Lösung 3.3:5b

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Complete the square of the left-hand side:
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Complete the square of the left-hand side,
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{{Displayed math||<math>\begin{align}
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\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(\frac{2-i}{2}\Bigr)^2+3-i &= 0\,,\\[5pt]
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\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(1-i+\frac{1}{4}i^2\Bigr)+3-i&=0\,,\\[5pt]
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\Bigl(z-\frac{2-i}{2}\Bigr)^2-1+i+\frac{1}{4}+3-i&=0\,,\\[5pt]
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\Bigl(z-\frac{2-i}{2}\Bigr)^2+\frac{9}{4}&=0\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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Taking the square root then gives that the solutions are
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& \left( z-\frac{2-i}{2} \right)^{2}-\left( \frac{2-i}{2} \right)^{2}+3-i=0 \\
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& \left( z-\frac{2-i}{2} \right)^{2}-\left( 1-i+\frac{1}{4}i^{2} \right)+3-i=0 \\
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& \left( z-\frac{2-i}{2} \right)^{2}-1+i+\frac{1}{4}+3-i=0 \\
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& \left( z-\frac{2-i}{2} \right)^{2}+\frac{9}{4}=0 \\
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& \\
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\end{align}</math>
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Taking the root then gives that the solutions are
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<math>z-\frac{2-i}{2}=\pm \frac{3}{2}i\quad \Leftrightarrow \quad z=\left\{ \begin{array}{*{35}l}
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1+i \\
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1-2i\text{ } \\
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\end{array} \right.</math>
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Finally, we substitute the solutions into the equation and check that it is satisfied:
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<math>\begin{align}
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& z=1+i:\quad z^{2}-\left( 2-i \right)z+\left( 3-i \right) \\
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& =\left( 1+i \right)^{2}-\left( 2-i \right)\left( 1+i \right)+3-i \\
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& =1+2i+i^{2}-\left( 2+2i-i-i^{2} \right)+3-i \\
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& =1+2i-1-2-i-1+3-i=0, \\
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\end{align}</math>
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{{Displayed math||<math>z-\frac{2-i}{2} = \pm\frac{3}{2}\,i\quad \Leftrightarrow \quad z=\left\{ \begin{align}
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&1+i\,,\\
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&1-2i\,\textrm{.}
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\end{align}\right.</math>}}
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Finally, we substitute the solutions into the equation and check that it is satisfied
<math>\begin{align}
<math>\begin{align}
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& z=1-2i:\quad z^{2}-\left( 2-i \right)z+\left( 3-i \right) \\
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z={}\rlap{1+i:}\phantom{1-2i:}{}\quad z^2-(2-i)z+(3-i)
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& =\left( 1-2i \right)^{2}-\left( 2-i \right)\left( 1-2i \right)+3-i \\
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&= (1+i)^2-(2-i)(1+i)+3-i\\[5pt]
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& =1-4i+4i^{2}-\left( 2-4i-i+2i^{2} \right)+3-i \\
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&= 1+2i+i^2-(2+2i-i-i^2)+3-i\\[5pt]
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& =1-4i-4-2+5i+2+3-i=0. \\
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&= 1+2i-1-2-i-1+3-i\\[5pt]
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&=0\,,\\[10pt]
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z=1-2i:\quad z^2-(2-i)z+(3-i)
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&= (1-2i)^2-(2-i)(1-2i)+3-i\\[5pt]
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&= 1-4i+4i^2-(2-4i-i+2i^2)+3-i\\[5pt]
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&= 1-4i-4-2+5i+2+3-i\\[5pt]
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&= 0\,\textrm{.}
\end{align}</math>
\end{align}</math>

Version vom 15:32, 30. Okt. 2008

Complete the square of the left-hand side,

\displaystyle \begin{align}

\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(\frac{2-i}{2}\Bigr)^2+3-i &= 0\,,\\[5pt] \Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(1-i+\frac{1}{4}i^2\Bigr)+3-i&=0\,,\\[5pt] \Bigl(z-\frac{2-i}{2}\Bigr)^2-1+i+\frac{1}{4}+3-i&=0\,,\\[5pt] \Bigl(z-\frac{2-i}{2}\Bigr)^2+\frac{9}{4}&=0\,\textrm{.} \end{align}

Taking the square root then gives that the solutions are

\displaystyle z-\frac{2-i}{2} = \pm\frac{3}{2}\,i\quad \Leftrightarrow \quad z=\left\{ \begin{align}

&1+i\,,\\ &1-2i\,\textrm{.} \end{align}\right.

Finally, we substitute the solutions into the equation and check that it is satisfied

\displaystyle \begin{align} z={}\rlap{1+i:}\phantom{1-2i:}{}\quad z^2-(2-i)z+(3-i) &= (1+i)^2-(2-i)(1+i)+3-i\\[5pt] &= 1+2i+i^2-(2+2i-i-i^2)+3-i\\[5pt] &= 1+2i-1-2-i-1+3-i\\[5pt] &=0\,,\\[10pt] z=1-2i:\quad z^2-(2-i)z+(3-i) &= (1-2i)^2-(2-i)(1-2i)+3-i\\[5pt] &= 1-4i+4i^2-(2-4i-i+2i^2)+3-i\\[5pt] &= 1-4i-4-2+5i+2+3-i\\[5pt] &= 0\,\textrm{.} \end{align}

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