Lösung 3.3:5a
Aus Online Mathematik Brückenkurs 2
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Version vom 10:48, 11. Mär. 2009
Even if the equation contains complex numbers as coefficients, we treat is as an ordinary second-degree equation and solve it by completing the square taking the square root.
We complete the square on the left-hand side,
| \displaystyle \begin{align}
(z-(1+i))^2-(1+i)^2+2i-1 &= 0\,,\\[5pt] (z-(1+i))^2-(1+2i+i^2)+2i-1&=0\,,\\[5pt] (z-(1+i))^2-1-2i+1+2i-1 &= 0\,,\\[5pt] (z-(1+i))^2-1 &= 0\,\textrm{.} \end{align} |
Now, we see that the equation has the solutions
| \displaystyle z-(1+i) = \pm 1\quad \Leftrightarrow \quad z=\left\{ \begin{align}
&2+i\,,\\ &i\,\textrm{.} \end{align}\right. |
We test the solutions,
\displaystyle \begin{align} z=2+i:\quad z^2-2(1+i)z+2i-1 &= (2+i)^2 - 2(1+i)(2+i)+2i-1\\[5pt] &= 4+4i+i^2-2(2+i+2i+i^2)+2i-1\\[5pt] &= 4+4i-1-4-6i+2+2i-1\\[5pt] &= 0\,,\\[10pt] z={}\rlap{i:}\phantom{2+i:}{}\quad z^2-2(1+i)z+2i-1 &= i^2-2(1+i)i+2i-1\\[5pt] &= -1-2(i+i^2)+2i-1\\[5pt] &= -1-2i+2+2i-1\\[5pt] &= 0\,\textrm{.} \end{align}
