Solve this

Let, $(a+i b)^{2}=5+12 i$

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

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=5+12 i$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=5+12 i$

Now, separating real and complex parts, we get

$\Rightarrow a^{2}-b^{2}=5 \ldots \ldots \ldots \ldots \ldots$ eq. 1

$\Rightarrow 2 \mathrm{ab}=12 \ldots \ldots . . \mathrm{q} .2$

$\Rightarrow \mathrm{a}=\frac{6}{b}$

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

$\Rightarrow\left(\frac{6}{b}\right)^{2}-b^{2}=5$

$\Rightarrow 36-b^{4}=5 b^{2}$

$\Rightarrow b^{4}+5 b^{2}-36=0$

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

$\Rightarrow b^{2}=-9$ or $b^{2}=4$

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

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

Therefore, $a=3$ or $a=-3$

Hence the square root of the complex no. is $3+2 i$ and $-3-2 i$.