complex analysis - Finding real and imaginary parts of $frac{1-e^{2i pi x}}{R left(1-e^{frac {2i pi x}{R}} right)}$

I have a function given by

$$\frac{1-e^{2i \pi x}}{R \left(1-e^{\frac {2i \pi x}{R}} \right)}$$

Using Euler's formula, I expand into real and complex components:

$$\frac{1-\cos 2 \pi x-i\sin2 \pi x}{R \left(1-\cos \frac{2 \pi x}{R}-i\sin\frac{2 \pi x}{R} \right)}$$

But for some reason, here I come unstuck. It seems obvious to me that the real part should be

$$\frac{1-\cos 2 \pi x}{R \left(1-\cos \frac{2 \pi x}{R} \right)}$$

but this appears not to be the case. In fact, it's pretty obvious from the plots below (with $R=3$), that I'm wrong:

And it's not even as simple as multiplying through by $\frac{1}{R}$, as this plot charting the variance after multiplying through shows:

Could someone please explain what I'm doing wrong?

(I'd also appreciate help with the complex component.)

Answer

You are right in your use of Euler's formula to convert to rectangular coordinates, but you seemed to just divide the real part of the numerator by the real part of the denominator. To rectify the issue, you should multiply by the complex conjugate of the denominator on top and bottom (so that way the denominator becomes a real scalar that can be applied to both the imaginary and real parts of the numerator):

$$F = \frac{1-\cos 2 \pi x-i\sin2 \pi x}{R \left(1-\cos \frac{2 \pi x}{R}-i\sin\frac{2 \pi x}{R} \right)} \cdot \frac{1-\cos \frac{2 \pi x}{R}+i\sin\frac{2 \pi x}{R}}{1-\cos \frac{2 \pi x}{R}+i\sin\frac{2 \pi x}{R}}$$

Letting the following substitutions take place:

$\quad a = 1 - \cos2\pi x$

$\quad b = \sin 2\pi x$

$\quad c = 1 - \cos \frac{2\pi x}{R}$

$\quad d = \sin\frac{2\pi x}{R}$

Then we get:

$$F = \frac{a-bi}{R(c-di)} \cdot \frac{c+di}{c+di} = \frac{ac+bd + i(ad-bc)}{R(c^2 + d^2)} $$

Thus:

$$Re(F) = \frac{ac+bd}{R(c^2 + d^2)}$$

Appropriate trig identities and algebraic manipulation can massage the above into a nicer form if you desire.