Solve this

$\sqrt{4 i}$


Let, $(a+i b)^{2}=0+4 i$

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

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=0+4 i$

Since $i^{2}=-1$

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

Now, separating real and complex parts, we get

$\Rightarrow a^{2}-b^{2}=0$  …………..eq.1

$\Rightarrow 2 \mathrm{ab}=4$ …….. eq.2

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

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


$\Rightarrow 4-\mathrm{b}^{4}=0$

$\Rightarrow b^{4}=4$

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

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

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

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

Therefore, $a=\sqrt{2}$ or $a=-\sqrt{2}$

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



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