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 $x,y$ .

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

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:

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

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 $3-4i$ .

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

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.