Aus Online Mathematik Brückenkurs 2

If we use \displaystyle w=z-1 as a new unknown and move the term \displaystyle 4 over to the right-hand side, we have a binomial equation,

\displaystyle w^4=-4\,\textrm{.}

We can solve this equation in the usual way by using polar form and de Moivre's formula. We have

\displaystyle \begin{align} w &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt] -4 &= 4(\cos\pi + i\sin\pi)\,, \end{align}

and the equation becomes

\displaystyle r^4(\cos 4\alpha + i\sin 4\alpha) = 4(\cos\pi + i\sin\pi)\,\textrm{.}

The only way that both sides can be equal is if the magnitudes agree and the arguments do not differ by anything other than a multiple of \displaystyle 2\pi,

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

which gives us that

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

For \displaystyle n=0$,$ 1$$, \displaystyle 2 and \displaystyle 3, the argument \displaystyle \alpha assumes the four different values

\displaystyle \frac{\pi}{4}, \displaystyle \quad\frac{3\pi}{4}, \displaystyle \quad\frac{5\pi}{4}\quadand\displaystyle \quad\frac{7\pi}{4}\,,

and for other values of \displaystyle n we obtain values of \displaystyle \alpha which are equal to those above, apart from multiples of \displaystyle 2\pi. Thus, we have four solutions,

\displaystyle w=\left\{\begin{align} &\sqrt{2}\Bigl(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\Bigr)\\[5pt] &\sqrt{2}\Bigl(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Bigr)\\[5pt] &\sqrt{2}\Bigl(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\Bigr)\\[5pt] &\sqrt{2}\Bigl(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}\Bigr) \end{align}\right. = \left\{\begin{align} 1+i\,,&\\[5pt] -1+i\,,&\\[5pt] -1-i\,,&\\[5pt] 1-i\,\textrm{,} \end{align}\right.

and the original variable z is

\displaystyle z=\left\{\begin{align} &2+i\,,\\[5pt] &i\,,\\[5pt] &-i\,,\\[5pt] &2-i\,\textrm{.} \end{align}\right.