Question

If $z = x - iy $ and $z^{\frac{1}{3}} = a + ib$, then $\frac{\left(\frac{x}{a}+\frac{y}{b}\right)}{a^{2}+b^{2}} = $

• A. -2
• B. -1
• C. 1
• D. 2

Answer 1

Answer: (A)

Given,
$z =x-i y$ and $z^{\frac{1}{3}}=a+i b $
$ z^{1 / 3} =a+i b $
Take cube on both sides, we get
$\left(z^{1 / 3}\right)^{3}=(a+i b)^{3}$
$\Rightarrow z=a^{3}+(i b)^{3}+3 a^{2} i b+3 a(i b)^{2}$
$\Rightarrow z=a^{3}+i^{3} b^{3}+3 a^{2} i b+3 a b^{2} i^{2}$
$\Rightarrow z=a^{3}-i b^{3}+3 a^{2} b i-3 a b^{2} $
$\left[\because i^{2}=-1\right]$
$\Rightarrow z=\left(a^{3}-3 a b^{2}\right)-i\left(b^{3}-3 a^{2} b\right)$
$\Rightarrow x-i y=\left(a^{3}-3 a b^{2}\right)-i\left(b^{3}-3 a^{2} b\right) $
$[\because z=x-i y]$
Here, $x=a^{3}-3 a b^{2}$ and $y=b^{3}-3 a^{2} b$
Now, $ \frac{\left(\frac{x}{a}+\frac{y}{b}\right)}{a^{2}+b^{2}} =\frac{\left(\frac{a^{3}-3 a b^{2}}{a}+\frac{b^{3}-3 a^{2} b}{b}\right)}{a^{2}+b^{2}} $
$=\frac{a^{2}-3 b^{2}+b^{2}-3 a^{2}}{a^{2}+b^{2}} $
$=\frac{-2 a^{2}-2 b^{2}}{a^{2}+b^{2}}=\frac{-2\left(a^{2}+b^{2}\right)}{a^{2}+b^{2}}=-2$