Lösung 3.3:3b

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When we complete the square, we replace all <math>z</math>-terms in the second-degree expression with a quadratic term which contains <math>z</math>, according to the formula
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When we complete the square, we replace all
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{{Displayed math||<math>z^2+az = \Bigl(z+\frac{a}{2}\Bigr)^2 - \Bigl(\frac{a}{2}\Bigr)^2\,\textrm{.}</math>}}
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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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In our case, we set <math>a=3i</math> in order to complete the square,
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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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{{Displayed math||<math>\begin{align}
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z^2+3iz-\frac{1}{4}
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&= \Bigl(z+\frac{3}{2}\,i\Bigr)^2 - \Bigl(\frac{3}{2}\,i\Bigr)^2 - \frac{1}{4}\\[5pt]
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In our case, we set
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&= \Bigl(z+\frac{3}{2}\,i\Bigr)^2 - \frac{9}{4}\cdot (-1) - \frac{1}{4}\\[5pt]
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<math>a=\text{3}i\text{ }</math>
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&= \Bigl(z+\frac{3}{2}\,i\Bigr)^2+2\,\textrm{.}
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in order to complete the square:
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\end{align}</math>}}
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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>
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Version vom 13:33, 30. 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 = \Bigl(z+\frac{a}{2}\Bigr)^2 - \Bigl(\frac{a}{2}\Bigr)^2\,\textrm{.}

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

\displaystyle \begin{align}

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

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