# Solve this

Question:

$\sqrt{1-i}$

Solution:

Let, $(a+i b)^{2}=1-i$

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

$\Rightarrow a^{2}+(b i)^{2}+2 a b i=1-i$

Since $i^{2}=-1$

$\Rightarrow a^{2}-b^{2}+2 a b i=1-i$

Now, separating real and complex parts, we get

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

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

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

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

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

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

$\Rightarrow 4 b^{4}+4 b^{2}-1=0$

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

$\Rightarrow b^{2}=\frac{-4 \pm \sqrt{32}}{8}$

As b is real no. so, $b^{2}=\frac{-4+4 \sqrt{2}}{8}$

$b^{2}=\frac{-1+\sqrt{2}}{2}$

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

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

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

and $\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{-1+\sqrt{2}}{2}}$ i.