Lösung 3.3:3b

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When we complete the square, we replace all
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<math>z</math>
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-terms in the second-degree expression with a quadratic term which contains
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<math>z</math>, according to the formula
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<math>z^{2}+az=\left( z+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
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In our case, we set
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<math>a=\text{3}i\text{ }</math>
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in order to complete the square:
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<math>\begin{align}
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& z^{2}+3iz-\frac{1}{4}=\left( z+\frac{3}{2}i \right)^{2}-\left( \frac{3}{2}i \right)^{2}-\frac{1}{4} \\
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& =\left( z+\frac{3}{2}i \right)^{2}-\frac{9}{4}\left( -1 \right)-\frac{1}{4} \\
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& =\left( z+\frac{3}{2}i \right)^{2}+2 \\
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\end{align}</math>

Version vom 10:02, 28. Okt. 2008


When we complete the square, we replace all \displaystyle z -terms in the second-degree expression with a quadratic term which contains \displaystyle z, according to the formula


\displaystyle z^{2}+az=\left( z+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}


In our case, we set \displaystyle a=\text{3}i\text{ } in order to complete the square:


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

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