Question

Find the square root of i.

Or

Solve the quadratic equation $x^2+x+\frac{1}{\sqrt{2}}=0$

Answer 1

Let $\sqrt{i}=x+iy.$

On squaring both side , we get

$i=(x+iy{)}^2$

$\Rightarrow i=x^2-y^2+2xyi$

On equating the real and imaginary parts, we get

$x^2-y^2=0$ and $2xy=1$

Now,$x^2+y^2=\sqrt{(x^2-y^2{)}^2+(2xy{)}^2}$

$\Rightarrow x^2+y^2=\sqrt{0+(1{)}^2}$

$\Rightarrow x^2+y^2=1$

Solving equations $x^2-y^2=0$ adn $x^2+y^2=1$, we get

$x=\pm \frac{1}{\sqrt{2}}$and $y=\pm \frac{1}{\sqrt{2}}$

Hence, the square root of i is $\pm (\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)$

Or

The given equation is $x^2+x+\frac{1}{\sqrt{2}}=0,$ comparing this equation with $a{x}^2+bx+c=0,$ we get a=1,b=1 and $c=\frac{1}{\sqrt{2}}$

Substituting these values in

$\alpha =\frac{-b+\sqrt{b^2-4ac}}{2a}$

and $\beta =\frac{-b-\sqrt{b^2-4ac}}{2a}$

Then,we get $\alpha =\frac{-1+\sqrt{1-2\sqrt{2}}}{2}$

and $\beta ==\frac{-1+\sqrt{1-2\sqrt{2}}}{2}$

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

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