Solve this


$\sqrt{-4-3 i}$



Let, $(a+i b)^{2}=-4-3 i$

Now using, $(a+b)^{2}=a^{2}+b^{2}+2 a b$

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=-4-3 i$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=-4-3 i$

Now, separating real and complex parts, we get

$\Rightarrow a^{2}-b^{2}=-4 \ldots \ldots \ldots \ldots \ldots$ eq. 1

$\Rightarrow 2 \mathrm{ab}=-3 \ldots \ldots .$ eq. 2

$\Rightarrow \mathrm{a}=-\frac{3}{2 b}$

Now, using the value of a in eq.1, we get

$\Rightarrow\left(-\frac{3}{2 b}\right)^{2}-b^{2}=-4$

$\Rightarrow 9-4 b^{4}=-16 b^{2}$

$\Rightarrow 4 b^{4}-16 b^{2}-9=0$

Simplify and get the value of $b^{2}$, we get,

$\Rightarrow b^{2}=\frac{9}{2}$ or $b^{2}=-2$

As $b$ is real no. so, $b^{2}=\frac{9}{2}$

$\mathrm{b}=\frac{3}{\sqrt{2}}$ or $\mathrm{b}=-\frac{3}{\sqrt{2}}$

Therefore, $a=\frac{1}{-\sqrt{2}}$ or $a=\frac{1}{\sqrt{2}}$

Hence the square root of the complex no. is $-\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{2}} i$ and $\frac{1}{\sqrt{2}}-\frac{3}{\sqrt{2}} i$.


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