# How do you find the two square roots of -1 + irad3?

Jul 27, 2018

$\pm \left(\frac{1}{\sqrt{2}} + i \sqrt{\frac{3}{2}}\right) = \pm \frac{1 + i \sqrt{3}}{\sqrt{2}}$.

#### Explanation:

The desired square roots can be easily foung using

D'Moivre's Theorem.

Here is an Aliter.

Suppose that, $z = x + i y = \sqrt{- 1 + i \sqrt{3}} , \left(x , y \in \mathbb{R}\right)$.

$\therefore {z}^{2} = {\left(x + i y\right)}^{2} = \left(- 1 + i \sqrt{3}\right)$.

But, ${\left(x + i y\right)}^{2} = {x}^{2} + 2 i x y + {i}^{2} {y}^{2} = \left({x}^{2} - {y}^{2}\right) + 2 i x y$.

$\therefore {\left(x + i y\right)}^{2} = - 1 + i \sqrt{3}$,

$\Rightarrow \left({x}^{2} - {y}^{2}\right) + i \left(2 x y\right) = - 1 + i \sqrt{3.}$

Comparing real & imaginary parts, we have,

${x}^{2} - {y}^{2} = - 1 \mathmr{and} 2 x y = \sqrt{3} , \text{ so that,}$

${\left({x}^{2} + {y}^{2}\right)}^{2} = {\left({x}^{2} - {y}^{2}\right)}^{2} + 4 {x}^{2} {y}^{2}$,

$= {\left(- 1\right)}^{2} + {\left(\sqrt{3}\right)}^{2}$.

$\Rightarrow {\left({x}^{2} + {y}^{2}\right)}^{2} = 4 , \text{ giving, }$

${x}^{2} + {y}^{2} = + 2. \ldots . . \left[\because , x , y \in \mathbb{R}\right]$.

Solving this with ${x}^{2} - {y}^{2} = - 1 ,$ we get,

$x = \pm \frac{1}{\sqrt{2}} , \mathmr{and} , y = \frac{\sqrt{3}}{2 x} = \pm \sqrt{\frac{3}{2}}$.

Thus, the desired square roots are given by,

$x + i y = \pm \left(\frac{1}{\sqrt{2}} + i \sqrt{\frac{3}{2}}\right) = \pm \frac{1 + i \sqrt{3}}{\sqrt{2}}$.

$\textcolor{g r e e n}{\text{Enjoy Maths.!}}$