Complex numbers expression

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  • #1
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Homework Statement


If n is an integer which leaves remainder one when divided by three, then (1+√3i)n + (1-√3i)n equals

a) -2n+1
b) 2n+1
c) -(-2)n
d) -2n


The Attempt at a Solution



Multiply and divide each of the two expression inside bracket by 2.
2n(cosπ/3 + isinπ/3)n + 2n(cosπ/3-isinπ/3)n
On simplifiying, the expression equals
2n(2cosnπ/3)=2n(-1)=-2n which is same as (d)

The correct option is (c). I have no idea where I went wrong.
 

Answers and Replies

  • #2
Dick
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Put n=1 into 2^n(2cosnπ/3). cos(pi/3) isn't equal to (-1/2). It's +1/2. Can you take it from there?
 
  • #3
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n=1 does not give 1 as a remainder when divided by 3 does it?
I put n=4 in the above expression. The answer from your idea is not even there in the options.
 
  • #4
eumyang
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n=1 does not give 1 as a remainder when divided by 3 does it?
Uh, yes it does. 1 divided by 3 is 0 with a remainder of 1.
 
  • #5
Dick
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n=1 does not give 1 as a remainder when divided by 3 does it?
I put n=4 in the above expression. The answer from your idea is not even there in the options.

What idea? I'm just trying to get you to realize 2*cos(n*pi/3) isn't always equal to (-1) when n=1,4,7,10,...
 
  • #6
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What idea? I'm just trying to get you to realize 2*cos(n*pi/3) isn't always equal to (-1) when n=1,4,7,10,...

Sorry I did not get your point earlier. I realise that for n=1,7,13.... cos(npi/3) is +1/2. How do I use this point to get -(-2)n ?
 
Last edited:
  • #7
Dick
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Sorry I did not get your point earlier. I realise that for n=1,7,13.... cos(npi/3) is +1/2. How do I use this point to get -(-2n) ?

Ok, and for n=4,10,16,... it's -1/2. Can't you write that result as c*(-1)^n for some constant c?
 
  • #8
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2n(2cosnπ/3) = 2n[2*1/2*(-1)n+1] = -(-2)n
I wonder if there is some other easier method.
 
  • #9
SammyS
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Let n = 3k+1, where k is an integer.

[tex]\left(1+i\sqrt{3}\right)^{3}=\left(1-i\sqrt{3}\right)^{3}=-8=(-2)^3[/tex]
 
  • #10
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I figured out a much easier way.
(1+√3i)n + (1-√3i)n = 2n(-w2)n + 2n(-w)n = (-2n)(w2+w) = -(-2)n, where w is the complex cube root of unity.
 
  • #11
SammyS
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Let n = 3k+1, where k is an integer.

[tex]\left(1+i\sqrt{3}\right)^{3}=\left(1-i\sqrt{3}\right)^{3}=-8=(-2)^3[/tex]
This is what I had in mind for the above:

n = 3k+1 → 3k = n-1

[tex]\left(1+i\sqrt{3}\right)^{3k+1}=\left(1+i\sqrt{3}\right)\left(\left(1+i\sqrt{3}\right)^{3}\right)^k[/tex]
[tex]=\left(1+i\sqrt{3}\right)\left(-2\right)^{3k}[/tex]

[tex]=\left(1+i\sqrt{3}\right)\left(-2\right)^{n-1}[/tex]​
Similarly:
[tex]\left(1-i\sqrt{3}\right)^{3k+1}=\left(1-i\sqrt{3}\right)\left(-2\right)^{n-1}[/tex]

Adding these two results gives: -(-2)(n)
 
  • #12
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Nice one!
 

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