Lösung 3.3:4c
Aus Online Mathematik Brückenkurs 2
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&= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 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] | &= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt] | ||
| - | &= 0. | + | &= 0\,\textrm{.} |
\end{align}</math> | \end{align}</math> | ||
Version vom 14:44, 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\,\textrm{.} \end{align}
