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

If $z=x-i y$ and $z^{\frac{1}{3}}=a+i b$, then $\frac{\left(\frac{x}{a}+\frac{y}{b}\right)}{a^2+b^2}=$
  1. -2
  2. -1
  3. 1
  4. 2

Solution

Given, $ \begin{aligned} z & =x-i y \text { and } z^{\frac{1}{3}}=a+i b \\ z^{1 / 3} & =a+i b \end{aligned} $ Take cube on both sides, we get $ \begin{array}{ll} \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) \quad[\because z=x-i y] \end{array} $ Here, $x=a^3-3 a b^2$ and $y=b^3-3 a^2 b$ Now, $ \begin{aligned} \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 \end{aligned} $

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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