Rechnen mit komplexen Zahlen V

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Ulrich Görtz

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4 years, 5 months ago

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Ulrich Görtz 4 years, 1 month ago

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Ulrich Görtz 4 years, 5 months ago

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Ulrich Görtz 4 years, 6 months ago

Created this as a copy of Arithmetics of complex numbers V.

Name	Status	Author	Last Modified
Arithmetics of complex numbers V	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Complex numbers Ex Sheet 4	draft	Violeta CIT	19/10/2017 15:26
Blathnaid's copy of Arithmetics of complex numbers V	draft	Blathnaid Sheridan	19/11/2018 13:38
Maria's copy of Arithmetics of complex numbers V	draft	Maria Aneiros	25/05/2019 08:20
Rechnen mit komplexen Zahlen V	Ready to use	Ulrich Görtz	08/02/2021 19:59
Complex numbers raised to powers	Ready to use	Shaheen Charlwood	27/10/2022 18:53

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Question statement

Schreiben Sie die folgenden komplexen Zahlen $z$ in der Form $a+bi$ mit reellen Zahlen $a$ und $b$ .

Geben Sie $a$ und $b$ als Bruchzahlen oder als ganze Zahlen an.

Je nachdem, wieviel "Glück" Sie mit dem Zufallszahlengenerator haben, sind die Zahlen bei dieser Aufgabe ziemlich groß. Lassen Sie gegebenenfalls eine neue Aufgabe erzeugen oder nehmen Sie für die Rechnungen in den ganzen Zahlen, die auszuführen sind, einen Taschenrechner zur Hand...

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			Name	Type	Generated Value

s3	integer	-1
s2	integer	-1
s1	integer	1
m	integer	3635
s4	integer	-1
d1	integer	3
rem	integer	2
c1	integer	4
n	integer	14542
z1	number	2 - 4i
z2	number	-2 - 2i
z3	number	-2 - i

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-1

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$z=(\var{z1})^{\var{c1}}(\var{conj(z1)})^{\var{c1}}$

$z=\;\;$ Unnamed gap

Advice

Für jede komplexe Zahl $z=a+bi$ gilt

$(a+bi)(a-bi) =a^2+b^2$ .

Also haben wir

$\begin{eqnarray*}z&=&(\var{z1})^{\var{c1}}(\var{conj(z1)})^{\var{c1}}\\ &=&((\var{z1})(\var{conj(z1)}))^{\var{c1}}\\ &=&\simplify[]{({re(z1)}^2+{im(z1)}^2)^{c1}}\\ &=&\var{(re(z1)^2+im(z1)^2)^c1} \end{eqnarray*}$

Beachten Sie, dass $(\var{z2})^4=((\var{z2})^2)^2$ .

Weil $(\var{z2})^2=\simplify[std]{{z2^2}}$ , gilt

$(\var{z2})^4=(\simplify[std]{{z2^2}})^2=\simplify[std]{{z2^4}}$

c)
Es gilt

$\begin{eqnarray*} z&=&(\var{z3})^{\var{-d1}}\\ &=&\frac{1}{(\var{z3})^{\var{d1}}}\\ &=&\frac{(\var{conj(z3)})^{\var{d1}}}{(\var{z3})^{\var{d1}}(\var{conj(z3)})^{\var{d1}}}\\ &=&\frac{\var{conj(z3)^d1}}{\var{abs(z3)^(2*d1)}}\\ &=&\simplify[std]{{re(conj(z3^d1))}/{(abs(z3)^(2*d1))}+({im(conj(z3^d1))}/{round(abs(z3)^(2*d1))})*i} \end{eqnarray*}$
d)
Es ist

$i^2=-1,\;\;i^3=-i,\;\;i^4=1$ .

Also gilt für $n=4m+r,\;\;0\le r\le 3$ , dass

$i^n=i^{4m+r}=(i^4)^m \cdot i^r=i^r$
In diesem Fall ist

$\var{n}=4\cdot\var{m}+\var{rem}$ , also folgt

$i^{\var{n}}=i^{\var{rem}}=\simplify{i^{rem}}$

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Ulrich Görtz 4 years, 1 month ago

Ulrich Görtz 4 years, 5 months ago

Ulrich Görtz 4 years, 6 months ago

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Rechnen mit komplexen Zahlen V

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Ulrich Görtz 4 years, 1 month ago

Ulrich Görtz 4 years, 5 months ago

Ulrich Görtz 4 years, 6 months ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`