Q70 / 170 · Manipal (MET) 2019 · Mathematics
If $z=x+i y, z^{1 / 3}=a-i b$, then $\frac{x}{a}-\frac{y}{b}=k\left(a^2-b^2\right)$, where $k$ is equal to:
- A$1$
- B$2$
- C$3$
- Correct answerD$4$
Solution
$\begin{aligned}
& \text { Here, } z^{1 / 3}=a-i b \\
& \Rightarrow z=(a-i b)^3 \\
& \Rightarrow z=a^3-3 a^2(i b)+3 a(i b)^2-(i b)^3 \\
& \Rightarrow x+i y=\left(a^3-3 b^2 a\right)+i\left(-3 a^2 b+b^3\right)
\end{aligned}$
Equating real and imaginary parts, we get $x=a\left(a^2-3 b^2\right) \text { and } y=b\left(b^2-3 a^2\right)$
Now, $\frac{a\left(a^2-3 b^2\right)}{a}-\frac{b\left(b^2-3 a^2\right)}{b}=k\left(a^2-b^2\right)$ $\Rightarrow 4 a^2-4 b^2=k\left(a^2-b^2\right) \Rightarrow k=4$
Equating real and imaginary parts, we get $x=a\left(a^2-3 b^2\right) \text { and } y=b\left(b^2-3 a^2\right)$
Now, $\frac{a\left(a^2-3 b^2\right)}{a}-\frac{b\left(b^2-3 a^2\right)}{b}=k\left(a^2-b^2\right)$ $\Rightarrow 4 a^2-4 b^2=k\left(a^2-b^2\right) \Rightarrow k=4$
Hence, the correct answer is option: 'D'.