Complex Variables: Need help with Chebyshek poly and de Moivre Theorem

[CTID17]

Homework Statement

The nth order Chebyshev polynomial is defined by
T_n(x)= cos( n arccos(x) ) , n is a positive integer; -1<= x <= 1.
Using the de Moivre theorem, show that T_n(x) has the polynomial representation
T_n(x)= 1/2 [(x+sqrt(x²-1))ⁿ+(x-sqrt(x²-1))ⁿ]

The Attempt at a Solution

I really have no idea where to begin. Only thing i can come up is to try to simplify cos (n arccos(x)) , but i get stuck.

 

[Dick]

Apply deMoivre to z=exp(i*n*arccos(x))=exp(i*arccos(x))^n. You want the real part. How do get the real part of z using complex conjugate?

 

[CTID17]

Re(z) = (z+conjugate z)/2
z = [cos(arccos(x)) + isin(arccos(x))]^n
conjugate z = [cos(arccos(x)) - isin(arccos(x))]^n

now, do i simplify z into
z= [x+ isqrt(1-x^2)]^n
and
conjugate z = [x- isqrt(1-x^2)]^n
?
if i keep them in polar form, i get
(z+conjugate z )/2 = 1/2 [2 cos(n arccos(x))]

That's as far as i can go. I just can't see what to do.

 

[Dick]

Re(z) = (z+conjugate z)/2
z = [cos(arccos(x)) + isin(arccos(x))]^n
conjugate z = [cos(arccos(x)) - isin(arccos(x))]^n

now, do i simplify z into
z= [x+ isqrt(1-x^2)]^n
and
conjugate z = [x- isqrt(1-x^2)]^n
?
if i keep them in polar form, i get
(z+conjugate z )/2 = 1/2 [2 cos(n arccos(x))]

That's as far as i can go. I just can't see what to do.

You are basically done, aren't you? The question is writing sqrt(x^2-1) instead of i*sqrt(1-x^2). But x^2-1 is negative.

 

[CTID17]

Thank you!