Solve this




Let, $(a+i b)^{2}=0-8 i$

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

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=0-8 i$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=0-8 i$

Now, separating real and complex parts, we get

$\Rightarrow a^{2}-b^{2}=0$ ………..eq.1

$\Rightarrow 2 \mathrm{ab}=-8 \ldots \ldots \ldots$ eq. 2

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

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


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

$\Rightarrow b^{4}=16$

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

$\Rightarrow b^{2}=-4$ or $b^{2}=4$

As $b$ is real no. so, $b^{2}=4$

$b=2$ or $b=-2$

Therefore, $a=-2$ or $a=2$

Hence the square root of the complex no. is $-2+2 i$ and $2-2 i$.


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