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Lösung 3.1:2c

Aus Online Mathematik Brückenkurs 2

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K (Lösning 3.1:2c moved to Solution 3.1:2c: Robot: moved page)
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<center> [[Image:3_1_2c.gif]] </center>
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We start by expanding the quadratic in the numerator with the square rule:
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<math>\begin{align}\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}}\\
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&=\frac{4-4\sqrt{3}i-3}{1+i\sqrt{3}}\\
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&=\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}\end{align}</math>
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Then, we multiply top and bottom by the complex conjugate of the numerator:
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<math>\begin{align}\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})}\\
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&=\frac{1\cdot 1-1\cdot i\sqrt{3}-1\cdot 4\sqrt{3}i+ 4\sqrt{3}i\cdot i \sqrt{3}}{1^2-(i\sqrt{3})^2}\\
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&=\frac{1-i\sqrt{3}-4\sqrt{3}i+4(\sqrt{3})^2i^2}{1+(\sqrt{3})^2)}\\
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&=\frac{1-(\sqrt{3}+4\sqrt{3})i-4\cdot 3}{1+3}\\
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&=\frac{1-12-(1+4)\sqrt{3}i}{4}\\
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&=-\frac{11}{4}-\frac{5\sqrt{3}}{4}i.\end{align}</math>
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Version vom 10:07, 23. Sep. 2008

We start by expanding the quadratic in the numerator with the square rule:


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)211−1i3−143i+43ii3=1+(3)2)1−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“