Why do some numbers have roots of unity? Let's say I wanted to express sqrt(4i) in a + bi form. A cursory glance at WolframAlpha tells me it has not just a solution of 2e^(iPi/4), which I found, but also 2e^(i(-3Pi/4))
Why do roots of unity exist, and why do they exist in this case? How could I find the second solution? 
 A: $\mathbf{i}$ is a root of unity. Thus any root of $\mathbf{i}$ is a root of unity.
How do you usually find the other square root of a number given one of its square roots? Same thing applies here.
Alternatively, recall that the exponential has period $2 \pi \mathbf{i}$. If you wrote
$$ \mathbf{i} = e^{\mathbf{i} \pi / 2} $$
it's helpful to remember that you also have
$$ \mathbf{i} = e^{2 \pi \mathbf{i} n + \mathbf{i} \pi / 2} $$
for every integer $n$.
A: You're solving the equation $z^2=4i$. According to the Fundamental Theorem of Algebra, this equation has two complex roots. You can find them in many ways. 
The most elementary approach: assume $z=a+bi$, where $a$ and $b$ are real. Then $(a+bi)^2=(a^2-b^2)+2abi=4i$. Equating real and imaginary parts, you need $a^2-b^2=0$ and $2ab=4$. The first equation says either $a=b$ or $a=-b$. If $a=b$, then the second equation says $a^2=2$, whence $a=\pm\sqrt{2}$. If $a=-b$, the second equation says $a^2=-2$, which has no solutions (because we assumed $a$ is real). So the solutions are $\sqrt{2}+i\sqrt{2}$ and $-\sqrt{2}-i\sqrt{2}$. 
Alternatively, write $4i=4e^{i\pi/2}$ and $z=re^{i\theta}$, with $r$ a positive real number and $\theta\in[0,2\pi)$. Then $z^2=4i$ says $r^2e^{2i\theta}=4e^{i\pi/2}$. Conclude that $r=2$. To finish we need to find $\theta$ such that $e^{2i\theta}=e^{i\pi/2}$. Clearly $\theta=\pi/4$ works. But $e^{i\pi/2}=e^{i(\pi/2+2\pi)}$, so we can also take $2\theta=\pi/2+2\pi$, whence $\theta=\pi/4+\pi=5\pi/4$. Hence the two solutions are $2e^{i\pi/4}$ and $2e^{5i\pi/4}$.
A: Roots of unity are basically the idea of multiple square roots extended to any level of square roots. First, we have to understand the complex plane. You might be familiar with the equation $e^{i \pi} = -1$. This expression is equivalent to another, $(e^{i \theta} = \cos \theta + i\sin \theta)$. Based on the properties of complex numbers and exponents, $e^{i \theta_1}*e^{i \theta_2} = e^{i (\theta_1 + \theta_2)}$, From here, if I were to ask you for the cube root of $1$, we see that there must be a radius of $1$, and $3*\theta \equiv 0 mod 2 \pi$. For this equation, there are three solutions: $0, 2 \pi/3, 4 \pi/3$, which lead us to the roots of unity.
A: By definition $\zeta\in\Bbb C$ is a root of unity if there is $n\in\Bbb N$ so that $\zeta^n=1$. Roots of unity exist thanks to $e^{2\pi i}=1$ and the usual fact about exponentials that $(e^a)^b=e^{ab}$ so that $e^{2\pi i/n}$ is always an $n^{th}$ root of unity.
To see how you can get them all just note that $e^{2\pi i k/n}$ is also an $n^{th}$ root of unity for any $0\le k\le n-1$, and since the smallest $x>0$ so that $e^{ix}=1$ is $x=2\pi$ this means we give rise to all distinct values for $0\le k\le n$. So this is how you find them all in general.
A: $$\sqrt{4i}$$
$$= \sqrt{4e^{\pi i /2 + 2\pi n}}$$
$$= 2e^{\pi i/4 + \pi n}$$
$$= 2e^{\pi i/4}, 2e^{5\pi i/4}$$
There are two solutions to the square root of any number. There are three cube roots, four fourth roots, ect.
