Computation of roots in algebraic form.

The computation of $ n^{th}$ roots of a given complex number $ u$ is performed using the algebraic representation of a complex number only when $ n$ is a power of 2. In other cases the problem is translated into a system of two non linear polynomials equations which is rather hard to solve without using appropriate software.

We illustrate with examples a suitable method to compute square roots and roots of order 4 of a given com[plex number.

Example 1.5.10   Let $ u=-3+4i$ and $ z=x+iy$ , with real $ x,y$ .

$\displaystyle z^2=u \Leftrightarrow x^2-y^2+2ixy= -3+4i \Leftrightarrow \begin{...
...s} \Leftrightarrow \begin{cases}x^2-y^2 = -3 \ x^2+y^2 =5 \ 2xy=4 \end{cases}$    

The extra equation is equivalent to $ \vert z^2\vert=\vert u\vert$ . Solving this system for $ x$ and $ y$ , we have $ (x,y)=(1,2)$ or $ (x,y)=(-1,-2)$ . This means that $ u$ has two square roots, namely $ 1+2i$ and $ -1-2i$ .

By iterations with this method more, we can solve compute roots of order 4,8,16,etc.

Example 1.5.11   Solve the equation: $ z^4=-7-24i$ .

First we compute the square roots of $ -7-24i$ . Denote $ z=x+iy$ , where $ x,y \in \mathbb{R}$ . The following holds:

$\displaystyle z^2=-7-24i \Longleftrightarrow \begin{cases}x^2-y^2=-7\ x^2+y^2=...
...end{cases} \Longleftrightarrow \begin{cases}x^2=9 \ y^2=16\ xy<0 \end{cases}.$    

Thus, the square roots of the complex number $ -7-24i$ are $ 3-4i$ and $ -3+4i$ .

By the same method we compute the square roots of $ 3-4i$ .

$\displaystyle z^2=3-4i \Longleftrightarrow \begin{cases}x^2-y^2=3\ x^2+y^2=5\\...
...\end{cases} \Longleftrightarrow \begin{cases}x^2=4 \ y^2=1\ xy<0 \end{cases}.$    

Hence, the square roots of $ 3-4i$ are $ 2+i$ and $ -2-i$ . These are two of the four roots of order 4 of the number $ -7-24i$ .

We leave to the reader the task of computing the square roots of $ -3+4i$ . Finally, we found the four roots of order 4 of $ -7-24i$ ; they are the complex numbers $ 2+i$ , $ -1+2i$ , $ -2-i$ and $ 1-2i$ .

Example 1.5.12   In order to solve the equation $ z^8=-1$ , the work is decomposed into three steps:
  1. Compute the square roots of -1; these are $ i$ and $ -i$ ;
  2. Compute the square roots of $ i$ ; these are $ \pm \left(\frac {\sqrt{2}}{2}+ i \; \frac
{\sqrt{2}}{2}\right)$ . Then compute the square roots of $ -i$ ; they are equal to $ \pm \left(\frac {\sqrt{2}}{2}+ i \; \frac
{\sqrt{2}}{2}\right)$ . The reader can have a confirmation of the validity of these results by computing the roots using the polar method.

Noah Dana-Picard 2007-12-24