Lösung 3.3:4a
Aus Online Mathematik Brückenkurs 2
This is a typical binomial equation which we solve in polar form.
We write
| \displaystyle \begin{align}
z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt] i &= 1\,\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,, \end{align} |
and, on using de Moivre's formula, the equation becomes
| \displaystyle r^2(\cos 2\alpha + i\sin 2\alpha) = 1\Bigl(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\Bigr)\,\textrm{.} |
Both sides are equal when
| \displaystyle \left\{\begin{align}
r^2 &= 1\,,\\[5pt] 2\alpha &= \frac{\pi}{2}+2n\pi\,,\quad\text{(n is an arbitrary integer),} \end{align}\right. |
which gives that
| \displaystyle \left\{\begin{align}
r &= 1\,,\\[5pt] \alpha &= \frac{\pi}{4} + n\pi\,,\quad\text{(n is an arbitrary integer).} \end{align}\right. |
When \displaystyle n=0 and \displaystyle n=1, we get two different arguments for \displaystyle \alpha, whilst different values of \displaystyle n only give these arguments plus/minus a multiple of \displaystyle 2\pi.
The solutions to the equation are
| \displaystyle z=\left\{\begin{align}
&1\cdot\Bigl(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\Bigr)\\[5pt] &1\cdot\Bigl(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Bigr) \end{align}\right. = \left\{\begin{align} &\frac{1+i}{\sqrt{2}}\,,\\[5pt] &-\frac{1+i}{\sqrt{2}}\,\textrm{.} \end{align}\right. |
