How can I find the other solutions of $\sqrt{i}$? I'm trying to learn more about mathematical proofs, what's allowed, what's not, etc. I remembered seeing a video about $\sqrt{i}$ and wanted to tackle the problem myself. Here's what I came up with:
$\sqrt{i} = i^{1/2} = e^{\ln(i^{1/2})}=e^{\frac12\ln(i)}=e^{\frac12*i\pi/2}=e^{i\frac{\pi}4}$
$e^{i\theta}=\cos(\theta)+i\sin(\theta)$. It follows that:
$ e^{i\frac{\pi}4}=\cos(\frac{\pi}4)+i\sin(\frac{\pi}4) = \frac{\sqrt2}2+i\frac{\sqrt2}2$.
Squaring this quantity does in fact result in $i$. However, the periodicity of trigonometric functions seems to imply further solutions. How can they be found? Also, how can I improve my proof?
 A: Writing $\sqrt i$ is not a good idea, because you don't know which root you are talking about (there are two of them). With positive numbers, this is not an issue because we have the convention that $\sqrt a$ means the positive square root of the positive number $a$. 
When looking for $z$ with $z^2=i$, we may write $z=re^{ it}$. Since $|i|=1$, we easily see that $r=1$. Then 
$$
z^2=e^{2it}. 
$$
So $z^2=i$ precisely when $2t=\frac\pi2+2k\pi$. We may rewrite this as $$t=\frac\pi4+k\pi.$$
Only two choices of $k$ will produce a different $z$, namely when $k$ is even or odd. So we may take $$t_1=\frac\pi4, \ \ t_2=\frac\pi4+\pi=\frac{5\pi}4.$$ This gives us the two roots
$$
z_1=e^{i\pi/4},\ \ \ z_2=e^{5i\pi/4}.
$$
We can write this more explicitly, since $\sin\pi/4=\cos\pi4=\sqrt2/2$, as 
$$
z_1=\frac{\sqrt2}2+i\frac{\sqrt2}2,\ \ \ z_2=-\frac{\sqrt2}2-i\frac{\sqrt2}2. 
$$
One can save a bit of work by noticing that if $z_1$ is a root of $z_0$, then $-z_1$ is also a root; and that every complex number has precisely two roots, given by the even/odd choice of $k$ as above. 
A: High-school method: $\sqrt{i}=a+bi$ means $i=(a+bi)^2=a^2-b^2+2abi$ and we have two equations for $a$ and $b$: $a^2-b^2=0$, $2ab=1$; $a=\pm b$; $2b^2=1$, $b=\pm \frac{1}{\sqrt{2}}$ and we have two solutions (because $2ab=1$): $a=\frac{1}{\sqrt{2}}, b=\frac{1}{\sqrt{2}}$ and $a=-\frac{1}{\sqrt{2}}, b=-\frac{1}{\sqrt{2}}$
A: You  can introduce the periodicity of the exponential function writing:
$$
i=e^{i(\frac{\pi}{2}+2k\pi)}
$$
so that:
$$i^{\frac{1}{2}}=e^{\frac{i}{2}(\frac{\pi}{2}+2k\pi)}$$
that, for $k=0$ and $k=1$ gives the two solutions:
$$e^{\frac{\pi}{4}i}$$
$$e^{(\frac{\pi}{4}+\pi)i}=-e^{\frac{\pi}{4}i}$$
A: Hint: this problem is much easier if you think about the geometry of the map $z \mapsto z^2$ on the complex plane. In polar coordinates, we have $(r, \theta)^2 = (r^2, 2\theta)$. I.e. squaring squares the modulus and doubles the argument. With $i$ represented as $(1, \pi/2)$, you should be able to visualise the two solutions to $(1, 2\theta)^2 = (1, \pi/2)$. They arise from the two solutions to $2\theta = \pi/2$ in angle arithmetic ($\theta = \pi/4$ and $\theta = 5\pi/4$.)
A: For any complex number $z= re^{ix} = r(\cos x + i\sin x)$, it is true that $z = re^{i(x + 2k\pi )} = r(\cos(x + 2k\pi) + i\sin(x + 2k \pi)$.
So $i = 1*e^{i\frac {\pi}2} = e^{i(\frac {\pi}2 + 2k \pi)}$.
And $\sqrt{i} = e^{\frac 12(i\frac {\pi}2 + 2k \pi)}= e^{i(\frac {\pi}4 + k\pi)}$
As $k \pi \equiv (k+2)\pi$ for trig functions. The two solutions are $e^{i(\frac \pi 4)} = \cos {\frac \pi 4} + i \sin \frac \pi 4 = \frac{ \sqrt{2}}2 + i\frac {\sqrt{2}}2$ and $e^{i(\frac \pi 4 + \pi)} = \cos {(\frac \pi 4+\pi)} + i \sin (\frac \pi 4 +\pi)= -\frac{ \sqrt{2}}2 - i\frac {\sqrt{2}}2$.
Which really shouldn't surprise us.
For any $z\ne 0$ there are two square roots, one the negative of the other.  $z = r(\cos \theta + i\sin \theta)=re^{i\theta}$ then $\sqrt z = \pm\sqrt r( \cos \frac \theta 2 + i \sin \frac \theta 2)= \sqrt r(\pm  \cos \frac \theta 2 \pm i \sin \frac \theta 2) = \sqrt r(\cos (\frac \theta 2 \{+ \pi\} + i\sin(\frac \theta 2 \{+ \pi\}) = \sqrt r e^{i(\frac \theta 2 \{+ \pi\})}$; one the negative of the other.
