Lösung 3.3:2d

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If we use
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If we use <math>w=z-1</math> as a new unknown and move the term <math>4</math> over to the right-hand side, we have a binomial equation,
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<math>w=z-\text{1}</math>
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as a new unknown and move the term
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<math>\text{4}</math>
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over to the right-hand side, we have a binomial equation,
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<math>w^{4}=-4</math>
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{{Displayed math||<math>w^4=-4\,\textrm{.}</math>}}
We can solve this equation in the usual way by using polar form and de Moivre's formula. We have
We can solve this equation in the usual way by using polar form and de Moivre's formula. We have
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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w &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
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& w=r\left( \cos \alpha +i\sin \alpha \right) \\
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-4 &= 4(\cos\pi + i\sin\pi)\,,
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& -4=4\left( \cos \pi +i\sin \pi \right) \\
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\end{align}</math>}}
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\end{align}</math>
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and the equation becomes
and the equation becomes
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{{Displayed math||<math>r^4(\cos 4\alpha + i\sin 4\alpha) = 4(\cos\pi + i\sin\pi)\,\textrm{.}</math>}}
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<math>r^{4}\left( \cos 4\alpha +i\sin 4\alpha \right)=4\left( \cos \pi +i\sin \pi \right)</math>
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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 <math>2\pi</math>,
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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
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<math>2\pi </math>,
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<math>\left\{ \begin{array}{*{35}l}
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r^{4}=4 \\
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4\alpha =\pi +2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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{{Displayed math||<math>\left\{\begin{align}
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r^4 &= 4\,,\\[5pt]
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4\alpha &= \pi + 2n\pi\,,\quad\text{(n is an arbitrary integer),}
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\end{align} \right.</math>}}
which gives us that
which gives us that
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{{Displayed math||<math>\left\{\begin{align}
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r &= \sqrt[4]{4} = \sqrt{2}\,,\\[5pt]
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\alpha &= \frac{\pi}{4}+\frac{n\pi}{2}\,,\quad\text{(n is an arbitrary integer).}
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\end{align}\right.</math>}}
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<math>\left\{ \begin{array}{*{35}l}
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For <math>n=0<math>, <math>1</math>, <math>2</math> and <math>3</math>, the argument <math>\alpha</math> assumes the four different values
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r=\sqrt[4]{2}=\sqrt{2} \\
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\alpha =\frac{\pi }{4}+\frac{n\pi }{2}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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{{Displayed math||<math>\frac{\pi}{4}</math>, <math>\quad\frac{3\pi}{4}</math>, <math>\quad\frac{5\pi}{4}\quad</math>and<math>\quad\frac{7\pi}{4}\,,</math>}}
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for
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and for other values of <math>n</math> we obtain values of <math>\alpha</math> which are equal to those above, apart from multiples of <math>2\pi</math>. Thus, we have four solutions,
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<math>n=0,\ 1,\ 2</math>
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and
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<math>3</math>, the argument
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<math>\alpha </math>
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assumes the four different values
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<math>\frac{\pi }{4},\ \frac{3\pi }{4},\ \frac{5\pi }{4}</math>
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and
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<math>\frac{7\pi }{4}</math>
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and for different values of
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<math>n</math>
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we obtain values of
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<math>\alpha </math>
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which are equal to those above, apart from multiples of
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<math>2\pi </math>. Thus, we have four solutions,
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<math>w=\left\{ \begin{array}{*{35}l}
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\sqrt{2}\left( \cos {\pi }/{4}\;+i\sin {\pi }/{4}\; \right) \\
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\sqrt{2}\left( \cos {3\pi }/{4}\;+i\sin 3{\pi }/{4}\; \right) \\
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\sqrt{2}\left( \cos {5\pi }/{4}\;+i\sin 5{\pi }/{4}\; \right) \\
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\sqrt{2}\left( \cos 7{\pi }/{4}\;+i\sin 7{\pi }/{4}\; \right) \\
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\end{array} \right.=\left\{ \begin{array}{*{35}l}
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1+i \\
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-1+i \\
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-1-i \\
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1-i \\
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\end{array} \right.</math>
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{{Displayed math||<math>w=\left\{\begin{align}
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&\sqrt{2}\Bigl(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\Bigr)\\[5pt]
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&\sqrt{2}\Bigl(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Bigr)\\[5pt]
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&\sqrt{2}\Bigl(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\Bigr)\\[5pt]
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&\sqrt{2}\Bigl(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}\Bigr)
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\end{align}\right.
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=
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\left\{\begin{align}
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1+i\,,&\\[5pt]
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-1+i\,,&\\[5pt]
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-1-i\,,&\\[5pt]
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1-i\,\textrm{,}
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\end{align}\right.</math>}}
and the original variable z is
and the original variable z is
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{{Displayed math||<math>z=\left\{\begin{align}
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<math>z=\left\{ \begin{array}{*{35}l}
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&2+i\,,\\[5pt]
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2+i \\
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&i\,,\\[5pt]
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i \\
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&-i\,,\\[5pt]
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-i \\
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&2-i\,\textrm{.}
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2-i \\
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\end{align}\right.</math>}}
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\end{array} \right.</math>
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Version vom 12:58, 30. Okt. 2008

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.

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