Aus Truth-Quark
(Unterschied zwischen Versionen)
Cdek (Diskussion | Beiträge) (→a)) |
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Zeile 6: | Zeile 6: | ||
<math>M_1\colon=\left\lbrace \mathbb C \ni z=x+ \mathrm i \cdot y \colon 0 \le x \right\rbrace</math> Graue Fläche<br /> | <math>M_1\colon=\left\lbrace \mathbb C \ni z=x+ \mathrm i \cdot y \colon 0 \le x \right\rbrace</math> Graue Fläche<br /> | ||
<math>M_2\colon=\left\lbrace \mathbb C \ni z=x+ \mathrm i \cdot y \colon x \le -y \le -x \right\rbrace</math> Rote Fläche<br /> | <math>M_2\colon=\left\lbrace \mathbb C \ni z=x+ \mathrm i \cdot y \colon x \le -y \le -x \right\rbrace</math> Rote Fläche<br /> | ||
- | <math>M_3\colon=\left\lbrace \mathbb C \ni z=x+ \mathrm i \cdot y \colon y = -x \right\rbrace</math>Schwarze Gerade<br /> | + | <math>M_3\colon=\left\lbrace \mathbb C \ni z=x+ \mathrm i \cdot y \colon y = -x \right\rbrace</math> Schwarze Gerade<br /> |
- | <math>M_4\colon=\left\lbrace \mathbb C \ni z=x+ \mathrm i \cdot y \colon (x-2)^2 + y^2 = 1^2 \right\rbrace</math>Grüner Kreisrand<br /> | + | <math>M_4\colon=\left\lbrace \mathbb C \ni z=x+ \mathrm i \cdot y \colon (x-2)^2 + y^2 = 1^2 \right\rbrace</math> Grüner Kreisrand<br /> |
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[[Kategorie:Übungen_Analysis_I]] | [[Kategorie:Übungen_Analysis_I]] |
Version vom 20:22, 7. Mai 2010
Aufgabe 1
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Aufgabe 2
a)
Graue Fläche
Rote Fläche
Schwarze Gerade
Grüner Kreisrand
Please install Java to use this page.