Question
Find the square root of the following complex numbers:
3 + 2√10i
3 + 2√10i
✓
Answer
Let $\sqrt{3+2 \sqrt{10}} i=\mathrm{a}+\mathrm{bi}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{R}$ Squaring on both sides, we get
$
\begin{aligned}
& 3+2 \sqrt{ } 10 \mathrm{i}=(\mathrm{a}+\mathrm{bi})^2 \\
& 3+2 \sqrt{ } 10 \mathrm{i}=\mathrm{a}^2+\mathrm{b}^2 \mathrm{i}^2+2 \mathrm{abi} \\
& 3+2 \sqrt{ } 10 \mathrm{i}=\left(\mathrm{a}^2-\mathrm{b}^2\right)+2 \mathrm{abi} \ldots . . .\left[\mathrm{i}^2=-1\right]
\end{aligned}
$
Equating real and imaginary parts, we get
$
\begin{array}{ll}
a^2-b^2=3 \text { and } 2 a b=2 \sqrt{ } 10 \\
a^2-b^2=3 \text { and } b=\frac{\sqrt{10}}{a} \\
\therefore & a^2-\left(\frac{\sqrt{10}}{a}\right)^2=3 \\
\therefore & a^2-\frac{10}{a^2}=3 \\
\therefore & a^4-10=3 a^2 \\
\therefore & a^4-3 a^2-10=0 \\
\therefore & \left(a^2-5\right)\left(a^2+2\right)=0 \\
\therefore & a^2=5 \text { or } a^2=-2
\end{array}
$
But $\mathrm{a} \in \mathrm{R}$
$
\begin{array}{ll}
\therefore & a^2 \neq-2 \\
\therefore & a^2=5 \text { } \\
\therefore & a= \pm \sqrt{5}
\end{array}
$
When $a=\sqrt{5}, \mathrm{~b}=\frac{\sqrt{10}}{\sqrt{5}}=\sqrt{2}$
When $\mathrm{a}=-\sqrt{5}, \mathrm{~b}=\frac{\sqrt{10}}{-\sqrt{5}}=-\sqrt{2}$
$
\therefore \quad \sqrt{3+2 \sqrt{10}} \mathrm{i}= \pm(\sqrt{5}+\sqrt{2} \mathrm{i})
$
$
\begin{aligned}
& 3+2 \sqrt{ } 10 \mathrm{i}=(\mathrm{a}+\mathrm{bi})^2 \\
& 3+2 \sqrt{ } 10 \mathrm{i}=\mathrm{a}^2+\mathrm{b}^2 \mathrm{i}^2+2 \mathrm{abi} \\
& 3+2 \sqrt{ } 10 \mathrm{i}=\left(\mathrm{a}^2-\mathrm{b}^2\right)+2 \mathrm{abi} \ldots . . .\left[\mathrm{i}^2=-1\right]
\end{aligned}
$
Equating real and imaginary parts, we get
$
\begin{array}{ll}
a^2-b^2=3 \text { and } 2 a b=2 \sqrt{ } 10 \\
a^2-b^2=3 \text { and } b=\frac{\sqrt{10}}{a} \\
\therefore & a^2-\left(\frac{\sqrt{10}}{a}\right)^2=3 \\
\therefore & a^2-\frac{10}{a^2}=3 \\
\therefore & a^4-10=3 a^2 \\
\therefore & a^4-3 a^2-10=0 \\
\therefore & \left(a^2-5\right)\left(a^2+2\right)=0 \\
\therefore & a^2=5 \text { or } a^2=-2
\end{array}
$
But $\mathrm{a} \in \mathrm{R}$
$
\begin{array}{ll}
\therefore & a^2 \neq-2 \\
\therefore & a^2=5 \text { } \\
\therefore & a= \pm \sqrt{5}
\end{array}
$
When $a=\sqrt{5}, \mathrm{~b}=\frac{\sqrt{10}}{\sqrt{5}}=\sqrt{2}$
When $\mathrm{a}=-\sqrt{5}, \mathrm{~b}=\frac{\sqrt{10}}{-\sqrt{5}}=-\sqrt{2}$
$
\therefore \quad \sqrt{3+2 \sqrt{10}} \mathrm{i}= \pm(\sqrt{5}+\sqrt{2} \mathrm{i})
$
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