Lösung 3.3:2e
Aus Online Mathematik Brückenkurs 2
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Version vom 10:45, 11. Mär. 2009
If we treat the expression \displaystyle w=\frac{z+i}{z-i} as an unknown, we have the equation
| \displaystyle w^2=-1\,\textrm{.} |
We know already that this equation has roots
| \displaystyle w=\left\{\begin{align}
-i\,,&\\[5pt] i\,,& \end{align}\right. |
so \displaystyle z should satisfy one of the equation's
| \displaystyle \frac{z+i}{z-i}=-i\quad or \displaystyle \quad\frac{z+i}{z-i}=i\,\textrm{.} |
We solve these equations one by one.
- \displaystyle (z+i)/(z-i)=-i:
- Multiply both sides by \displaystyle z-i,
| \displaystyle z+i=-i(z-i)\,\textrm{.} |
- Move all the \displaystyle z-terms over to the left-hand side and all the constants to the right-hand side,
| \displaystyle z+iz=-1-i\,\textrm{.} |
- This gives
| \displaystyle z = \frac{-1-i}{1+i} = \frac{-(1+i)}{1+i} = -1\,\textrm{.} |
- \displaystyle (z+i)/(z-i)=i:
- Multiply both sides by \displaystyle z-i,
| \displaystyle z+i=i(z-i)\,\textrm{.} |
- Move all the \displaystyle z-terms over to the left-hand side and all the constants to the right-hand side,
| \displaystyle z-iz=1-i\,\textrm{.} |
- This gives
| \displaystyle z = \frac{1-i}{1-i} = 1\,\textrm{.} |
The solutions are therefore \displaystyle z=-1 and \displaystyle z=1\,.
