Lösung 3.3:3c

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If we take the minus sign out in front of the whole expression,
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<center> [[Image:3_3_3c.gif]] </center>
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<math>-\left( z^{2}+2iz-4z-1 \right)</math>
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and collect together the first-degree terms,
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<math>-\left( z^{2}+\left( -4+2i \right)z-1 \right)</math>
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we can then complete the square of the expression inside the outer bracket
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<math>\begin{align}
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& -\left( z^{2}+\left( -4+2i \right)z-1 \right)=-\left( \left( z+\frac{-4+2i}{2} \right)^{2}-\left( \frac{-4+2i}{2} \right)^{2}-1 \right) \\
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& =-\left( \left( z-2+i \right)^{2}-\left( -2+i \right)^{2}-1 \right) \\
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& =-\left( \left( z-2+i \right)^{2}-\left( -2 \right)^{2}+4i-i^{2}-1 \right) \\
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& =-\left( \left( z-2+i \right)^{2}-4+4i+1-1 \right) \\
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& =-\left( \left( z-2+i \right)^{2}-4+4i \right) \\
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& =-\left( z-2+i \right)^{2}+4-4i. \\
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\end{align}</math>

Version vom 08:47, 25. Okt. 2008

If we take the minus sign out in front of the whole expression,


\displaystyle -\left( z^{2}+2iz-4z-1 \right)


and collect together the first-degree terms,


\displaystyle -\left( z^{2}+\left( -4+2i \right)z-1 \right)


we can then complete the square of the expression inside the outer bracket


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

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