Polar form.

Let $ u=r ( \cos \theta + i \sin \theta)$ , where $ r>0$ and $ \theta \in \mathbb{R}$ .

$\displaystyle z^n=u \Leftrightarrow \begin{cases}\vert z^n\vert=r \ \operatorn...
...eratorname{arg}\nolimits (z)= \frac {\theta}{n} + \frac {2k \pi}{n} \end{cases}$    

The precise meaning of the last equation, namely of

$\displaystyle \operatorname{arg}\nolimits (z)= \frac {\theta}{n} + \frac {2k \pi}{n}$    

is that there exist exactly $ n$ distinct $ n^{th}$ roots for any non zero complex number $ u$ . You can check this by the following way: let $ k$ increase from 0 to $ n-1$ , you have distinct numbers. From $ k-n$ and further on, you get the same complex numbers again (you only added a multiple of $ 2 \pi$ to the argument...).

Example 1.5.2   Find the cubic roots of $ u=-1+i$ .

$ u=1-i = \sqrt{2} \left( \cos \frac {3 \pi}{4} = i \sin \frac {3 \pi}{4} \right)$ . Thus:

$\displaystyle z^3=-1+i \Leftrightarrow \begin{cases}\vert z\vert^3= \sqrt{2} \\...
...\operatorname{arg}\nolimits (z) =\frac {\pi}{4} + \frac {2k \pi}{3} \end{cases}$    

The cubic roots of $ u$ are:

$\displaystyle z_1$ $\displaystyle =\sqrt[6]{2} \left( \cos \frac {\pi}{4} + i \sin \frac {\pi}{4} \...
...eft( \frac {\sqrt{2}}{2} + i \frac {\sqrt{2}}{2} \right) =2^{-1/3} + i 2^{-1/3}$    
$\displaystyle z_2$ $\displaystyle = \sqrt[6]{2} \left( \cos \frac {11 \pi}{12} + i \sin \frac {11 \pi}{12} \right)$    
$\displaystyle z_3$ $\displaystyle = \sqrt[6]{2} \left( \cos \frac {19 \pi}{12} + i \sin \frac {19 \pi}{12} \right)$    

Example 1.5.3   Find the cubic roots of $ u=-2+2i \sqrt{3}$ .

$ u=-1+i \sqrt{3}= 8 \left( \cos \frac {2 \pi}{3} = i \sin \frac {2 \pi}{3} \right)$ . Thus:

$\displaystyle z^3=-1+i \sqrt{3} \Leftrightarrow \begin{cases}\vert z\vert^3= 8 ...
...peratorname{arg}\nolimits (z)= \frac {2 \pi}{9} + \frac {2k \pi}{3} \end{cases}$    

The cubic roots of $ u$ are:

$\displaystyle 2 \left( \cos \frac {2 \pi}{9} + i \sin \frac {2 \pi}{9} \right), \; 2 \left( \cos \frac {8 \pi}{9} + i \sin \frac {8 \pi}{9} \right) \;$   and $\displaystyle \; 2 \left( \cos \frac {14 \pi}{9} + i \sin \frac {14 \pi}{9} \right).$    

Noah Dana-Picard 2007-12-24