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 :
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