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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.


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Think of $i$ as a quarter-turn counter-clockwise. Square root of $i$ is a one-eighth turn or a "half turn plus one-eighth turn" because if you do either twice, you get a quarter-turn. $-i$ is a three-quarter turn counter-clockwise (= clockwise quarter-turn). – Mark Bennet Sep 23 '12 at 21:27
Instead of ´sqrt(9)´, etc. use ´\sqrt{9}´, etc. enclosed in $\$\,\, \$ $. For some basic information about writing math at this site see e.g. here, here, here and here. – Américo Tavares Sep 23 '12 at 21:27
In addition to other comments, this may also help: – Amzoti Sep 23 '12 at 21:28
$\sqrt{9}\ne (3,0)$. – Américo Tavares Sep 23 '12 at 21:34
up vote 1 down vote accepted

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 $$ \begin{align} 1 &= \sqrt{1} \\ &= \sqrt{(-1)(-1)} \\ &= \sqrt{-1}\sqrt{-1}\\ &= i\cdot i\\ &= -1. \end{align} $$

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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 $\displaystyle\pm\frac{1+i}{|1+i|} = \pm\frac{1+i}{\sqrt 2}$.

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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 $\left(z^{\frac{1}{n}}\right)_k=|z|^{\frac{1}{n}}\left(\cos(\frac{\arg{z}+2k\pi}{n})+i\sin(\frac{\arg{z}+2k\pi}{n})\right), \quad 0\leqslant k \leqslant n-1$.

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Be careful with sqare-roots. The is a "branch" issue and you can wind up with "multivalued" quantities if you are careless.

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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.

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As elements of $\mathbb{C}$ you only have two square roots. There are only two solutions to the equation $z^2 = -9i$. – Thomas Sep 23 '12 at 23:59
@Thomas: Then explain why $$\left(3 e^{i (-\pi/4 + k \pi)}\right)^2 = 9 e^{i (-\pi/2 + 2 k \pi)} = - 9 i$$ Since when does $i$ have a unique polar representation? – Rod Carvalho Sep 24 '12 at 0:00
(1) Remember that $e^{2\pi ik} = 1$ for all integers $k$. (2) A polynomial of degree $n$ has at most $n$ distinct roots in the complex numbers (Fundamental Theorem of Algebra). – Thomas Sep 24 '12 at 0:05
@Thomas: So, could we say that there are two solutions, and that these two solutions can be represented in infinitely many ways in polar form? – Rod Carvalho Sep 24 '12 at 0:16
That would IMO be perfectly fine. – Thomas Sep 24 '12 at 0:19

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