Lösung 3.1:2a
Aus Online Mathematik Brückenkurs 2
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Version vom 10:30, 11. Mär. 2009
A quotient of two complex numbers is calculated by multiplying the top and bottom of the fraction by the complex conjugate of the denominator,
| \displaystyle \frac{3-2i}{1+i} = \frac{3-2i}{1+i}\cdot\frac{1-i}{1-i}\,\textrm{.} |
Then, the formula for the difference of two squares gives that the new denominator is a real number,
| \displaystyle \begin{align}
\frac{3-2i}{1+i}\cdot\frac{1-i}{1-i} &= \frac{(3-2i)(1-i)}{(1+i)(1-i)}\\[5pt] &= \frac{(3-2i)(1-i)}{1^2-i^2}\\[5pt] &= \frac{(3-2i)(1-i)}{1+1}\\[5pt] &= \frac{(3-2i)(1-i)}{2}\,\textrm{.} \end{align} |
All that remains is to multiply together what is in the numerator,
| \displaystyle \begin{align}
\frac{(3-2i)(1-i)}{2} &= \frac{3\cdot 1 -3\cdot i - 2i\cdot 1 - 2i\cdot(-i)}{2}\\[5pt] &= \frac{3-3i-2i+2i^2}{2}\\[5pt] &= \frac{3-(3+2)i+2\cdot (-1)}{2}\\[5pt] &= \frac{1-5i}{2}\\[5pt] &= \frac{1}{2}-\frac{5}{2}\,i\,\textrm{.} \end{align} |
