy)=p^3+{(iq)}^3+3p^{2}qi+3p{q}^2{i}^2$ $\Rightarrow$ $(x-iy)=p^3-i{q}^3+3p^{2}qi-3p{q}^2$ $\Rightarrow$ $(x-iy)=(p^3-3p{q}^2)+i(3p^{2}q-q^3)$ On comparing both sides, we get $x=(p^3-3p{q}^2)$ and $-y=3p^{2}q-q^3$ $\Rightarrow$ $x=p(p^2-3q^2)$ and $y=q(q^2-3p^2)$ $\Rightarrow$ $\frac{x}{p}=(p^2-3q^2)$ and $\frac{y}{q}=(q^2-3p^2)$ Now, $\frac{x}{p}+\frac{y}{p}=p^2-3q^2+q^2-3p^2$ $\Rightarrow$ $\frac{x}{p}+\frac{y}{q}=-2p^2-2q^2$ $\Rightarrow$ $\frac{x}{p}+\frac{y}{q}=-2(p^2+q^2)$ $\Rightarrow$ $\frac{x/p+y/q}{(p^2+q^2)}=-2$