I don't understand $\sqrt{-9i}$. I try to visualise it on a graph, where x is real numbers and y is the imaginary numbers. 
$\sqrt{9} = (3,0)$ and $(-3,0)$.
$\sqrt{-9} = \sqrt{-1} \times \sqrt{9} = (0,3) $ and $(0,-3)$.
$\sqrt{9i}$ = 
$\sqrt{-9i}$ =
Basically, I have some trouble representing the numbers visually on the graph.
Thanks.
 A: So, basically you are looking for $\sqrt i$. Do you know the geometrical meaning of complex multiplication? The lengths are multiplied and the angles (counted from the right half of the real axis) are added.
If this is clear, a square root of a complex number with absolute value (length) $1$, means halfing the angle.
So, $\sqrt i$ has angle $45^\circ$ (or $(180+45)^\circ$) and has length $1$. So it is $\displaystyle\pm\frac{1+i}{|1+i|} = \pm\frac{1+i}{\sqrt 2}$.
A: First: one can (hear) talk about a square root. We might say that a number $a$ is a square root of $b$ is $a^2 = b$. In this sense both $3$ and $-3$ are square roots of $9$.
Second: Most of the time (IMO) when one comes across the radical sign $\sqrt{}$, then one is thinking about the square root also known as the principal square root. For the non-negative real numbers, the square root of $b\geq 0$ is then defined to the the unique positive number $a$ such that $a^2 = b$. Hence we say that the square root of $9$ is equal to $3$ and we write $\sqrt{9} = 3$. (Granted, one might consider the radical sign as denoting the set consisting of all the square roots of a number). Note that for this setup we think og $\sqrt{}$ as a function from $[0,\infty) \to [0,\infty)$.
For complex numbers we also can talk about a square root or the (principal) square root. For the square root of a complex number $z = re^{i\theta}$, with $r\geq 0$ and $-\pi < \theta \leq \pi$ one usually defined the square root as: $\sqrt{re^{i\theta}} = \sqrt{r}e^{i\theta/2}$. So with this definition we have $$\begin{align}
\sqrt{i} &= \sqrt{e^{i\pi/2}} = e^{i\pi/4} = \frac{1}{\sqrt{2}}(1+i) = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}  \\
\sqrt{-i} &= \sqrt{e^{i(-\pi/2)}} = e^{-i\pi/4} = -\frac{1}{\sqrt{2}}(1+i).
\end{align}
$$
And you would then get for example $\sqrt{9i} = \frac{3}{\sqrt{2}}(1+i)$.
Graphically you would then represent $\sqrt{9i}$ as the point $(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}})$
Note that with this definition certain familiar rules don't hold. You for example do not have that $\sqrt{ab} = \sqrt{w}\sqrt{z}$ for all complex numbers $w$ and $z$. If you did, then you would have
$$
\begin{align}
1 &= \sqrt{1} \\
  &= \sqrt{(-1)(-1)} \\
  &= \sqrt{-1}\sqrt{-1}\\
  &= i\cdot i\\
  &= -1.
\end{align}
$$
A: Try by represent complex number $-9i$ in trigonometrical form $z=|z|\left(\cos(\arg{z})+i\sin(\arg{z})\right)$, putting $z=-9i$, and then find square root, applying de Moivre's formula $\left(z^{\frac{1}{n}}\right)_k=|z|^{\frac{1}{n}}\left(\cos(\frac{\arg{z}+2k\pi}{n})+i\sin(\frac{\arg{z}+2k\pi}{n})\right), \quad 0\leqslant k \leqslant n-1$.
A: Be careful with sqare-roots. The is a "branch" issue and you can wind up with "multivalued" quantities if you are careless.
A: Note that
$$-9 i = 9 e^{i (-\pi/2 + 2 k \pi)}$$
where $k \in \mathbb{Z}$. Therefore, if we take the square root we otain
$$\sqrt{-9 i} = \sqrt{9 e^{i (-\pi/2 + 2 k \pi)}} = 3 e^{i (-\pi/4 + k \pi)}$$
and we can conclude that the solution set is infinite (but countable). The most notorious solutions are $3 e^{- i \pi/4}$ and $3 e^{i ( 3\pi/4)}$, since all the other solutions will fall on top of these two.
