De Moivre's theorem 3

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Frank Doheny

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8 years, 1 month ago

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Frank Doheny 8 years, 1 month ago

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Frank Doheny 8 years, 1 month ago

Gave some feedback: Ready to use

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Frank Doheny 8 years, 5 months ago

Created this as a copy of De Moivre's theorem 2.

Name	Status	Author	Last Modified
DeMoivre's theorem 1	Ready to use	Frank Doheny	04/04/2017 08:31
De Moivre's theorem 2	Ready to use	Frank Doheny	04/04/2017 08:33
De Moivre's theorem 3	Ready to use	Frank Doheny	04/04/2017 08:42
DeMoivre's Theorem 1	draft	Peter Johnston	07/04/2017 07:29
De Moivre's theorem - Square Root	draft	Peter Johnston	07/04/2017 07:27
Maria's copy of DeMoivre's Theorem 1	draft	Maria Aneiros	25/05/2019 08:24
Maria's copy of De Moivre's theorem - Square Root	draft	Maria Aneiros	11/06/2019 06:19

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Question statement

This question asks you to use De Moivre's theorem to find the cube roots of $Z=\var{x}+\var{y}j$

Give any introductory information the student needs.

			Name	Type	Generated Value

x	integer	2
y	integer	9
n	rational	1/3
theta	decimal	dec("1.352127380920954657189147941389812850984")
mod	number	9.2195444573
x2	number	1.8874659456
y2	number	0.9134013227
x3	number	-1.7347617221
y3	number	1.1778927963
x4	number	-0.1527042235
y4	number	-2.091294119

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$Z=\var{x}+\var{y}j$ has three cube roots.

$Z=\var{x2}+\var{y2}j$ and $Z=\var{x3}+\var{y3}j$ are two of the roots.

The third cube root is $Z=A+Bj$

Calculate $A$ and $B$

$A$ = Gap 0.

$B$ = Gap 1.

Advice

To evaluate $(\var{x}+\var{y}j)^{\frac{1}{3}}$ we must first express $Z=\var{x}+\var{y}j$ in polar form.

The modulus of $\var{x}+\var{y}j$ = $\sqrt{\var{x}^2+(\var{y})^2}=\sqrt{\simplify{{x^2}+{y}^2}}$

The argument of the complex number is given by $\theta=tan^{-1}\left(\frac{\var{y}}{\var{x}}\right)=\var{theta}$

According to De Moivre's theorem $Z^{\frac{1}{3}}=|Z|^{\frac{1}{3}}\left(cos(\frac{\theta+2n\pi}{3})+jsin(\frac{\theta+2n\pi}{3})\right)$

By letting $\theta$ take the values 0, 1 and 2 we find each of the three cube roots.

Below is the third cube root which can be found by setting $\theta=1$ .

$(\simplify{{x}+{y}j})^{\frac{1}{3}}=\left(\sqrt{\simplify{{x^2}+{y}^2}}\right)^{\frac{1}{3}}\left(cos(\simplify{{n}*({theta}+2*pi)})+jsin(\simplify{{n}*({theta}+2pi)})\right)$

= $\simplify{{x4}+{y4}j}$

$A=\var{x4}$ and $B=\var{y4}$

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Frank Doheny 8 years, 1 month ago

Frank Doheny 8 years, 1 month ago

Frank Doheny 8 years, 5 months ago

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De Moivre's theorem 3

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Frank Doheny 8 years, 1 month ago

Frank Doheny 8 years, 1 month ago

Frank Doheny 8 years, 5 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`