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Prove of complex numbers

  • Thread starter Theofilius
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  • #1
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Homework Statement



Hello! :smile:

a)Prove that [tex](a+bi)^n[/tex] and [tex](a-bi)^n[/tex], n [tex]\in \mathbb{N}[/tex] are conjugate complex numbers;

b)Prove that quotient of any two numbers from the set of [tex]\sqrt[n]{1}[/tex] is again number from the set of [tex]\sqrt[n]{1}[/tex]

c)Prove that reciproca value of any number from the set of [tex]\sqrt[n]{1}[/tex], is again number from the set [tex]\sqrt[n]{1}[/tex]

Homework Equations



[tex]z=r(cos\alpha+isin\alpha)[/tex]

[tex]\bar{z}=r(cos\alpha-isin\alpha)[/tex]

[tex]w_k=\sqrt[n]{r}(cos\frac{\alpha + 2k\pi}{n}+isin{\alpha + 2k\pi}{n}) ; k=0,1,2,...,n-1[/tex]

The Attempt at a Solution



a) [tex](a+bi)=r(cos\alpha+isin\alpha)[/tex]

[tex] (a+bi)^n=r^n(cosn\alpha+isinn\alpha)[/tex]

[tex] (a-bi)=r(cos\alpha-sin\alpha)[/tex]

[tex] (a-bi)^n=r^n(cosn\alpha-sinn\alpha)[/tex]

Is this enough to prove that they are conjugate complex numbers?

b)[tex]E_k=\sqrt[n]{1}=cos\frac{2k\pi}{n}+isin{2k\pi}{n} , k=0,1,2,3,...,n-1[/tex]

Should I make like this?

[tex]E_n_-_1=\sqrt[n]{1}=cos\frac{2(n-1)\pi}{n}+isin\frac{2(n-1)\pi}{n}[/tex]

What should I do next?

c)If I know how to prove b) I will prove c)

In this case just [tex](\sqrt[n]{1})^-^1=\sqrt[n]{1}[/tex], right?
 

Answers and Replies

  • #2
Defennder
Homework Helper
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a) [tex](a+bi)=r(cos\alpha+isin\alpha)[/tex]

[tex] (a+bi)^n=r^n(cosn\alpha+isinn\alpha)[/tex]

[tex] (a-bi)=r(cos\alpha-sin\alpha)[/tex]

[tex] (a-bi)^n=r^n(cosn\alpha-sinn\alpha)[/tex]

Is this enough to prove that they are conjugate complex numbers?
Multiply eqns 2 and 4 and you're done.

For b,c you don't have to resort to any fanciful complex number manipulation. Suppose a complex number [tex]w = \sqrt[n]{1}[/tex] That implies [tex]w^n = 1[/tex] Similarly consider a different nth complex root of unity, say z. z^n = 1. Quotient = w/z. How can we tell if w/z is also an nth root of unity?

c)

Again suppose [tex]w = \sqrt[n]{1}[/tex]. It's reciprocal is [tex]\frac{1}{w}[/tex]. What is [tex]\left (\frac{1}{w}\right)^n[/tex]
 
  • #3
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But I need to prove like I wrote at b)

First, let me apologise, I have some typo error in b), it should look like:

[tex]E_k=\sqrt[n]{1}=cos\frac{2k\pi}{n}+isin\frac{2k\pi}{n} , k=0,1,2,3,...,n-1[/tex]

In a) why I need to multiply them ?
 
  • #4
Defennder
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Well, okay I misread your answer for a). It's fine.

For b,c, what kind of proof are you looking for? Is there something unacceptable about the suggestion I gave?
 
  • #5
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I think your way of thinking is good, but probably I should get two general solutions from the set of solutions of [tex]\sqrt[n]{1}[/tex] and prove...
 
  • #6
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For b), switching to exponential form simplifies a lot.
 
  • #7
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I need to prove with complex numbers like the [tex]E_k[/tex] formula above. Please help!
 
  • #8
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Any help?
 
  • #9
HallsofIvy
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Suppose a and b are both "from the set [itex]^n\sqrt{1}[/tex]. What is (ab)n?
 
  • #10
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Maybe again [tex](ab)^n[/tex] is from the set of [tex]\sqrt[n]{1}[/tex]. I am not sure.
 
  • #11
tiny-tim
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Maybe again [tex](ab)^n[/tex] is from the set of [tex]\sqrt[n]{1}[/tex]. I am not sure.
Hint: [itex](ab)^n[/itex] = ababab … ababab = … ? :smile:
 
  • #12
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[tex]a^nb^n[/tex] ? :smile:
 
  • #13
tiny-tim
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[tex]a^nb^n[/tex] ? :smile:
Yes … but … why the question-mark?

Theofilius , you keep answering questions with a question … even when, as in this case, it's actually the complete answer.

ok … and [tex]a^nb^n[/tex] = … ?

:smile: … and don't answer with a question … ! :smile:
 
  • #14
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[tex]a^nb^n=[/tex]? I honesly don't know.. :smile: I don't even know what are you giving me that question for?
 
  • #15
tiny-tim
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[tex]a^nb^n=[/tex]? I honesly don't know.. :smile: I don't even know what are you giving me that question for?
Because of:
Suppose a and b are both "from the set [itex]^n\sqrt{1}[/tex]. What is (ab)n?
 
  • #16
[tex](ab)^n[/tex] is complex or real number. :smile:
 
  • #17
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I don't know, we are just going in circle.
 
  • #18
tiny-tim
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Suppose a and b are both "from the set [itex]^n\sqrt{1}[/itex]
So [itex]a^n\,=\,b^n\,=\,1[/itex]
 
  • #19
Is that the proof?
 
  • #20
tiny-tim
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Is that the proof?
Well, it needs to be put together into one continuous piece, and the actual question was about a/b rather than ab, but apart from that … yes! :smile:
 

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