Lösung 3.3:4d
Aus Online Mathematik Brückenkurs 2
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| - | To avoid having | + | To avoid having <math>z</math> in the denominator, we multiply both sides of the equation by <math>z</math>, |
| - | <math>z</math> | + | |
| - | in the denominator, we multiply both sides of the equation by | + | |
| - | <math>z</math> | + | |
| + | {{Displayed math||<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 | + | |
| - | <math>z=0 | + | |
| - | as a root. The old equation, for understandable reasons, have | + | |
| - | <math>z=0 | + | |
| - | as a solution. | + | |
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 | ||
| - | + | {{Displayed math||<math>\begin{align} | |
| - | <math>\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}</math>}} | |
| - | \end{align}</math> | + | |
| - | + | ||
This gives that the equation has solutions | This gives that the equation has solutions | ||
| - | + | {{Displayed math||<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>}} | |
| - | <math>z=\frac{1}{4}+i\frac{\sqrt{15}}{4}</math> | + | |
| - | and | + | |
| - | <math>z=\frac{1}{4}-i\frac{\sqrt{15}}{4}</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. | ||
We substitute the solutions into the original equations to assure ourselves that we have calculated correctly. | We substitute the solutions into the original equations to assure ourselves that we have calculated correctly. | ||
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| - | <math>\begin{align} | ||
| - | & z=\frac{1}{4}-i\frac{\sqrt{15}}{4}:\quad \text{LHS}=\frac{1}{z}+z= \\ | ||
| - | & =\frac{1}{\frac{1}{4}-i\frac{\sqrt{15}}{14}}+\frac{1}{4}-i\frac{\sqrt{15}}{4} \\ | ||
| - | & =\frac{\frac{1}{4}+i\frac{\sqrt{15}}{4}}{\left( \frac{1}{4}-i\frac{\sqrt{15}}{4} \right)\left( \frac{1}{4}+i\frac{\sqrt{15}}{4} \right)}+\frac{1}{4}-i\frac{\sqrt{15}}{4} \\ | ||
| - | & =\frac{\frac{1}{4}+i\frac{\sqrt{15}}{4}}{\frac{1}{16}+\frac{15}{16}}+\frac{1}{4}-i\frac{\sqrt{15}}{4} \\ | ||
| - | & =\frac{1}{4}+i\frac{\sqrt{15}}{4}+\frac{1}{4}-i\frac{\sqrt{15}}{4}=\frac{1}{2}=\text{RHS} \\ | ||
| - | \end{align}</math> | ||
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| - | |||
<math>\begin{align} | <math>\begin{align} | ||
| - | + | z=\frac{1}{4}-i\,\frac{\sqrt{15}}{4}:\quad \text{LHS} | |
| - | & =\frac{1}{\ | + | &= \frac{1}{\dfrac{1}{4}-i\dfrac{\sqrt{15}}{14}} + \frac{1}{4} - i\frac{\sqrt{15}}{4}\\[5pt] |
| - | & =\frac{\frac{1}{4}-i\frac{\sqrt{15}}{4}}{\ | + | &= \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{\ | + | &= \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}=\frac{1}{2}=\text{RHS} | + | &= \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}</math> | \end{align}</math> | ||
Version vom 15:07, 30. Okt. 2008
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}
