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Lösung 3.3:2d

Aus Online Mathematik Brückenkurs 2

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

w4=−4.

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

w−4=r(cos+isin)=4(cos+isin)

and the equation becomes

r4(cos4+isin4)=4(cos+isin).

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 2,

r44=4=+2n(n is an arbitrary integer), 

which gives us that

r=44=2=4+2n(n is an arbitrary integer).

For n=01, 2 and 3, the argument assumes the four different values

4, 43, 45and47

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.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:2d“