(no subject)
On one of my Complex Variables homework problems, I am asked to establish two identities (where x is between 0 and 2*pi, n is an integer, and z is a complex number). First I had to prove
(1) 1 + z + z2 + ... + zn = (zn+1 - 1)/(z - 1)
Fine, I figured that out. Now I need to use the above and De Moivre's formula to show the following:
(2) 1 + cos(x) + cos(2x) + ... cos(nx) = [sin((n+½)x)/(2sin(x/2)) + ½]
(3) sin(x) + sin(2x) + ... sin(nx) = [½*cot(x/2) - cos((n+½)x)/(2sin(x/2))]
(4) ei*n*x = cos(nx) + i*sin(nx) is De Moivre's formula.
What I've got so far is that, if I let z=eix and I plug it in equation (1), I get for the left hand side:
1 + ei*x + e2*i*x + e3*i*x + ... + en*i*x = [...]
Replacing each of the exponentials with (4) gives:
1 + (cos(x) + i*sin(x)) + (cos(2x) + i*sin(2x)) + (cos(3x) + i*sin(3x)) + ... + (cos(nx) + i*sin(nx)) = [...]
Regrouping into real and imaginary gives
(5) (1 + cos(x) + cos(2x) + cos(3x) + ... + cos(nx) ) + i*(sin(x) + sin(2x) + sin(3x) + ... + sin(nx)) = [...]
I can already recognize that the right-hand side of the equation is going to have a real and imaginary part, so the real part of the RHS of (5) will need to equal the real part of the LHS of (5), and likewise with the imaginary parts. I can see that I'm going to simultaneously establish both (2) and (3). What I can't seem to make out is how to get the RHS to look like the RHS of (2) or (3).
If I plug z=eix into the RHS of (1), I am left with an ugly, complex numerator and denominator. If I multiply that fraction by (the complex conjugate of the denominator divided by that same value), I get a denominator of (2 - 2cosx).
My numerator is now ugly multiplied by ugly. I can't seem to determine the right sequence of trigonometric substitutions to make the real part of the ugliness look like the RHS of (2) and the imaginary part look like the RHS of (3). Please help?
(1) 1 + z + z2 + ... + zn = (zn+1 - 1)/(z - 1)
Fine, I figured that out. Now I need to use the above and De Moivre's formula to show the following:
(2) 1 + cos(x) + cos(2x) + ... cos(nx) = [sin((n+½)x)/(2sin(x/2)) + ½]
(3) sin(x) + sin(2x) + ... sin(nx) = [½*cot(x/2) - cos((n+½)x)/(2sin(x/2))]
(4) ei*n*x = cos(nx) + i*sin(nx) is De Moivre's formula.
What I've got so far is that, if I let z=eix and I plug it in equation (1), I get for the left hand side:
1 + ei*x + e2*i*x + e3*i*x + ... + en*i*x = [...]
Replacing each of the exponentials with (4) gives:
1 + (cos(x) + i*sin(x)) + (cos(2x) + i*sin(2x)) + (cos(3x) + i*sin(3x)) + ... + (cos(nx) + i*sin(nx)) = [...]
Regrouping into real and imaginary gives
(5) (1 + cos(x) + cos(2x) + cos(3x) + ... + cos(nx) ) + i*(sin(x) + sin(2x) + sin(3x) + ... + sin(nx)) = [...]
I can already recognize that the right-hand side of the equation is going to have a real and imaginary part, so the real part of the RHS of (5) will need to equal the real part of the LHS of (5), and likewise with the imaginary parts. I can see that I'm going to simultaneously establish both (2) and (3). What I can't seem to make out is how to get the RHS to look like the RHS of (2) or (3).
If I plug z=eix into the RHS of (1), I am left with an ugly, complex numerator and denominator. If I multiply that fraction by (the complex conjugate of the denominator divided by that same value), I get a denominator of (2 - 2cosx).
My numerator is now ugly multiplied by ugly. I can't seem to determine the right sequence of trigonometric substitutions to make the real part of the ugliness look like the RHS of (2) and the imaginary part look like the RHS of (3). Please help?