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Complex Numbers(p-1) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Complex Numbers(p-1)4 Marks
Question
Find the square root of the following complex numbers:
-8 – 6i
✓

Answer

Let $\sqrt{-8-6 i}=\mathrm{a}+\mathrm{bi}$, where $\mathrm{a}, \mathrm{b} \in \mathrm{R}$ Squaring on both sides, we get
$
\begin{aligned}
-8-6 i & =(a+b i)^2 \\
-8-6 i & =a^2+b^2 i^2+2 a b i \\
-8-6 i & =\left(a^2-b^2\right)+2 a b i \quad . . . .\left[\because i^2=-1\right]
\end{aligned}
$
Equating real and imaginary parts, we get
$
\begin{array}{ll}
& a^2-b^2=-8 \text { and } 2 a b=-6 \\
\therefore & a^2-b^2=-8 \text { and } b=\frac{-3}{a} \\
\therefore & a^2-\left(-\frac{3}{a}\right)^2=-8 \\
\therefore & a^2-\frac{9}{a^2}=-8 \\
\therefore & a^4-9=-8 a^2 \\
\therefore & a^4+8 a^2-9=0 \\
\therefore & \left(a^2+9\right)\left(a^2-1\right)=0 \\
\therefore & a^2=-9 \text { or } a^2=1
\end{array}
$
But $a \in R$
$
\begin{array}{ll}
\therefore & \mathrm{a}^2 \neq-9 \\
\therefore & \mathrm{a}^2=1 \text { } \\
\therefore & \mathrm{a}= \pm 1
\end{array}
$
When $a=1, b=\frac{-3}{1}=-3$
When $\mathrm{a}=-1, \mathrm{~b}=\frac{-3}{-1}=3$
$
\therefore \quad \sqrt{-8-6 i}= \pm(1-3 i)
$

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