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Lösung 3.3:2c

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
We write <math>z</math> and the right-hand side <math>-1-i</math> in polar form
We write <math>z</math> and the right-hand side <math>-1-i</math> in polar form
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
z &= r(\cos\alpha + i\sin\alpha)\,,\\[5pt]
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{.}
-1-i &= \sqrt{2}\Bigl(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\Bigr)\,\textrm{.}
Zeile 8: Zeile 8:
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>}}
+
{{Abgesetzte Formel||<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>}}
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}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
r^5 &= \sqrt{2}\,,\\[5pt]
r^5 &= \sqrt{2}\,,\\[5pt]
5\alpha &= \frac{5\pi}{4} + 2n\pi\,,\quad\text{(n is an arbitrary integer).}
5\alpha &= \frac{5\pi}{4} + 2n\pi\,,\quad\text{(n is an arbitrary integer).}
Zeile 21: Zeile 21:
This gives that
This gives that
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{{Displayed math||<math>\left\{\begin{align}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}
r &= \sqrt[5]{2} = \bigl(2^{1/2}\bigr)^{1/5} = 2^{1/10}\,,\\[5pt]
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).}
\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).}
Zeile 28: Zeile 28:
If we investigate the argument <math>\alpha</math> more closely, we see that it assumes essentially only five different values,
If we investigate the argument <math>\alpha</math> more closely, we see that it assumes essentially only five different values,
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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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{{Abgesetzte Formel||<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>}}
since these angle values then repeat to within a multiple of <math>2\pi</math>.
since these angle values then repeat to within a multiple of <math>2\pi</math>.
Zeile 34: Zeile 34:
In summary, the solutions are
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>}}
+
{{Abgesetzte Formel||<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>}}
for <math>n=0</math>, <math>1</math>, <math>2</math>, <math>3</math> and
for <math>n=0</math>, <math>1</math>, <math>2</math>, <math>3</math> and

Version vom 13:11, 10. Mär. 2009

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

z−1−i=r(cos+isin)=2cos45+isin45. 

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

r5(cos5+isin5)=2cos45+isin45. 

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

r55=2=45+2n(n is an arbitrary integer).

(The arguments 5 and 54 can differ by a multiple of 2 and still correspond to the same complex number.)

This gives that

r=52=21215=2110=5145+2n=4+52n(n is an arbitrary integer).

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

4, 4+52, 4+54, 4+56 and 4+58

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

In summary, the solutions are

z=2110cos4+52n+isin4+52n 

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

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