Solve this

Let, $(a+i b)^{2}=1+4^{\sqrt{3}} i$

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

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=1+4^{\sqrt{3}} i$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=1+4^{\sqrt{3}} i$

Now, separating real and complex parts, we get

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

$\Rightarrow 2 \mathrm{ab}=4^{\sqrt{3}} \ldots \ldots . \mathrm{eq} .2$

$\Rightarrow a=\frac{2 \sqrt{3}}{b}$

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

$\Rightarrow\left(\frac{2 \sqrt{3}}{b}\right)^{2}-b^{2}=1$

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

$\Rightarrow b^{4}+b^{2}-12=0$

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

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

As $b$ is real no. so, $b^{2}=3$

$b=\sqrt{3}$ or $b=-\sqrt{3}$

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

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