So to prove that |sin(nTheta/2)/sin(Theta/2)|<=n for theta!={0,+/-2pi,+/-4pi,...} and n a positive integer I'm told that I can assign z=e^iTheta and replace the left side of the inequality with |(1-z^n)/(1-z)|. After that I just divide to get |1+z+z^2+...z^n-1| which is less than the sum of the magnitudes by the triangle inequality and thusly less than or equal to n. My problem is that I don't see why |sin(nTheta/2)/sin(Theta/2)|=|(1-z^n)/(1-z)| is true.

I don't see how 1-z could equal sin(Theta/2) because 1-z could clearly be complex while sin(Theta/2) is real, so I guess the relationship must be strictly between the magnitudes. (Since the |a/b|=|a|/|b|, right?) And so all I need is to figure out why |1-z|=|sin(Theta/2)| (I'm hoping the numerator follows from that)

Any advice or corrections would be appreciated,

Thanks