Lösung 3.1:4b
Aus Online Mathematik Brückenkurs 2
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Version vom 10:32, 11. Mär. 2009
If we divide both sides by \displaystyle 2-i, we obtain \displaystyle z by itself on the left-hand side,
| \displaystyle z=\frac{3+2i}{2-i}\,\textrm{.} |
It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,
| \displaystyle \begin{align}
z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt] &= \frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt] &= \frac{6+3i+4i-2}{4+1}\\[5pt] &= \frac{4+7i}{5}\\[5pt] &= \frac{4}{5}+\frac{7}{5}\,i\,\textrm{.} \end{align} |
Also, we substitute \displaystyle z=\tfrac{4}{5}+\tfrac{7}{5}i into the original equation to assure ourselves that we have calculated correctly,
| \displaystyle \begin{align}
\text{LHS} &= (2-i)z\\[5pt] &= (2-i)\Bigl(\frac{4}{5}+\frac{7}{5}\,i\bigr)\\[5pt] &= 2\cdot\frac{4}{5} + 2\cdot\frac{7}{5}\,i - i\cdot\frac{4}{5} - i\cdot\frac{7}{5}\,i\\[5pt] &= \frac{8}{5} + \frac{14}{5}\,i - \frac{4}{5}\,i + \frac{7}{5}\\[5pt] &= \frac{8+7}{5} + \frac{14-4}{5}\,i\\[5pt] &= \frac{15}{5} + \frac{10}{5}\,i\\[5pt] &= 3+2i\\[5pt] &= \text{RHS.}\end{align} |
