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Lösung 3.3:3c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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If we take the minus sign out in front of the whole expression,
If we take the minus sign out in front of the whole expression,
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{{Displayed math||<math>-\bigl(z^2+2iz-4z-1\bigr)\,,</math>}}
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<math>-\left( z^{2}+2iz-4z-1 \right)</math>
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and collect together the first-degree terms,
and collect together the first-degree terms,
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{{Displayed math||<math>-\bigl(z^2+(-4+2i)z-1\bigr)\,,</math>}}
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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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we can then complete the square of the expression inside the outer bracket
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<math>\begin{align}
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{{Displayed math||<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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-\bigl(z^2+(-4+2i)z-1\bigr)
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& =-\left( \left( z-2+i \right)^{2}-\left( -2+i \right)^{2}-1 \right) \\
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&= -\Bigl(\Bigl(z+\frac{-4+2i}{2}\Bigr)^2-\Bigl(\frac{-4+2i}{2}\Bigr)^2-1\Bigr)\\[5pt]
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& =-\left( \left( z-2+i \right)^{2}-\left( -2 \right)^{2}+4i-i^{2}-1 \right) \\
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&= -\bigl((z-2+i)^2-(-2+i)^2-1\bigr)\\[5pt]
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& =-\left( \left( z-2+i \right)^{2}-4+4i+1-1 \right) \\
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&= -\bigl((z-2+i)^2-(-2)^2+4i-i^2-1\bigr)\\[5pt]
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& =-\left( \left( z-2+i \right)^{2}-4+4i \right) \\
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&= -\bigl((z-2+i)^2-4+4i+1-1\bigr)\\[5pt]
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& =-\left( z-2+i \right)^{2}+4-4i. \\
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&= -\bigl((z-2+i)^2-4+4i\bigr)\\[5pt]
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\end{align}</math>
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&= -(z-2+i)^2+4-4i\,\textrm{.}
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\end{align}</math>}}

Version vom 13:45, 30. Okt. 2008

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

−z2+2iz−4z−1 

and collect together the first-degree terms,

−z2+(−4+2i)z−1 

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

−z2+(−4+2i)z−1=−z+2−4+2i2−2−4+2i2−1=−(z−2+i)2−(−2+i)2−1=−(z−2+i)2−(−2)2+4i−i2−1=−(z−2+i)2−4+4i+1−1=−(z−2+i)2−4+4i=−(z−2+i)2+4−4i.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:3c“