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Universität/Hochschule J Komplexe Zahlen Betrag Rechnung
LukasNiessen
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Hallo!

Ich verstehe folgenden Schritt in einer Rechnung nicht:

Sei $z \in \IC$ und $x \in \IC-\{0\}$.
Weiter sei $z \bar{z} = |z|^2 = 1$ und $z = \frac{x}{\bar{x}}$.

Es heißt nun, man dürfe annehmen, dass $x \bar{x} = |x|^2 = 1$ gilt.

Das sehe ich aber nicht, wieso muss das gelten?

Danke!


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Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄
\(\endgroup\)


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Diophant
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Hallo,

hier ist mir ein Denkfehler passiert, sorry.


Gruß, Diophant


[Verschoben aus Forum 'Analysis' in Forum 'Komplexe Zahlen' von Diophant]



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Kezer
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Ohne Kontext ist das schwer zu sagen. In dem Argument scheint das Skalieren von $x$ keine Auswirkung zu haben, man ersetzt dort $x$ durch $\frac{x}{|x|}$.

[Die Antwort wurde vor Beitrag No.1 begonnen.]


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The difference between the novice and the master is that the master has failed more times than the novice has tried. ~ Koro-Sensei



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Triceratops
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Wenn man sowas schreibt wie "wir dürfen annehmen, dass ...", dann bezieht sich das immer auf eine Behauptung, die zu zeigen ist. Hier hast du diese aber nicht angegeben. Insofern ist die Frage derzeit unvollständig.



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LukasNiessen
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Danke euch!

Ich dachte der Kontext wäre irrelevant, da die Folgerung allgemein gültig wäre.

Hier ist der Kontext:
Man soll Theorem 90 von Hilbert (hier) bezüglich $\IC / \IR$ untersuchen.

---

Man sieht schnell, dass die Erweiterung zyklisch von der Ordnung 2 ist und die Galois-Gruppe von der komplexen Konjugation erzeugt wird.
Außerdem gilt für die Norm:

$N_{\IC / \IR}(z) = z \bar{z} = |z|^2$ wobei $z \in \IC$.

Wenn nun $N_{\IC / \IR}(z) = 1$ gilt, dann folgt mit Hilberts Satz:

Es gibt $x \in \IC^*: z = \frac{x}{\bar{x}}$.

Hier wird nun gesagt: Man dürfe $x \bar{x} = |x|^2 = 1$ annehmen.
Gefolgert wird dann:

$z = x^2$, also, dass x eine Quadratwurzel von z ist.

---

Ich verstehe alles, außer, dass man $x \bar{x} = |x|^2 = 1$ annehmen darf.

Danke!


-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄
\(\endgroup\)


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LukasNiessen
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Dabei seit: 30.09.2019
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Okay, ich glaube ich hab's!

Man darf x auf die Länge 1 skalieren, denn es geht lediglich um die Existenz eines x ungleich 0, welches $z = \frac{x}{\bar{x}}$ erfüllt.

Und angenommen, ein solches x erfüllt obiges, dann tut es auch die auf Länge 1 skalierte "Version" $x'$, denn:

$\frac{x}{\bar{x}}\bar{(\frac{x}{\bar{x}})}^{-1} = \frac{x}{\bar{x}} \cdot (\bar{x} \frac{1}{|x|})^{-1} = \frac{x}{\bar{x}}$

Ich sehe gerade, obige Formel sieht unleserlich aus.

Ich habe auf jeden Fall nur $\frac{x'}{\bar{x'}}$ berechnet und gesehen, dass $\frac{x}{\bar{x}}$ rauskommt, also damit auch gleich z ist.

Danke!


-----------------
Beste Grüße, Lukas Nießen
PS: Schreibt mir gerne 😄
\(\endgroup\)


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