Solve this

Let, $(a+i b)^{2}=7-30^{\sqrt{2}}_{i}$

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

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=7-30^{\sqrt{2}}_{i}$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=7-30^{\sqrt{2}} i$

Now, separating real and complex parts, we get

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

$\Rightarrow 2 \mathrm{ab}=30^{\sqrt{2}} \ldots \ldots . . \mathrm{eq} .2$

$\Rightarrow a=\frac{15 \sqrt{2}}{b}$

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

$\Rightarrow\left(\frac{15 \sqrt{2}}{b}\right)^{2}-b^{2}=7$

$\Rightarrow 450-b^{4}=7 b^{2}$

$\Rightarrow b^{4}+7 b^{2}-450=0$

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

$\Rightarrow b^{2}=-25$ or $b^{2}=18$

As $\mathrm{b}$ is real no. so, $\mathrm{b}^{2}=18$

$\mathrm{b}=3 \sqrt{2}$ or $\mathrm{b}=-3 \sqrt{2}$

Therefore, $a=5$ or $a=-5$

Hence the square root of the complex no. is $5+3 \sqrt{2}_{i}$ and $-5-3 \sqrt{2}_{i}$.