Lösung 3.3:4c

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We complete the square on the left-hand side:
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We complete the square on the left-hand side,
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{{Displayed math||<math>\begin{align}
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(z+1)^2-1^2+3 &= 0\,,\\[5pt]
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(z+1)^2+2 &= 0\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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Taking the root now gives <math>z+1=\pm i\sqrt{2}</math>, i.e. <math>z=-1+i\sqrt{2}</math> and <math>z=-1-i\sqrt{2}</math>.
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& \left( z+1 \right)^{\text{2}}-1^{2}+3=0 \\
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& \left( z+1 \right)^{\text{2}}+2=0 \\
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\end{align}</math>
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Taking the root now gives
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<math>z+1=\pm i\sqrt{2}</math>
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i.e.
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<math>z=-1+i\sqrt{2}</math>
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and
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<math>z=-1-i\sqrt{2}</math>.
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We test the solutions in the equation to ascertain that we have calculated correctly.
We test the solutions in the equation to ascertain that we have calculated correctly.
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<math>\begin{align}
 
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& z=-1+i\sqrt{2}:\quad z^{2}+2z+3=\left( -1+i\sqrt{2} \right)^{2}+2\left( -1+i\sqrt{2} \right)+3 \\
 
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& =\left( -1 \right)^{2}-2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2+2i\sqrt{2}+3 \\
 
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& =1-2\centerdot i\sqrt{2}-2-2+2i\sqrt{2}+3=0, \\
 
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\end{align}</math>
 
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<math>\begin{align}
<math>\begin{align}
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& z=-1-i\sqrt{2}:\quad z^{2}+2z+3=\left( -1-i\sqrt{2} \right)^{2}+2\left( -1-i\sqrt{2} \right)+3 \\
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z=-1+i\sqrt{2}:\quad z^2+2z+3
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& =\left( -1 \right)^{2}+2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2-2i\sqrt{2}+3 \\
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&= \bigl(-1+i\sqrt{2}\,\bigr)^2 + 2\bigl(-1+i\sqrt{2}\bigr) + 3\\[5pt]
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& =1+2\centerdot i\sqrt{2}-2-2-2\sqrt{2}i+3=0, \\
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&= (-1)^2 - 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 + 2i\sqrt{2} + 3\\[5pt]
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&= 1-2\cdot i\sqrt{2}-2-2+2i\sqrt{2}+3\\[5pt]
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&= 0,\\[10pt]
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z={}\rlap{-1-i\sqrt{2}:}\phantom{-1+i\sqrt{2}:}{}\quad z^2+2z+3
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&= \bigl(-1-i\sqrt{2}\,\bigr)^2 + 2\bigl(-1-i\sqrt{2}\,\bigr) + 3\\[5pt]
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&= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt]
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&= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt]
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&= 0,
\end{align}</math>
\end{align}</math>

Version vom 14:43, 30. Okt. 2008

We complete the square on the left-hand side,

\displaystyle \begin{align}

(z+1)^2-1^2+3 &= 0\,,\\[5pt] (z+1)^2+2 &= 0\,\textrm{.} \end{align}

Taking the root now gives \displaystyle z+1=\pm i\sqrt{2}, i.e. \displaystyle z=-1+i\sqrt{2} and \displaystyle z=-1-i\sqrt{2}.

We test the solutions in the equation to ascertain that we have calculated correctly.

\displaystyle \begin{align} z=-1+i\sqrt{2}:\quad z^2+2z+3 &= \bigl(-1+i\sqrt{2}\,\bigr)^2 + 2\bigl(-1+i\sqrt{2}\bigr) + 3\\[5pt] &= (-1)^2 - 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 + 2i\sqrt{2} + 3\\[5pt] &= 1-2\cdot i\sqrt{2}-2-2+2i\sqrt{2}+3\\[5pt] &= 0,\\[10pt] z={}\rlap{-1-i\sqrt{2}:}\phantom{-1+i\sqrt{2}:}{}\quad z^2+2z+3 &= \bigl(-1-i\sqrt{2}\,\bigr)^2 + 2\bigl(-1-i\sqrt{2}\,\bigr) + 3\\[5pt] &= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt] &= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt] &= 0, \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:4c“