Complex inequalities

Can you use demoivre for the denominator in a) & b)

Yes, but this is what happens:

Not sure about the first 2/3, but the last 1/3 seems about right. What's stopping you finish?
Tbh it's probably easier to represent in polar form at the start and use that throughout. 2^13 is obviously the magnitude of the numerator.

I tried to do that, but I cannot manage the values in cos and sin. I tried to work it out in polar form, but I thought it would be easier to do it like this.

The solution is supposed to be 8192i in this one. I'd appreciate any help!

You've made an awful lot of work for yourself there by not using exponential form

The bracket in the denominator is clearly $e^{-i\pi/12}$ so its 8th power is therefore $e^{-2i\pi/3}$

The bracket in the numerator is almost a well known combination of trig values, but is multiplied by 2. So if you take out a factor of 2, you can rewrite that bracket as $e^{-i\pi/k}$ where k is an integer. The numerator then becomes $2^{13} \cdot e^{-13i\pi/k}$.

Conveniently, 2^13 = 8192 which is what you want, and that exponential simplifies after remembering that $e^{-2i\pi}$ = 1.

You should then be able to combine the remaining exponential on the numerator with the one on the denominator