Lösung 3.3:5b

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Complete the square of the left-hand side:
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<center> [[Image:3_3_5b.gif]] </center>
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<math>\begin{align}
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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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<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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& =\left( 1-2i \right)^{2}-\left( 2-i \right)\left( 1-2i \right)+3-i \\
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& =1-4i+4i^{2}-\left( 2-4i-i+2i^{2} \right)+3-i \\
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& =1-4i-4-2+5i+2+3-i=0. \\
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\end{align}</math>

Version vom 11:30, 25. Okt. 2008

Complete the square of the left-hand side:


\displaystyle \begin{align} & \left( z-\frac{2-i}{2} \right)^{2}-\left( \frac{2-i}{2} \right)^{2}+3-i=0 \\ & \left( z-\frac{2-i}{2} \right)^{2}-\left( 1-i+\frac{1}{4}i^{2} \right)+3-i=0 \\ & \left( z-\frac{2-i}{2} \right)^{2}-1+i+\frac{1}{4}+3-i=0 \\ & \left( z-\frac{2-i}{2} \right)^{2}+\frac{9}{4}=0 \\ & \\ \end{align}


Taking the root then gives that the solutions are


\displaystyle z-\frac{2-i}{2}=\pm \frac{3}{2}i\quad \Leftrightarrow \quad z=\left\{ \begin{array}{*{35}l} 1+i \\ 1-2i\text{ } \\ \end{array} \right.


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


\displaystyle \begin{align} & z=1+i:\quad z^{2}-\left( 2-i \right)z+\left( 3-i \right) \\ & =\left( 1+i \right)^{2}-\left( 2-i \right)\left( 1+i \right)+3-i \\ & =1+2i+i^{2}-\left( 2+2i-i-i^{2} \right)+3-i \\ & =1+2i-1-2-i-1+3-i=0, \\ \end{align}



\displaystyle \begin{align} & z=1-2i:\quad z^{2}-\left( 2-i \right)z+\left( 3-i \right) \\ & =\left( 1-2i \right)^{2}-\left( 2-i \right)\left( 1-2i \right)+3-i \\ & =1-4i+4i^{2}-\left( 2-4i-i+2i^{2} \right)+3-i \\ & =1-4i-4-2+5i+2+3-i=0. \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:5b“