MHB Discrepency with book's answer to characteristic function

[Dustinsfl]

Find the characteristic function for the PMF \(p_X[k] = \frac{1}{5}\) for \(k = -2, -1,\ldots, 2\).

The characteristic function can be found with
\begin{align*}
\phi_X(\omega) &= E[\exp(i\omega X)]\\
&= \frac{1}{5}\sum_kke^{i\omega k}\\
&= \frac{1}{5}\big(-2e^{-2i\omega} - e^{-i\omega} +
2e^{2i\omega} + e^{i\omega}\big)\\
&= \frac{2i}{5}\bigg(\frac{e^{i\omega} - e^{-i\omega}}{2i} +
\frac{e^{2i\omega} - e^{-2i\omega}}{i}\bigg)\\
&= \frac{2i}{5}\big(1 + 4\cos(x)\big)\sin(\omega)
\end{align*}

However, the book says the answer is \(\frac{1}{5}(1 + 2\cos(\omega) + 2\cos(2\omega))\).

 

[Siron]

The problem is your definition of the characteristic function. For a discrete random variable the characteristic function is defined as
$$\phi_X(\omega) = \sum_{k} e^{i\omega k} \mathbb{P}(X = k) $$
in this case we get
$$= \frac{1}{5} \sum_{k} e^{i \omega k} = \frac{1}{5} \left(e^{-2i\omega}+e^{-i\omega}+1+e^{i \omega}+e^{2i\omega}\right)$$

Now, use the fact that $e^{i \omega} = \cos \omega + i \sin \omega$ and simplify.