Question
Find the square root of the following complex numbers:
2(1 – √3i)

Answer

Let $\sqrt{2(1-\sqrt{3} i)}=\mathrm{a}+\mathrm{bi}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{R}$
Squaring on both sides, we get
$
\begin{aligned}
& 2(1-\sqrt{ } 3 i)=(a+b i)^2 \\
& 2(1-\sqrt{ } 3 i)=a^2+b^2 i^2+2 a b i \\
& 2-2 \sqrt{ } 3 i=\left(a^2-b^2\right)+2 a b i \quad \ldots . .\left[\because i^2=-1\right]
\end{aligned}
$
Equating real and imaginary parts, we get
$
\begin{array}{ll}
& a^2-b^2=2 \text { and } 2 a b=-2 \sqrt{3} \\
\therefore & a^2-b^2=2 \text { and } b=-\frac{\sqrt{3}}{a} \\
\therefore & a^2-\left(-\frac{\sqrt{3}}{a}\right)^2=2 \\
& \text { } \\
\therefore \quad & a^2-\frac{3}{a^2}=2 \\
\therefore & a^4-3=2 a^2 \\
\therefore & a^4-2 a^2-3=0 \\
\therefore & \left(a^2-3\right)\left(a^2+1\right)=0 \\
\therefore & a^2=3 \text { or } a^2=-1 \\
& B a \in R
\end{array}
$
$
\begin{array}{ll}
\therefore & a^2-\frac{3}{a^2}=2 \\
\therefore & a^4-3=2 a^2 \\
\therefore & a^4-2 a^2-3=0 \\
\therefore & \left(a^2-3\right)\left(a^2+1\right)=0 \\
\therefore & a^2=3 \text { or } a^2=-1
\end{array}
$
But $\mathrm{a} \in \mathrm{R}$
$
\begin{array}{ll}
\therefore & \mathrm{a}^2 \neq-1 \\
\therefore & \mathrm{a}^2=3 \\
\therefore & \mathrm{a}= \pm \sqrt{3}
\end{array}
$
When $\mathrm{a}=\sqrt{3}, \mathrm{~b}=\frac{-\sqrt{3}}{\sqrt{3}}=-1$
When $\mathrm{a}=-\sqrt{3}, \mathrm{~b}=\frac{-\sqrt{3}}{-\sqrt{3}}=1$
$
\therefore \quad \sqrt{2(1-\sqrt{3} \mathrm{i})}= \pm(\sqrt{3}-\mathrm{i})
$

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