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,
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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{{Displayed math||<math>z=\frac{3+2i}{2-i}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>z=\frac{3+2i}{2-i}\,\textrm{.}</math>}}
It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the denominator,
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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{{Abgesetzte Formel||<math>\begin{align}
z &= \frac{(3+2i)(2+i)}{(2-i)(2+i)}\\[5pt]
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{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\[5pt]
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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,
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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{{Abgesetzte Formel||<math>\begin{align}
\text{LHS}
\text{LHS}
&= (2-i)z\\[5pt]
&= (2-i)z\\[5pt]

Version vom 13:06, 10. 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}

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