Lösung 3.1:4b

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If we divide both sides by <math>2-i</math>, we obtain <math>z</math> by itself on the left-hand side,
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If we divide both sides by 2-i, we obtain z by itself on the left-hand side:
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{{Displayed math||<math>z=\frac{3+2i}{2-i}\,\textrm{.}</math>}}
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<math>z=\frac{3+2i}{2-i}.</math>
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It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,
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{{Displayed math||<math>\begin{align}
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z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt]
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&= \frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt]
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&= \frac{6+3i+4i-2}{4+1}\\[5pt]
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&= \frac{4+7i}{5}\\[5pt]
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&= \frac{4}{5}+\frac{7}{5}\,i\,\textrm{.}
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\end{align}</math>}}
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It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the numerator:
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Also, we substitute <math>z=\tfrac{4}{5}+\tfrac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}z&= \frac{(3+2i)(2+i)}{(2-i)(2+i)}=\frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\
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\text{LHS}
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&=\frac{6+3i+4i-2}{4+1}=\frac{4+7i}{5}=\frac{4}{5}+\frac{7}{5}i\end{align}</math>
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&= (2-i)z\\[5pt]
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&= (2-i)\Bigl(\frac{4}{5}+\frac{7}{5}\,i\bigr)\\[5pt]
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&= 2\cdot\frac{4}{5} + 2\cdot\frac{7}{5}\,i - i\cdot\frac{4}{5} - i\cdot\frac{7}{5}\,i\\[5pt]
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Also, we substitute <math>z=\frac{4}{5}+\frac{7}{5}i</math> into the original equation to assure ourselves that we have calculated correctly:
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&= \frac{8}{5} + \frac{14}{5}\,i - \frac{4}{5}\,i + \frac{7}{5}\\[5pt]
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&= \frac{8+7}{5} + \frac{14-4}{5}\,i\\[5pt]
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&= \frac{15}{5} + \frac{10}{5}\,i\\[5pt]
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<math>\begin{align}LHS &= (2-i)z=(2-i)(\frac{4}{5}+\frac{7}{5}i)\\
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&= 3+2i\\[5pt]
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&=2\cdot\frac{4}{5}-i\cdot\frac{4}{5}+2\cdot \frac{7}{5}i -i\cdot \frac{7}{5}i\\
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&= \text{RHS.}\end{align}</math>}}
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&=\frac{8}{5}-\frac{4}{5}i+\frac{14}{5}i+\frac{7}{5} = \frac{8+7}{5}+\frac{14-4}{5}i\\
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&=\frac{15}{5}+\frac{10}{5}i=3+2i=RHS.\end{align}</math>
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Version vom 07:27, 30. Okt. 2008

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}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:4b“