Sum of Series Question (image) - Complex Numbers

Question

The series $P$ and $Q$ are defined for $0<\theta<\pi$ as

$$P=1+\cos\theta+\cos2\theta+\cos3\theta+\ldots+\cos12\theta$$

$$Q=\sin\theta+\sin2\theta+\sin3\theta+\ldots+\sin12\theta$$

Show that $\displaystyle P+\mathrm iQ=\frac{\mathrm e^{6\mathrm i\theta}(\mathrm e^{\frac{13\mathrm i\theta}2}-\mathrm e^{-\frac{13\mathrm i\theta}2})}{\mathrm e^{\frac{\mathrm i\theta}2}-\mathrm e^{-\frac{\mathrm i\theta}2}}$.

We have

$$P=1+\cos\theta+\cos2\theta+\cos3\theta+\ldots+\cos12\theta$$

$$Q=\sin\theta+\sin2\theta+\sin3\theta+\ldots+\sin12\theta$$

So that

$$P+iQ=1+\mathrm e^{i\theta}+\mathrm e^{2i\theta}+\mathrm e^{3i\theta}+\ldots+\mathrm e^{11i\theta}+\mathrm e^{12i\theta}$$

Answer 1

$$=\dfrac{e^{i13\theta}-1}{e^{i\theta}-1}=\dfrac{e^{i13\theta/2}(e^{i13\theta/2}-e^{-i13\theta/2})}{ e^{i\theta/2}(e^{i\theta/2}-e^{-i\theta/2})}$$

Using Intuition behind euler's formula we can actually simplify further

$$2i\sin y=e^{iy}-e^{-iy}$$