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Math Help - De Moivre's Theorem and Nth Roots of Unity Question

  1. #1
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    De Moivre's Theorem and Nth Roots of Unity Question

    I've been having difficulty with the second and third parts of this question, any help would be appreciated.

    Consider the complex number z = cis ([2Pi(k)]/5) for any integer k such that z does not equal 1.

    a) Show that z^n + z^-n = 2cos([2Pi(k)(n)]/5) for any integer n.

    I was able to verify this result, the next part is where I'm a bit a confused:

    b) Show that z^5 = 1. Hence, show that 1+z+z^2+z^3+z^4 = 0

    Using the earlier result, I substituted z^5 in for 1 so that

    z^n + z^5/z^n = 2cos([2Pi(k)(n)]/5)

    substituting values for n, I was able to come up with the pairs:

    z^4 + z^1 and z^2 +z^3

    I'm now stuck, although placing the respective values given by the earlier result into my calculator gives me the desired outcome, I am looking for a way to prove the result purely through algebra.

    I also noticed that the given equation is in a geometric progression with initial term 1 and common ratio z:

    Using the sum formula:

    S = 1(1-r^n)/(1-r)

    Sum of first 5 terms :
    = 1(1-z^5)/(1-z)
    =1(0)/1-z
    =0

    This gives the desired outcome, but I am looking to get the same result using the earlier parts of the question - I haven't had any luck.

    c) Find the value of b, given that:

    (z+z^-n)^2 + (z^2 + z^-2)^2 = b

    I have tried using the earlier result and then using trig identities to try to solve for b, but I can't seem to solve the equation.

    Sorry if this is too many questions, but I am also a bit confused about the following question:

    If n is a positive integer, show that (1+i)^n + (1-i)^n = 2^((n/2)+1) * cos([Pi(n)]/4)

    I was able to simplify this to 2cos([Pi(n)]/4) but I do not really understand how the exponent ((n/2)+1) comes about


    Thanks in advance
    Last edited by Wandering; April 16th 2011 at 12:45 PM.
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  2. #2
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    Hi

    z=e^{i2kpi/5} therefore z^5=e^{i2kpi}=1

    Quote Originally Posted by Wandering View Post
    I also noticed that the given equation is in a geometric progression with initial term 1 and common ratio z:

    Using the sum formula:

    S = 1(1-r^n)/(1-r)

    Sum of first 5 terms :
    = 1(1-z^5)/(1-z)
    =1(0)/1-z
    =0

    This gives the desired outcome, but I am looking to get the same result using the earlier parts of the question - I haven't had any luck.
    This is a good way to proceed
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  3. #3
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    Quote Originally Posted by Wandering View Post
    I've been having difficulty with the second and third parts of this question, any help would be appreciated.

    Consider the complex number z = cis ([2Pi(k)]/5) for any integer k such that z does not equal 1.

    a) Show that z^n + z^-n = 2cos([2Pi(k)(n)]/5) for any integer n.

    I was able to verify this result, the next part is where I'm a bit a confused:

    b) Show that z^5 = 1. Hence, show that 1+z+z^2+z^3+z^4 = 0

    Using the earlier result, I substituted z^5 in for 1 so that

    z^n + z^5/z^n = 2cos([2Pi(k)(n)]/5)

    substituting values for n, I was able to come up with the pairs:

    z^4 + z^1 and z^2 +z^3

    I'm now stuck, although placing the respective values given by the earlier result into my calculator gives me the desired outcome, I am looking for a way to prove the result purely through algebra.
    I really have no idea what you are doing with those. Do you know the fact that \left(cis(\left(\frac{2\pi k}{m}\right)\right)^n= cis\left\frac{2\pi kn}{m}\right). That's really the whole point. If you do not know that, then try proving it, say by induction on n. With that formula, take m= n= 5.

    I also noticed that the given equation is in a geometric progression with initial term 1 and common ratio z:

    Using the sum formula:

    S = 1(1-r^n)/(1-r)

    Sum of first 5 terms :
    = 1(1-z^5)/(1-z)
    =1(0)/1-z
    =0

    This gives the desired outcome, but I am looking to get the same result using the earlier parts of the question - I haven't had any luck.
    Of course, z^5- 1= (z- 1)(z^4+ z^3+ z^2+ z+ 1). The fact that z^n- 1= (z- 1)(z^{n-1}+ z^{n-2}+ \cdot\cdot\cdot+ z^2+ z+ 1) is a pretty basic algebra result.

    c) Find the value of b, given that:

    (z+z^-n)^2 + (z^2 + z^-2)^2 = b

    I have tried using the earlier result and then using trig identities to try to solve for b, but I can't seem to solve the equation.
    You've already proved that z+ z^{-n}= 2 cis(\frac{2\pi k}{n} so ](z+ z^{-n})^2= (2 cis(\frac{2\pi k}{n}))^2= 4 cis(\frac{4\pi k}{n})[/tex]
    and you are given that z= cis(\frac{2\pi k}{5}) so that z^2= cis(\frac{4\pi k}{5}) and z^{-2}= cis(-4\pi k}{5}.

    Sorry if this is too many questions, but I am also a bit confused about the following question:

    If n is a positive integer, show that (1+i)^n + (1-i)^n = 2^((n/2)+1) * cos([Pi(n)]/4)

    I was able to simplify this to 2cos([Pi(n)]/4) but I do not really understand how the exponent ((n/2)+1) comes about
    The exponent ((n/2)+ 1) "comes about" because you are given it- that's all you need to know for this. Of course, 1+ i= 2^{1/2}cis(\pi/4) and 1- i= 2^{1/2}cis(-\pi/4) so that (1+ i)^n= 2^{n/2}cis(n\pi/2) and (i- 1)^n= 2^{n/2}cis(-n\pi/2). Now, think about what happens when n is even and when it is odd.


    Thanks in advance
    Last edited by HallsofIvy; April 17th 2011 at 11:47 AM.
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  4. #4
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    on part b), if z^5 = 1, then z is a root of the polynomial x^5 - 1. now 1 is also a root of this polynomial, so (x - 1) is a factor.

    what do you get when you divide: (x^5 - 1)/(x - 1) = ?

    (in other words, 0 is an integer, but z = cis(0) isn't very interesting. what about the other 4 values for k? they should be the roots of...?)

    on part d) what is |1+i|? what is |(1+i)^n|? how about |1-i|, and |(1-i)^n|? might this have something to do with a common factor the sum (1+i)^n + (1-i)^n has?
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