Further evaluating $|1 - \exp^{-ika}|^2$ for $a \in [0, 1]$ and $k$ in $\mathbb R$

Question

I've been calculating the fourier transformation for

$$\psi(x) = \begin{cases} \dfrac 1 {\sqrt{a}} & \text{for } 0 \le x \le a \\[6pt] 0 & \text{else} \end{cases}$$

With

$$\psi'(k) = \frac{1}{\sqrt{2 \pi}} \int_{0}^{a} \psi(x) \exp^{-ikx} dx$$

I quickly found that

$$\psi'(k) = \frac{1}{i \sqrt{2 \pi a k^2}} (1 - \exp^{-ika})$$

Now I want to calculate $|\psi'(k)|^2$ but I don't know how to proceed further than this:

$$ |\psi'(k)|^2 = \frac{1}{2 \pi a k^2} |1 - \exp^{-ika}|^2$$

How can I continue crunching on $|1 - \exp^{-ika}|^2$?
I know that $a \in [0, 1]$. So if $\exp$ was real I'd safely argue that $0 < \exp^{-ka} \le 1$ assuming $k \ge 0$. That would only help me continue for that case though - other than that I'm clueless.

Thanks for your help in advance!

Answer 1

\begin{align} \vert 1 - e^{ika} \vert^2 & = (1 - e^{ika}) \overline{(1 - e^{ika})} = (1 - e^{ika})(1 - e^{-ika}) \\[6pt] & = 1 - e^{-ika} - e^{ika} + e^{ika}e^{-ika} = 2 - (e^{ika} + e^{-ika}) = 2 - 2 \cos(ka); \tag 1 \end{align}

I think that's about as far as anyone can take it.

Answer 2

Use $|z|^2 = z \bar{z} $, where $\bar{z}$ is the complex conjugate of $z$.

$$|1 - e^{-ika}|^2 = (1 -e^{-ika})(1 - e^{+ika}) = 2 - e^{ika} - e^{-ika} = 2 - 2 \cos (ka) $$