Solve this


$\sqrt{-11-60 i}$


Let, $(a+i b)^{2}=-11-60 i$

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

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=-11-60 i$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=-11-60 i$

Now, separating real and complex parts, we get

$\Rightarrow a^{2}-b^{2}=-11 \ldots \ldots \ldots \ldots \ldots$ eq. 1

$\Rightarrow 2 a b=-60 \ldots \ldots . .$ eq. 2

$\Rightarrow \mathrm{a}=-\frac{30}{b}$

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


$\Rightarrow 900-b^{4}=-11 b^{2}$

$\Rightarrow b^{4}-11 b^{2}-900=0$

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

$\Rightarrow b^{2}=36$ or $b^{2}=-25$

as $b$ is real no. so, $b^{2}=36$

$b=6$ or $b=-6$

Therefore, $a=-5$ or $a=5$

Hence the square root of the complex no. is -5 + 6i and 5 – 6i.


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