Computation of roots in algebraic form

Computation of roots in algebraic form

The computation of $n^{th}$ roots of a given complex number

is performed using the algebraic representation of a complex number only when

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

and

, with real

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

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

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

Example 1.5.11 Solve the equation:

First we compute the square roots of . Denote , 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 are and .

By the same method we compute the square roots of .

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

Hence, the square roots of are and . These are two of the four roots of order 4 of the number .

We leave to the reader the task of computing the square roots of . Finally, we found the four roots of order 4 of ; they are the complex numbers , , and .

Example 1.5.12 In order to solve the equation

, the work is decomposed into three steps:

Compute the square roots of -1; these are and ;
Compute the square roots of ; these are $\pm \left(\frac {\sqrt{2}}{2}+ i \; \frac {\sqrt{2}}{2}\right)$ . Then compute the square roots of ; 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.