Lösung 3.1:2c
Aus Online Mathematik Brückenkurs 2
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| Zeile 1: | Zeile 1: | ||
We start by expanding the quadratic in the numerator, | We start by expanding the quadratic in the numerator, | ||
| - | {{ | + | {{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, | ||
| - | {{ | + | {{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,
| \displaystyle \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}}\\[5pt] &= \frac{4-4\sqrt{3}i-3}{1+i\sqrt{3}}\\[5pt] &= \frac{1-4\sqrt{3}i}{1+i\sqrt{3}}\,\textrm{.} \end{align} |
Then, we multiply top and bottom by the complex conjugate of the numerator,
| \displaystyle \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})}\\[5pt] &= \frac{1\cdot 1-1\cdot i\sqrt{3}-4\sqrt{3}i\cdot 1+ 4\sqrt{3}i\cdot i \sqrt{3}}{1^2-(i\sqrt{3})^2}\\[5pt] &= \frac{1-i\sqrt{3}-4\sqrt{3}i+4(\sqrt{3})^2i^2}{1+(\sqrt{3})^2}\\[5pt] &= \frac{1-(\sqrt{3}+4\sqrt{3})i-4\cdot 3}{1+3}\\[5pt] &= \frac{1-12-(1+4)\sqrt{3}i}{4}\\[5pt] &= -\frac{11}{4}-\frac{5\sqrt{3}}{4}i\,\textrm{.} \end{align} |
