Lösung 3.3:4d
Aus Online Mathematik Brückenkurs 2
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To avoid having <math>z</math> in the denominator, we multiply both sides of the equation by <math>z</math>, | To avoid having <math>z</math> in the denominator, we multiply both sides of the equation by <math>z</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>1+z^2=\frac{1}{2}\,z\,\textrm{.}</math>}} |
This multiplication could possibly introduce a spurious root if it turns out that the new equation has <math>z=0</math> as a root. The old equation, for understandable reasons, doesn't have <math>z=0</math> as a solution. | This multiplication could possibly introduce a spurious root if it turns out that the new equation has <math>z=0</math> as a root. The old equation, for understandable reasons, doesn't have <math>z=0</math> as a solution. | ||
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If we move the terms over to the left-hand side and complete the square, we get | If we move the terms over to the left-hand side and complete the square, we get | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
z^2 - \frac{1}{2}\,z + 1 &= 0\,,\\[5pt] | z^2 - \frac{1}{2}\,z + 1 &= 0\,,\\[5pt] | ||
\Bigl(z-\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 + 1 &= 0\,,\\[5pt] | \Bigl(z-\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 + 1 &= 0\,,\\[5pt] | ||
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This gives that the equation has solutions | This gives that the equation has solutions | ||
| - | {{ | + | {{Abgesetzte Formel||<math>z=\frac{1}{4}+i\,\frac{\sqrt{15}}{4}\quad</math> and <math>\quad z=\frac{1}{4}-i\,\frac{\sqrt{15}}{4}\,\textrm{.}</math>}} |
None of these solutions are equal to zero, so these are also solutions to the original equation. | None of these solutions are equal to zero, so these are also solutions to the original equation. | ||
Version vom 13:13, 10. Mär. 2009
To avoid having \displaystyle z in the denominator, we multiply both sides of the equation by \displaystyle z,
| \displaystyle 1+z^2=\frac{1}{2}\,z\,\textrm{.} |
This multiplication could possibly introduce a spurious root if it turns out that the new equation has \displaystyle z=0 as a root. The old equation, for understandable reasons, doesn't have \displaystyle z=0 as a solution.
If we move the terms over to the left-hand side and complete the square, we get
| \displaystyle \begin{align}
z^2 - \frac{1}{2}\,z + 1 &= 0\,,\\[5pt] \Bigl(z-\frac{1}{4}\Bigr)^2 - \Bigl(\frac{1}{4}\Bigr)^2 + 1 &= 0\,,\\[5pt] \Bigl(z-\frac{1}{4}\Bigr)^2 + \frac{15}{16} &= 0\,\textrm{.} \end{align} |
This gives that the equation has solutions
| \displaystyle z=\frac{1}{4}+i\,\frac{\sqrt{15}}{4}\quad and \displaystyle \quad z=\frac{1}{4}-i\,\frac{\sqrt{15}}{4}\,\textrm{.} |
None of these solutions are equal to zero, so these are also solutions to the original equation.
We substitute the solutions into the original equations to assure ourselves that we have calculated correctly.
\displaystyle \begin{align} z=\frac{1}{4}-i\,\frac{\sqrt{15}}{4}:\quad \text{LHS} &= \frac{1}{\dfrac{1}{4}-i\dfrac{\sqrt{15}}{14}} + \frac{1}{4} - i\frac{\sqrt{15}}{4}\\[5pt] &= \frac{\dfrac{1}{4}+i\dfrac{\sqrt{15}}{4}}{\Bigl(\dfrac{1}{4} - i\dfrac{\sqrt{15}}{4}\,\Bigr)\Bigl(\dfrac{1}{4} + i\,\dfrac{\sqrt{15}}{4}\Bigr)} + \frac{1}{4} - i\frac{\sqrt{15}}{4}\\[5pt] &= \frac{\dfrac{1}{4}+i\dfrac{\sqrt{15}}{4}}{\dfrac{1}{16}+\dfrac{15}{16}}+\frac{1}{4}-i\frac{\sqrt{15}}{4}\\[5pt] &= \frac{1}{4} + i\frac{\sqrt{15}}{4} + \frac{1}{4} - i\frac{\sqrt{15}}{4}\\[5pt] &= \frac{1}{2}\\[5pt] &= \text{RHS,}\\[10pt] z={}\rlap{\frac{1}{4}+i\frac{\sqrt{15}}{4}:}\phantom{\frac{1}{4}-i\,\frac{\sqrt{15}}{4}:}{}\quad \text{LHS} &= \frac{1}{\dfrac{1}{4}+i\dfrac{\sqrt{15}}{14}} + \frac{1}{4} + i\frac{\sqrt{15}}{4}\\[5pt] &= \frac{\dfrac{1}{4}-i\dfrac{\sqrt{15}}{4}}{\Bigl(\dfrac{1}{4}+i\dfrac{\sqrt{15}}{4}\Bigr)\Bigl(\dfrac{1}{4}-i\dfrac{\sqrt{15}}{4}\Bigr)} + \frac{1}{4} + i\frac{\sqrt{15}}{4}\\[5pt] &= \frac{\dfrac{1}{4}-i\dfrac{\sqrt{15}}{4}}{\dfrac{1}{16}+\dfrac{15}{16}} + \frac{1}{4} + i\frac{\sqrt{15}}{4}\\[5pt] &= \frac{1}{4} - i\frac{\sqrt{15}}{4} + \frac{1}{4} + i\frac{\sqrt{15}}{4}\\[5pt] &= \frac{1}{2}\\[5pt] &= \text{RHS.} \end{align}
