Arithmetics of complex numbers V

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Newcastle University Mathematics and Statistics

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Ulrich Görtz 4 years, 4 months ago

Wants to merge their version, Ulrich's copy of Arithmetics of complex numbers V .

In part d) the expected answer should be {i^rem} not {i^n} because the latter, while "mathematically" the same, leads to floating point number issues when revealing the answers (it shows something like, e.g., "Expected answer 1 - 5047/2104i*10^(-12)" instead of "1" (in case rem==4).

I hope that this is the appropriate way to submit this comment.

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

Created this.

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

There are 15 other versions that do you not have access to.

Question statement

Express the following complex numbers $z$ in the form $a+bi$ .

Input $a$ and $b$ as fractions and not as decimals.

Give any introductory information the student needs.

			Name	Type	Generated Value

s3	integer	1
s2	integer	-1
s1	integer	1
m	integer	2126
s4	integer	1
d1	integer	2
rem	integer	0
c1	integer	5
n	integer	8504
z1	number	2 + 5i
z2	number	-4 + 2i
z3	number	-2 + 2i

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Parts

Part a) Gap-fill
1. Gap Mathematical expression
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Part b) Gap-fill
1. Gap Mathematical expression
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Part c) Gap-fill
1. Gap Mathematical expression
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Part d) Gap-fill
1. Gap Mathematical expression
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$z=(\var{z1})^{\var{c1}}(\var{conj(z1)})^{\var{c1}}$

$z=\;\;$ Unnamed gap

Advice

Note that for a complex number $z=a+bi$ we have:

$z\overline{z}=|z|^2=a^2+b^2$ .

But since $\var{conj(z1)}=\overline{\var{z1}}$ we have:

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

Note that $(\var{z2})^4=((\var{z2})^2)^2$ .

Since $(\var{z2})^2=\simplify[std]{{z2^2}}$ we have:

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

c)
We have

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

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

So if $n=4m+r,\;\;0\le r\le 3$ we have

$i^n=i^{4m+r}=(i^4)^m \times i^r=i^r$
Hence since

$\var{n}=4\times\var{m}+\var{rem}$ we have:

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

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Ulrich Görtz 4 years, 4 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Arithmetics of complex numbers V

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Ulrich Görtz 4 years, 4 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`