Question

If $z=x+iy$ and $z^{1/3}=a-ib$, then $\frac{x}{a}-\frac{y}{b}=k(a^2-b^2)$, where $k$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$z=x+iy$

$z^{1/3}=a-ib$

$(x+iy{)}^{1/3}=a-ib$

$x+iy=a^3+i{b}^3-3a^{2}bi-3a{b}^2$ $[\because (a-b{)}^3=a^3-b^3-3ab(a-b)]$

Equating real parts

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

Equating imaginary parts

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

$\frac{x}{a}-\frac{y}{b}=k(a^2-b^2)$..........given

putting the value of $x$ and $y$ we get,

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

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

$\therefore k=4$