Lösung 3.2:2b

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<center> [[Image:3_2_2b-1(2).gif]] </center>
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The inequality <math>0\leq \mathrm{Re}z \leq \mathrm{Im}z \leq 1</math> is actually several inequalities:
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<center> [[Image:3_2_2b-2(2).gif]] </center>
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<math>\begin{align}
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0 &\leq \mathrm{Re}z \leq 1,\\
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0 &\leq \mathrm{Im}z \leq 1,\end{align}
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\mathrm{Re}z \leq \mathrm{Im}z.
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</math>
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The first two inequalities in this list define the unit square in the complex number plane.
[[Image:3_2_2_b1.gif|center]]
[[Image:3_2_2_b1.gif|center]]
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The last inequality says that the real part of <math>z</math> should be less than or equal to the imaginary part of <math>z</math>, I.e. <math>z</math> should lie to the left of the line <math>y=x</math> if <math>x=\mathrm{Re} z</math> and <math>y = \mathrm{Im} z</math>.
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[[Image:3_2_2_b2.gif|center]]
[[Image:3_2_2_b2.gif|center]]
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All together, the inequalities define the region which the unit square and the half-plane have in common: a triangle with corner points at <math>0</math>, <math>i</math> and <math>1+i</math>.
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[[Image:3_2_2_b3.gif|center]]
[[Image:3_2_2_b3.gif|center]]
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Version vom 10:49, 3. Okt. 2008