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

• Apr 16th 2011, 11:42 AM
Wandering
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

• Apr 17th 2011, 08:10 AM
running-gag
Hi

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

Quote:

Originally Posted by Wandering
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
• Apr 17th 2011, 11:21 AM
HallsofIvy
Quote:

Originally Posted by Wandering
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.

Quote:

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.

Quote:

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}.

Quote:

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.

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