Solve this
Question:

$\sqrt{i}$

Solution:

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

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

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

Since $i^{2}=-1$

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

Now, separating real and complex parts, we get

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

$\Rightarrow 2 \mathrm{ab}=1$ …….. eq.2

$\Rightarrow \mathrm{a}=\frac{1}{2 b}$

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

$\Rightarrow\left(\frac{1}{2 b}\right)^{2}-b^{2}=0$

$\Rightarrow 1-4 b^{4}=0$

$\Rightarrow 4 b^{2}=1$

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

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

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

$\mathrm{b}=\frac{1}{\sqrt{2}}$ or $\mathrm{b}=-\frac{1}{\sqrt{2}}$

Therefore,$a=\frac{1}{\sqrt{2}}$ or $a=-\frac{1}{\sqrt{2}}$

Hence the square root of the complex no. is $\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$ i and $\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}$ i.

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.