Solve this
Question:

$\sqrt{-2+2 \sqrt{3} i}$

Solution:

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

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

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

Since $i^{2}=-1$

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

Now, separating real and complex parts, we get

$\Rightarrow a^{2}-b^{2}=-2 \ldots \ldots \ldots \ldots \ldots e q \cdot 1$

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

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

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

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

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

$\Rightarrow b_{4}-2 \mathrm{~b}^{2}-3=0$

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

$\Rightarrow b^{2}=-1$ or $b^{2}=3$

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

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

Therefore, a= 1 or a= -1

Hence the square root of the complex no. is $1+\sqrt{3}_{i}$ and $-1-\sqrt{3}_{i}$.

 

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