Lösung 3.3:2c

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We write
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We write <math>z</math> and the right-hand side <math>-1-i</math> in polar form
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<math>z\text{ }</math>
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and the right-hand side
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<math>\text{-1-}i</math>
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in polar form
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<math>\begin{align}
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& z=r\left( \cos \alpha +i\sin \alpha \right) \\
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& \text{-1-}i=\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right) \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
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-1-i &= \sqrt{2}\Bigl(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\Bigr)\,\textrm{.}
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\end{align}</math>}}
Using de Moivre's formula, the equation can now be written as
Using de Moivre's formula, the equation can now be written as
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{{Displayed math||<math>r^5(\cos 5\alpha + i\sin 5\alpha) = \sqrt{2}\Bigl(\cos \frac{5\pi}{4} + i\sin\frac{5\pi}{4}\Bigr)\,\textrm{.}</math>}}
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<math>r^{5}\left( \cos 5\alpha +i\sin 5\alpha \right)=\sqrt{2}\left( \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4} \right)</math>
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If we identify the magnitude and argument on both sides, we get
If we identify the magnitude and argument on both sides, we get
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{{Displayed math||<math>\left\{\begin{align}
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r^5 &= \sqrt{2}\,,\\[5pt]
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5\alpha &= \frac{5\pi}{4} + 2n\pi\,,\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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(The arguments <math>5\alpha</math> and <math>5\pi/4</math> can differ by a multiple of <math>2\pi</math> and still correspond to the same complex number.)
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r^{5}=\sqrt{2} \\
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5\alpha =\frac{5\pi }{4}+2n\pi \quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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(The arguments
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<math>5\alpha </math>
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and
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<math>\frac{5\pi }{4}</math>
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can differ by a multiple of
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<math>2\pi </math>
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and still correspond to the same complex number.)
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This gives that
This gives that
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{{Displayed math||<math>\left\{\begin{align}
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r &= \sqrt[5]{2} = \bigl(2^{1/2}\bigr)^{1/5} = 2^{1/10}\,,\\[5pt]
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\alpha &= \frac{1}{5}\Bigl(\frac{5\pi}{4}+2n\pi\Bigr) = \frac{\pi}{4} + \frac{2n\pi}{5}\,,\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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If we investigate the argument <math>\alpha</math> more closely, we see that it assumes essentially only five different values,
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r=\sqrt[5]{2}=\left( 2^{{1}/{2}\;} \right)^{{1}/{5}\;}=2^{{1}/{10}\;} \\
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\alpha =\frac{1}{5}\left( \frac{5\pi }{4}+2n\pi \right)=\frac{\pi }{4}+\frac{2n\pi }{5}\quad \left( n\text{ an arbitrary integer} \right)\text{ } \\
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\end{array} \right.</math>
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If we investigate the argument
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<math>\alpha </math>
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more closely, we see that it assumes essentially only five different values,
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<math>\frac{\pi }{4},\ \frac{\pi }{4}+\frac{2\pi }{5},\ \frac{\pi }{4}+\frac{4\pi }{5},\ \frac{\pi }{4}+\frac{6\pi }{5}</math>
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and
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<math>\ \frac{\pi }{4}+\frac{8\pi }{5}</math>
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since these angle values then repeat to within a multiple of
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<math>2\pi </math>.
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In summary, the roots of the equation are
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<math>z=2^{{1}/{10}\;}\left( \cos \left( \frac{\pi }{4}+\frac{2n\pi }{5} \right)+i\sin \left( \frac{\pi }{4}+\frac{2n\pi }{5} \right) \right)</math>
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{{Displayed math||<math>\frac{\pi}{4}</math>, <math>\quad\frac{\pi}{4}+\frac{2\pi}{5}</math>, <math>\quad\frac{\pi}{4}+\frac{4\pi}{5}</math>, <math>\quad\frac{\pi}{4}+\frac{6\pi}{5}\quad</math> and <math>\quad\frac{\pi}{4}+\frac{8\pi}{5}</math>}}
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for
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since these angle values then repeat to within a multiple of <math>2\pi</math>.
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<math>n=0,\ 1,\ 2,\ 3</math>
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and
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<math>4</math>
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In summary, the solutions are
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{{Displayed math||<math>z = 2^{1/10}\,\Bigl(\cos\Bigl(\frac{\pi}{4}+\frac{2n\pi}{5}\Bigr) + i\sin\Bigl(\frac{\pi}{4}+\frac{2n\pi}{5}\Bigr)\Bigr)\,,</math>}}
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for <math>n=0</math>, <math>1</math>, <math>2</math>, <math>3</math> and
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<math>4</math>.
[[Image:3_3_2_c.gif|center]]
[[Image:3_3_2_c.gif|center]]

Version vom 12:34, 30. Okt. 2008

We write \displaystyle z and the right-hand side \displaystyle -1-i in polar form

\displaystyle \begin{align}

z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt] -1-i &= \sqrt{2}\Bigl(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\Bigr)\,\textrm{.} \end{align}

Using de Moivre's formula, the equation can now be written as

\displaystyle r^5(\cos 5\alpha + i\sin 5\alpha) = \sqrt{2}\Bigl(\cos \frac{5\pi}{4} + i\sin\frac{5\pi}{4}\Bigr)\,\textrm{.}

If we identify the magnitude and argument on both sides, we get

\displaystyle \left\{\begin{align}

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

(The arguments \displaystyle 5\alpha and \displaystyle 5\pi/4 can differ by a multiple of \displaystyle 2\pi and still correspond to the same complex number.)

This gives that

\displaystyle \left\{\begin{align}

r &= \sqrt[5]{2} = \bigl(2^{1/2}\bigr)^{1/5} = 2^{1/10}\,,\\[5pt] \alpha &= \frac{1}{5}\Bigl(\frac{5\pi}{4}+2n\pi\Bigr) = \frac{\pi}{4} + \frac{2n\pi}{5}\,,\quad\text{(n is an arbitrary integer).} \end{align}\right.

If we investigate the argument \displaystyle \alpha more closely, we see that it assumes essentially only five different values,

\displaystyle \frac{\pi}{4}, \displaystyle \quad\frac{\pi}{4}+\frac{2\pi}{5}, \displaystyle \quad\frac{\pi}{4}+\frac{4\pi}{5}, \displaystyle \quad\frac{\pi}{4}+\frac{6\pi}{5}\quad and \displaystyle \quad\frac{\pi}{4}+\frac{8\pi}{5}

since these angle values then repeat to within a multiple of \displaystyle 2\pi.

In summary, the solutions are

\displaystyle z = 2^{1/10}\,\Bigl(\cos\Bigl(\frac{\pi}{4}+\frac{2n\pi}{5}\Bigr) + i\sin\Bigl(\frac{\pi}{4}+\frac{2n\pi}{5}\Bigr)\Bigr)\,,

for \displaystyle n=0, \displaystyle 1, \displaystyle 2, \displaystyle 3 and \displaystyle 4.

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