Lösung 3.3:3c

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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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,
-
{{Displayed math||<math>-\bigl(z^2+2iz-4z-1\bigr)\,,</math>}}
+
{{Abgesetzte Formel||<math>-\bigl(z^2+2iz-4z-1\bigr)\,,</math>}}
and collect together the first-degree terms,
and collect together the first-degree terms,
-
{{Displayed math||<math>-\bigl(z^2+(-4+2i)z-1\bigr)\,,</math>}}
+
{{Abgesetzte Formel||<math>-\bigl(z^2+(-4+2i)z-1\bigr)\,,</math>}}
we can then complete the square of the expression inside the outer bracket,
we can then complete the square of the expression inside the outer bracket,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
-\bigl(z^2+(-4+2i)z-1\bigr)
-\bigl(z^2+(-4+2i)z-1\bigr)
&= -\Bigl(\Bigl(z+\frac{-4+2i}{2}\Bigr)^2-\Bigl(\frac{-4+2i}{2}\Bigr)^2-1\Bigr)\\[5pt]
&= -\Bigl(\Bigl(z+\frac{-4+2i}{2}\Bigr)^2-\Bigl(\frac{-4+2i}{2}\Bigr)^2-1\Bigr)\\[5pt]

Version vom 13:12, 10. Mär. 2009

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

\displaystyle -\bigl(z^2+2iz-4z-1\bigr)\,,

and collect together the first-degree terms,

\displaystyle -\bigl(z^2+(-4+2i)z-1\bigr)\,,

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

\displaystyle \begin{align}

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

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