Does the square root function behave the same when calculating complex roots? This is incredibly basic, yet I cannot for the life of me confirm it. Let's say in the topic of complex numbers we are told to calculate the roots of:
$
z^4 = \sqrt 9
$
Are we to assume $ z^4 = 3 $ and calculate 4 roots or should we proceed with " $ z^8 = 9 $ ", and calculate 8 roots ? Since we are not operating in real numbers, the fact that a negative square root can exist in this instance confused me. Do we just take it as +3 since the square root function is specific to non-negative real numbers? I know the professor wanted us to calculate 4 roots, but I would like to make sure.

$z^4=\sqrt9$ which implies: $z^4=3$ and $z^4=-3$
Solving: $z^4=3$
Let $u=z^2$, then $u^{2}=3 \rightarrow u=\pm \sqrt 3$
Now, $z^2=\sqrt3$ and $z^2=-\sqrt3$. Thus, there are four roots $$z=\pm\sqrt[4]{3}i \,\,\,\&\,\,\, z=\pm\sqrt[4]{3}$$
Solving $z^4=-3$
Let $u=z^2$, thus $u^{2}=-3 \rightarrow u=\sqrt3i \,\,\,\&\,\,\, u=-\sqrt3 i$
Hence, $z^2=\sqrt3i$ and $z^2=-\sqrt3 i$
Let $z=x+iy$, for $x,y \in \mathbb{R}$. Then $z^2=\left(x^2-y^2\right)+i(2xy)$.
Case 1: $z^2=\sqrt 3 i$
Now, the real part equals the real part, and the imaginary part equals the imaginary part. Hence, $$x^2-y^2=0\\ x=\pm y\\x=y$$ and $$2xy=\sqrt 3 \\2x^2=\sqrt 3 \\x=\pm\sqrt\frac{\sqrt 3}{2}\\ x=\pm\frac{\sqrt[4]{3}\sqrt{2}}{2}$$ and plugging back in these $x$-values into $2xy=\sqrt3$, we find that for $x=\frac{\sqrt[4]{3}\sqrt{2}}{2}$, $y=-\sqrt[4]{\frac{3}{4}}$ and for $x=-\frac{\sqrt[4]{3}\sqrt{2}}{2}$, $y=\sqrt[4]{\frac{3}{4}}$.
Hence writing them in terms of $z$, we get: $$z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i$$
Case 2:
$z^2=-\sqrt 3 i$. Similarly, for $x,y \in \mathbb{R}$ $x^2-y^2=0\\ x=\pm y\\ x=-y$ and $2xy=-\sqrt 3\\-2x^2=-\sqrt 3\\ \therefore x= \pm \frac{\sqrt[4]{3}\sqrt{2}}{2}$ corresponding to $y=\pm \sqrt[4]{\frac{3}{4}}$. Hence the solution in terms of $z$ are: $$z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i$$
Overall, there are $8$ roots to this equation: $$z=\pm\sqrt[4]{3}i \,\,\,\&\,\,\, z=\pm\sqrt[4]{3} \,\,\,\&\,\,\, z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i \,\,\,\&\,\,\,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i $$
