Quadratische Gleichung in den komplexen Zahlen

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Name	Status	Author	Last Modified
Quadratische Gleichung in den komplexen Zahlen	Ready to use	Ulrich Görtz	22/11/2020 11:28
Quadratic equation in the complex numbers	draft	Ulrich Görtz	20/11/2020 21:07

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Question statement

Lösen Sie die folgenden quadratischen Gleichungen in den komplexen Zahlen. Die Gleichungen sind so ausgewählt, dass Sie die Lösungen in der Form $a+bi$ für Bruchzahlen (bzw. ganze Zahlen) $a$ und $b$ angeben können. Geben Sie Ihre Antwort ohne Klammern ein.

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			Name	Type	Generated Value

p	decimal	dec("-5")
q	decimal	dec("31.25")
c1	decimal	dec("2.5") - dec("5")*i
c2	decimal	dec("2.5") + dec("5")*i
c	rational	5/2
d	rational	5
cb	rational	-2/3
db	rational	-1
c1b	decimal	dec("-6.666666666666666666666666666666666666667e-1") + i
c2b	decimal	dec("-6.666666666666666666666666666666666666667e-1") - i
pb	decimal	dec("1.333333333333333333333333333333333333333")
qb	decimal	dec("1.444444444444444444444444444444444444445")

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-c1-c2

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Parts

Part a) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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Part b) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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Gegeben ist die Gleichung $\simplify[all, fractionNumbers]{x^2 + {p}*x + {q}} = 0$ .

Die Lösungen sind Gap 0. und Gap 1..

Advice

Sie können diese quadratischen Gleichungen wie Gleichungen über den reellen Zahlen behandeln, nur dass Sie die Wurzel aus einer negativen Zahl ziehen müssen. Das ist in den komplexen Zahlen möglich: $\sqrt{-a} = i\sqrt{a}$ (für $a\in\mathbb R_{> 0}$ ).

Im ersten Fall schreiben Sie die Gleichung um als $\simplify[fractionNumbers]{(x+{p/2})^2 - {abs(p)/2}^2 + {q}}$ , also sind die Lösungen $\var[fractionnumbers]{-p/2} \pm i\sqrt{\var[fractionNumbers]{-(p/2)^2+q}}$ , also $\var[fractionNumbers]{c1}$ und $\var[fractionNumbers]{c2}$ .

Im zweiten Fall schreiben Sie die Gleichung um als $\simplify[fractionNumbers]{(x+{pb/2})^2 - {abs(pb)/2}^2 + {qb}}$ , also sind die Lösungen $\var[fractionnumbers]{-pb/2} \pm i\sqrt{\var[fractionNumbers]{-(pb/2)^2+qb}}$ , also $\var[fractionNumbers]{c1b}$ und $\var[fractionNumbers]{c2b}$ .

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This question is used in the following exams:

Online-Test I by Ulrich Görtz in Lineare Algebra 1.
Komplexe Zahlen by Ulrich Görtz in Lineare Algebra 1.
quadratic equations by Mikel Artester in Mikel's workspace.
Woche 4: Komplexe Zahlen kartesische Darstellung by Felix Lenders in Mathematik 2 LSE.
Anna-Maria's copy of Komplexe Zahlen by Anna-Maria Pippich in Vorkurs Konstanz.

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Quadratische Gleichung in den komplexen Zahlen

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Ulrich Görtz 4 years, 5 months ago

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Quadratische Gleichung in den komplexen Zahlen

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

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Comment

Checkpoint description

Ulrich Görtz 4 years, 5 months ago

Ulrich Görtz 4 years, 5 months ago

Ulrich Görtz 4 years, 5 months ago

Error in variable testing condition

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Generated value: decimal

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Gap-fill

Generated value: `decimal`