Question

Find the square root of $1+i\sqrt{3}$.

Answer 1

Let $\sqrt{1+\sqrt{3}i}=z=x+iy$

$\Rightarrow 1+\sqrt{3}i=x^2-y^2+i.2xy$

$\therefore x^2-y^2=1$.....(i)

$\therefore 2xy=\sqrt{3}$....(ii)

$(x^2+y^2{)}^2=(x^2-y^2{)}^2+4x^2{y}^2$

$\Rightarrow (x^2+y^2{)}^2=1+3$

$\Rightarrow x^2+y^2=2$....(iii)

(i)+(iii) $2x^2=3x=2\sqrt{\frac{3}{2}}$

$\therefore y=\pm \frac{1}{\sqrt{2}}$

But $xy>0$

$\therefore \sqrt{1+\sqrt{3}i}=\sqrt{\frac{3}{2}}+\frac{1}{\sqrt{2}}i$

or

$\sqrt{1+\sqrt{3}i}=-\sqrt{\frac{3}{2}}-\frac{1}{\sqrt{2}}i$