Calculating a root of a complex number with euler formula Let $z= 1+i$.
The polar form of $z^{1/5}$ can be easily calculated: $$z = 2^{1 \over 10}\left\{\cos({\pi/4 + 2\pi k \over 5}) + i\cdot  \sin({\pi/4 + 2\pi k \over 5})\right\}$$
the "principal root" as wolframalpha denotes is: $$e^{i\pi \over 20}$$
How can one "jump" easily to the next root, which is: $$e^{9i\pi \over 20}$$
 A: Firstly, Wolfram alpha gave for:
cos(((pi/4))/5))*2^(1/10)
(which is what I guess you tested too)
about 1.058
I suspect you made a little error with the bracketing (or maybe Wolfram is being silly). Either way, never refrain from putting in brackets to make ext a sure, it can't hurt.
This number definitely can't be a negative, as the cosine angle is between $0$ and $pi/4$ and the tenth root of 2 is taken as the modulus (so the root is always positive). Really, what you got must have been due to an error of some form.
To make the answers "jump", you just need to think of the whole system as an equation. So what is the variable that should, in multiple ways, satisfy the whole equation?? It's $k$. So we set up the whole equation with $k$, as it's sole variable:
$ 1 + i = 2^{1 \over 10}\left\{\cos({\pi/4 + 2\pi k \over 5}) + i\cdot  \sin({\pi/4 + 2\pi k \over 5})\right\}$
This will solve for $k$, and Wolfram will spit out a general form, hence, giving you a means to easily obtain the following roots.
Have a nice day !
A: Just increment $k$? So, the first ("principal") root is when $k = 0$, the second is when $k = 1$ and so on, until $k = 4$.
$z^{1/5}_0 = 2^{\frac{1}{10}} e^{\frac{\pi/4 + 2\pi * 0}{5}}$
$z^{1/5}_1 = 2^{\frac{1}{10}} e^{\frac{\pi/4 + 2\pi * 1}{5}}$
$z^{1/5}_2 = 2^{\frac{1}{10}} e^{\frac{\pi/4 + 2\pi * 2}{5}}$
$\dots$
A: The $n$-th roots of a complex number $z = re^{i\varphi}$ are $$
  \alpha_k = \sqrt[n]{r}e^{i\frac{\varphi}{n} + i\frac{2\pi}{n}k} \quad\text{ where } k \in \{0,1,\ldots,n-1\}
$$
because then $$
  \alpha_k^n = re^{i\varphi + i2\pi k} = re^{i\varphi} \text{.}
$$
(Remember that $e^{i2\pi k} = 1$ for $k \in \mathbb{Z}$)
Thus, the $n$ complex $n$-th roots of $z$ always form a regular polygon whose vertices lie on the circle with radius $\sqrt[n]{|z|}$ around 0.
You can get from one root to the next by multiplying with $e^{i\frac{2\pi}{n}}$, i.e. you have $$
  \alpha_{n+1} = \alpha_n e^{i\frac{2\pi}{n}} \text{.}
$$
One way to define the principal $n$-th root is to pick that $\alpha_n$ for which the angle between $\alpha_n$ and the positive real axis is smallest. In that case, principal roots always have positive imaginary part, and for $n \geq 4$ also positive real part. 
