$$W = \frac{1+i}{\sqrt{2}}$$
I need to find 5th root of $W$ where $Z^5 = W$
$\theta$ is: $$\frac{\pi}{20} + \frac{2\pi k}{5}$$
I always thought You need to plug in $K = 0, \pm 1, \pm 2,\ldots$ to find the solution.
However, prof did this differently and wrote $K = 0, 1, 2, 3, 4,\ldots$.
Therefore my question, does it matter which of the two options you pick?
On
For the periodicity of circular function is the same if you use $k\in \Bbb N$ or $k\in \Bbb Z$, it depends on your preferences. Sine and cosine are defined for every real numbers, both positive and negative and they have the same periodicity both on the positive and negative part of the real line. Just to clarify I give you some examples:
$\cos (\pi)=\cos (-\pi)=-1$
$\sin (\frac {5\pi} 4)=\sin (-\frac {3\pi} 4)=-\frac 1 {\sqrt 2}$
And so on, using the unit circle to count angles you count positive anticlock-way and negative clock-way. My personal advice is to use only positive angles, which looks more clare to me.
$$W=\frac{1+i}{\sqrt{2}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i=e^{\frac{1}{4}\pi}$$