Question

Find
$\sqrt{i}+\sqrt{-i}$

Answer 1

$\sqrt{i}+\sqrt{(-i)}=\sqrt{i}+\sqrt{(-1)i}$

$=\sqrt{i}+\sqrt{\left(i^2\right)i}=\sqrt{i}+i\sqrt{i}$

$=\sqrt{i}(1+i)$

for $\sqrt{i}$

$x+iy=\sqrt{i}$

squaring both the sides,

$x^2-y^2+2xiy=(\sqrt{i}{)}^2=i$

$x^2-y^2=0$ & $2xiy=i$

$x=y$ & $xy=\frac{1}{2}$

$\therefore x=\frac{1}{\sqrt{2}}$ & $y=\frac{1}{\sqrt{2}}$ OR $x=-\frac{1}{\sqrt{2}}$ & $y=-\frac{1}{\sqrt{2}}$

are the solutions.

So

$\sqrt{i}+\sqrt{-i}=\sqrt{i}(1+i)=(1+i)(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$

$=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+2\frac{i}{\sqrt{2}}=\sqrt{2}i$

$Ans=\sqrt{2}i$