Fourth complex roots of $i$ Solve $z^4 = i$. I cannot figure out why the angle of $i$ is $\frac{\pi}{2}$ and how to determine the values of $k$. If someone could show step-by-step that would be great!
Thanks.
 A: for $z=x+iy$ you can write its Polar Representation $z=re^{i\theta‎}=r(\cos\theta+i\sin\theta)=\color{red}{r\cos\theta}+i\color{blue}{r\sin\theta}$(respectively real part and imaginary part red and blue) where $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}\frac{y}{x} $ for $x>0$ and $\theta=tan^{-1}\frac{y}{x}+\pi $ for $x<0$.
For $w = \rho e^{i \phi}$ that $w^n = z$,   $\rho = r^{1/n}$, $\phi = \frac{\theta}{n} + \frac{2\pi k}{n}$ for $k=0, 1, ..,n-1$.
here for $z=i$, we have $r=\sqrt{0^2+1^2}=1$ and $\theta=tan^{-1}\frac{1}{0}=\frac{\pi}{2} $
thus answers will be $w=(1)^\frac{1}{4}e^{i(\frac{\frac{\pi}{2}}{4} + \frac{2\pi k}{4})}, k=0,1,2,3$
A: So, first know that whenever one does complex arithmetic, the trick is almost always to write the numbers in the for $re^{i\theta}$. So, how to do this? 
The easy way to see it is draw the unit circle, put your pencil at the point  1 (right hand side of the circle on the x-axis). Now your on the positive real line which has angle 0 (radians). Ok, tracing out the circle counter clockwise i.e. walking counter-clockwise around the unit circle, our angle increases from 0 to 2$\pi$ (radians, or 360 degrees if you like) when you've made one complete rotation. Notice when you're 1/4 of the way around you're at the top of the circle (aka (0,1) or $i$). This means that $i$ has angle one fourth of 2$\pi$ or $\pi/2$. But wait there's more! This also means that when you've completed one full circle, you're back at the point 1, and it must also have angle $2\pi$ in addition to 0. Walking around again we see $i$ also have angle $2\pi$ + one fourth of another rotation = $2\pi + \pi/2 = 5\pi/2$.  Iteratively, $i$ can has angle $k\times\frac{\pi}{2}$ for $k = ..., 1,5,9, ...$. 
More generally, every complex number has infinitely many angle representations. 
If you like trig, here's an algorithm to produce that intuition.


*

*Well, first write an arbitrary number $x+iy$ as $(x,y)$, in other words switch from thinking of the complex plane to $\mathbb{R}^2$. (for $i$ we get (0,1) ).  

*Now, $r$ is the distance of the point $(x,y)$ to the origin $(0,0)$. So we see $r = \sqrt{x^2+y^2}$ Draw the picture, it's just the Pythagorean Theorem. (for $i$ we get $ r = \sqrt{0^2+1^2} = 1$).  

*Next, using the same trig class from the back of our brain, we see $\theta = \arctan(y/x)$. Aha! there's the trick. Recall that arctan doesn't satisfy the vertical line test in we pick a branch! we have to choose that output will be in $[0,2\pi)$, or $[2\pi, 4\pi)$, etc.. for $i$ we see the angle is $\arctan(\frac{1}{0}$ (undefined) $=\frac{\pi}{2}$ if we pick the usual branch. 


Ok, let's solve the problem.  If we write $i$ in this form we get $z^4 = e^{i\pi/2}$, but also  $z^4 = e^{i5\pi/2}$, and $z^4 = e^{i9\pi/2}$, etc... In general, $z^4 = e^{i (4k+1)\pi/2}$ for integer $k.$ 
Taking everything to the $1/4$ power gives solutions $z = e^{i (4k+1)\pi/8}$. (Notice there are only 4 solutions ($k = 0,1,2,3)$ b/c if you plug in $k = -1$ that solutions $\theta$ is $2\pi$ difference from the angle for solution $ k = 3$-- there the same number, and those are two different angle representations. 
A: Are you familiar with Argand diagrams, if so I recommend making a quick sketch. On an argand diagram the complex vector $i$ points straight up (it lies on the imaginary axis in the positive direction) this is why the angle is $\frac{\pi}{2}$.
Write $z^4=x+iy=0+i$ where $x,y \in \mathbb{R}$.
$\arg(z^4)=\tan^{-1}\left(\cfrac{y}{x}\right)$. In your case $x=0$ and $y=1$. So $\arg(i)=\tan^{-1}\left(\cfrac{1}{0}\right)=\tan^{-1}(\infty)=\cfrac{\pi}{2}=\theta$. This is the angle the complex vector makes with the real-axis on an Argand diagram. Obviously taking the inverse tan of infinity is not the best way to do it as $\infty$ is undefined so making a sketch on an Argand Diagram makes more sense. You can also see this by looking at the $\tan x$ graph and noting that there is an asymptote where $x=\frac{\pi}{2}$.
First rewrite $z^4 =i$ as $z^4 - i =0$ and this time $\arg(z^4)=2k\pi+\cfrac{\pi}{2}$ as any integer $k$ of $2\pi$ yields the same result. Then one way of finding the solutions is to use $$z^n=(x+iy)^n=r^ne^{in\theta}=r\left(\cos(n\theta)+i\sin (n\theta)\right)$$ which is known as De Moivres Theorem. But if you have not come across this it might be best to look it up. 
In your case $r=|i|=1$. This is the magnitude of the complex vector.
So we have 
$$z=r^{\frac{1}{n}}\left(\cos(\theta)+i\sin (\theta)\right)^\frac{1}{n}=
1^{\frac{1}{4}}\left(\cos\left(\cfrac{\pi}{2}+2k\pi\right)+i\sin \left(\cfrac{\pi}{2}+2k\pi\right)\right)^\frac{1}{4}=
\cos\left(\cfrac{\frac{\pi}{2}+2k\pi}{4}\right)+i\sin \left(\cfrac{\frac{\pi}{2}+2k\pi}{4}\right)$$ and this is valid for $k=0,1,2,3$ as there are $4$ roots to the equation. So now you simply substitute these values of $k$ into the formula to get the $4$ solutions. 
A: $$z^4 = i \Longleftrightarrow$$
$$z^4 = |i| e^{\arg(i)i} \Longleftrightarrow$$
$$z^4 = 1 e^{\frac{\pi}{2}i} \Longleftrightarrow$$
$$z^4 = e^{\frac{\pi}{2}i} \Longleftrightarrow$$
$$z = \left(e^{\left(\frac{\pi}{2}+2\pi k\right)i}\right)^{\frac{1}{4}} \Longleftrightarrow$$
$$z = e^{{\frac{1}{4}}\left(\frac{\pi}{2}+2\pi k\right)i} \Longleftrightarrow$$
$$z = e^{\frac{\pi + 4\pi k}{8}i}$$
With $k\in\mathbb{Z}$ and $k:0-3$
So the solutions are:
$$z_0=e^{\frac{\pi + 4\pi \cdot 0}{8}i}=e^{\frac{\pi}{8}i}$$
$$z_1=e^{\frac{\pi + 4\pi \cdot 1}{8}i}=e^{\frac{5\pi}{8}i}$$
$$z_2=e^{\frac{\pi + 4\pi \cdot 2}{8}i}=e^{-\frac{7\pi}{8}i}$$
$$z_3=e^{\frac{\pi + 4\pi \cdot 3}{8}i}=e^{-\frac{3\pi}{8}i}$$
