Aus Online Mathematik Brückenkurs 2

An equation of the type "\displaystyle z^{n} = \text{a complex number}" is called a binomial equation and these are usually solved by going over to polar form and using de Moivre's formula.

We start by writing \displaystyle z and \displaystyle 1 in polar form

\displaystyle \begin{align} z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt] 1 &= 1(\cos 0 + i\sin 0)\,\textrm{.} \end{align}

The equation then becomes

\displaystyle r^4(\cos 4\alpha + i\sin 4\alpha) = 1\,(\cos 0 + i\sin 0)\,,

where we have used de Moivre's formula on the left-hand side. In order that both sides are equal, they must have the same magnitude and the same argument to within a multiple of \displaystyle 2\pi, i.e.

\displaystyle \left\{\begin{align} r^{4} &= 1\,,\\[5pt] 4\alpha &= 0+2n\pi\,,\quad (\text{n is an arbitrary integer})\,\textrm{.} \end{align}\right.

This means that

\displaystyle \left\{\begin{align} r &= 1\,,\\[5pt] \alpha &= \frac{n\pi}{2}\,,\quad \text{(n is an arbitrary integer).} \end{align}\right.

The solutions are thus (in polar form)

\displaystyle z = 1\cdot\Bigl(\cos\frac{n\pi}{2} + i\sin\frac{n\pi}{2}\Bigr)\,,\quad\text{for }n=0,\ \pm 1,\ \pm 2,\ldots

but observe that the argument on the right-hand side essentially takes only four different values \displaystyle 0, \displaystyle \pi/2, \displaystyle \pi and \displaystyle 3\pi/2\,, because other values of \displaystyle n give some of these values plus/minus a multiple of \displaystyle 2\pi\,.

The equation's solutions are therefore

\displaystyle z=\left\{\begin{align} &1\cdot(\cos 0 + i\sin 0)\,,\\[5pt] &1\cdot(\cos (\pi/2) + i\sin (\pi/2))\,,\\[5pt] &1\cdot(\cos \pi + i\sin \pi)\,,\\[5pt] &1\cdot(\cos (3\pi/2) + i\sin (3\pi/2))\,, \end{align}\right. = \left\{ \begin{align} 1\,,&\\[5pt] i\,,&\\[5pt] -1\,,&\\[5pt] -i\,\textrm{.}& \end{align}\right.

Note: If we mark these solutions on the complex number plane, we see that they are corners in a regular quadrilateral.