Question

Find the square root of $x+i\sqrt{x^4+x^2+1}$, where $i=\sqrt{-1}$.

Answer 1

Now,

$x+i\sqrt{x^4+x^2+1}$

$=\frac{1}{2}(2x+2i\sqrt{x^4+x^2+1})$

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

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

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

$=\frac{1}{2}\left\{\right(\sqrt{x^2+x+1}{)}^2+(i\sqrt{x^2-x+1}{)}^2+2i\sqrt{(x^2+x+1)(x^2-x+1)}\}$

$=\frac{1}{2}\left\{\right(\sqrt{x^2+x+1}+i\sqrt{x^2-x+1}{)}^2\}$

Now, square root of $x+i\sqrt{x^4+x^2+1}$ is $\pm \frac{1}{\sqrt{2}}(\sqrt{x^2+x+1}+i\sqrt{x^2-x+1})$