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Lösung 3.1:2c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
We start by expanding the quadratic in the numerator,
We start by expanding the quadratic in the numerator,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{(2-i\sqrt{3})^2}{1+i\sqrt{3}}
\frac{(2-i\sqrt{3})^2}{1+i\sqrt{3}}
&= \frac{2^2-2\cdot 2\cdot i\sqrt{3}+(i\sqrt{3})^2}{1+i\sqrt{3}}\\[5pt]
&= \frac{2^2-2\cdot 2\cdot i\sqrt{3}+(i\sqrt{3})^2}{1+i\sqrt{3}}\\[5pt]
Zeile 10: Zeile 10:
Then, we multiply top and bottom by the complex conjugate of the numerator,
Then, we multiply top and bottom by the complex conjugate of the numerator,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}
\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}
&= \frac{(1-4\sqrt{3}i)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}\\[5pt]
&= \frac{(1-4\sqrt{3}i)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}\\[5pt]

Version vom 13:06, 10. Mär. 2009

We start by expanding the quadratic in the numerator,

1+i3(2−i3)2=1+i322−22i3+(i3)2=1+i34−43i−3=1+i31−43i.

Then, we multiply top and bottom by the complex conjugate of the numerator,

1+i31−43i=(1+i3)(1−i3)(1−43i)(1−i3)=12−(i3)211−1i3−43i1+43ii3=1+(3)21−i3−43i+4(3)2i2=1+31−(3+43)i−43=41−12−(1+4)3i=−411−453i.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:2c“