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Lösung 3.3:5a

Aus Online Mathematik Brückenkurs 2

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Even if the equation contains complex numbers as coefficients, we treat is as an ordinary second-degree equation and solve it by completing the square taking the root.
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<center> [[Image:3_3_5a.gif]] </center>
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We complete the square on the left-hand side:
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<math>\begin{align}
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& \left( z-\left( 1+i \right) \right)^{2}-\left( 1+i \right)^{2}+2i-1=0 \\
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& \left( z-\left( 1+i \right) \right)^{2}-\left( 1+2i+i^{2} \right)+2i-1=0 \\
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& \left( z-\left( 1+i \right) \right)^{2}-1-2i+1+2i-1=0 \\
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& \left( z-\left( 1+i \right) \right)^{2}-1=0 \\
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\end{align}</math>
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Now, we see that the equation has the solutions
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<math>z-\left( 1+i \right)=\pm 1\quad \Leftrightarrow \quad \left\{ \begin{array}{*{35}l}
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2+i \\
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i\text{ } \\
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\end{array} \right.</math>
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We test the solutions:
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<math>\begin{align}
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& z=2+i:\quad z^{2}-2\left( 1+i \right)z+2i-1 \\
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& =\left( 2+i \right)^{2}-2\left( 1+i \right)\left( 2+i \right)+2i-1 \\
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& =4+4i+i^{2}-2\left( 2+i+2i+i^{2} \right)+2i-1 \\
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& =4+4i-1-4-6i+2+2i-1=0 \\
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& \\
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\end{align}</math>
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<math>\begin{align}
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& z=i:\quad z^{2}-2\left( 1+i \right)z+2i-1 \\
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& =i^{2}-2\left( 1+i \right)i+2i-1 \\
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& =-1-2\left( i+i^{2} \right)+2i-1 \\
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& =-1-2i+2+2i-1=0 \\
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\end{align}</math>

Version vom 11:09, 25. Okt. 2008

Even if the equation contains complex numbers as coefficients, we treat is as an ordinary second-degree equation and solve it by completing the square taking the root.

We complete the square on the left-hand side:


z−1+i2−1+i2+2i−1=0z−1+i2−1+2i+i2+2i−1=0z−1+i2−1−2i+1+2i−1=0z−1+i2−1=0


Now, we see that the equation has the solutions


z−1+i=12+ii  


We test the solutions:


z=2+i:z2−21+iz+2i−1=2+i2−21+i2+i+2i−1=4+4i+i2−22+i+2i+i2+2i−1=4+4i−1−4−6i+2+2i−1=0


z=i:z2−21+iz+2i−1=i2−21+ii+2i−1=−1−2i+i2+2i−1=−1−2i+2+2i−1=0

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