Solve this


$\sqrt{-15-8 i}$


Let, $(a+i b)^{2}=-15-8 i$

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

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=-15-8 i$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=-15-8 i$

Now, separating real and complex parts, we get

$\Rightarrow a^{2}-b^{2}=-15$ …………..eq.1

$\Rightarrow 2 a b=-8$ …….. eq.2

$\Rightarrow \mathrm{a}=-\frac{4}{b}$

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


$\Rightarrow 16-\mathrm{b}^{4}=-15 \mathrm{~b}^{2}$

$\Rightarrow \mathrm{b}^{4}-15 \mathrm{~b}^{2}-16=0$

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

$\Rightarrow b^{2}=16$ or $b^{2}=-1$

As $b$ is real no. so, $b^{2}=16$

$b=4$ or $b=-4$

Therefore, $a=-1$ or $a=1$

Hence the square root of the complex no. is $-1+4 i$ and $1-4 i$.


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