Find the square root of the following complex numbers: 1 + 4√3i - Vidyadip

Question

Find the square root of the following complex numbers:
1 + 4√3i

✓

Answer

Let $\sqrt{1}+4 \sqrt{3} i=\mathrm{a}+\mathrm{bi}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{R}$ Squaring on both sides, we get
$
\begin{aligned}
& 1+4 \sqrt{ } 3 i=(a+b i)^2 \\
& 1+4 \sqrt{ } 3 i=a^2+b^2 i^2+2 a b i \\
& 1+4 \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=1 \text { and } 2 a b=4 \sqrt{3} \\
\therefore \quad & a^2-b^2=1 \text { and } b=\frac{2 \sqrt{3}}{a} \\
\therefore \quad & a^2-\left(\frac{2 \sqrt{3}}{a}\right)^2=1 \\
\therefore \quad & a^2-\frac{12}{a^2}=1 \\
\therefore \quad & a^4-12=a^2 \\
\therefore \quad & a^4-a^2-12=0 \\
\therefore \quad & \left(a^2-4\right)\left(a^2+3\right)=0 \\
\therefore \quad & a^2=4 \text { or } a^2=-3 \\
\therefore \quad & \text { But } a \in R \\
\therefore \quad & a^2 \neq-3=4 \text { } \\
\therefore \quad & a= \pm 2 \\
& \text { When } a=2, b=\frac{2 \sqrt{3}}{2}=\sqrt{3} \\
& \text { When } a=-2, b=\frac{2 \sqrt{3}}{-2}=-\sqrt{3} \\
\therefore \quad & \sqrt{1+4 \sqrt{3} i}= \pm(2+\sqrt{3} i)
\end{array}
$

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