The square roots of the complex number $(-5-12 \mathrm{i})$ are

The square roots of the complex number $(-5-12 \mathrm{i})$ are
  1. $\pm(2-3 i)$
  2. $\pm(3+2 \mathrm{i})$
  3. $\pm(2+3 i)$
  4. $\pm(3-2 \mathrm{i})$

Solution

$\begin{aligned} & \text { Let } \sqrt{-5-12 i}=a+i b \quad \ldots[a, b \in R] \\ & \therefore(a+i b)^2=-5-12 i \\ & \therefore\left(a^2-b^2\right)+i(2 a b)=-5-i(12) \\ & \therefore a^2-b^2=-5 \text { and } 2 a b=-12 \Rightarrow a b=-6 \Rightarrow b=\frac{-6}{a} \\ & \therefore a^2-\left(\frac{-6}{a}\right)^2=-5 \\ & \therefore a^4-36=-5 a^2 \Rightarrow a^4+5 a^2-36=0 \\ & \therefore\left(a^2+9\right)\left(a^2-4\right)=0 \Rightarrow a^2=4 \quad \ldots[a \in R] \\ & \therefore a= \pm 2 \Rightarrow b=\frac{-6}{ \pm 2}=\mp 3 \\ & \therefore \sqrt{-5-12 i}= \pm(2-3 i)\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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