Solve this

Let, $(a+i b)^{2}=-7+24 i$

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

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=-7+24 i$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=-7+24 i$

Now, separating real and complex parts, we get

$\Rightarrow a^{2}-b^{2}=-7 \ldots \ldots \ldots \ldots \ldots$ eq. 1

$\Rightarrow 2 \mathrm{ab}=24 \ldots \ldots . . \mathrm{eq} .2$

$\Rightarrow a=\frac{12}{b}$

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

$\Rightarrow\left(\frac{12}{b}\right)^{2}-b^{2}=-7$

$\Rightarrow 144-b^{4}=-7 b^{2}$

$\Rightarrow b_{4}-7 b^{2}-144=0$

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

$\Rightarrow b^{2}=-9$ or $b^{2}=16$

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

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

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

Hence the square root of the complex no. is $3+4 \mathrm{i}$ and $-3-4 \mathrm{i}$.