Solve this

Let, $(a+i b)^{2}=3+4^{\sqrt{7}} i$

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

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=3+4^{\sqrt{7}} i$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=3+4^{\sqrt{7}} i$

now, separating real and complex parts, we get

$\Rightarrow a^{2}-b^{2}=3 \ldots \ldots \ldots \ldots \ldots e q \cdot 1$

$\Rightarrow 2 \mathrm{ab}=4^{\sqrt{7}} \ldots \ldots \ldots$ eq.2

$\Rightarrow a=\frac{2 \sqrt{7}}{b}$

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

$\Rightarrow\left(\frac{2 \sqrt{7}}{b}\right)^{2}-b^{2}=3$

$\Rightarrow 12-b^{4}=3 b^{2}$

$\Rightarrow b^{4}+3 b^{2}-28=0$

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

$\Rightarrow b^{2}=-7$ or $b^{2}=4$

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

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

Therefore, $a=\sqrt{7}$ or $a=-\sqrt{7}$

Hence the square root of the complex no. is ${ }^{\sqrt{7}}+2 i$ and $-\sqrt{7}-2 i$.