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
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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{{Displayed math||<math>z^2+az = \Bigl(z+\frac{a}{2}\Bigr)^2 - \Bigl(\frac{a}{2}\Bigr)^2\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>z^2+az = \Bigl(z+\frac{a}{2}\Bigr)^2 - \Bigl(\frac{a}{2}\Bigr)^2\,\textrm{.}</math>}}
In our case, we set <math>a=3i</math> in order to complete the square,
In our case, we set <math>a=3i</math> in order to complete the square,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
z^2+3iz-\frac{1}{4}
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 - \Bigl(\frac{3}{2}\,i\Bigr)^2 - \frac{1}{4}\\[5pt]

Version vom 13:12, 10. Mär. 2009

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“