Magnitude of difference of two complex exponentials

Question

I am in a Fourier Transforms course, and I ran into some trouble understanding how my professor went from this step in his solution: $$\int_{-\pi}^\pi \left|{\frac{1}{\sqrt{2\pi}} e^{imx} - \frac{1}{\sqrt{2\pi}} e^{inx}}\right|^2\mathrm dx$$ to this step: $$\int_{-\pi}^{\pi} \frac{1}{2\pi} \left(1-e^{i(m-n)x}-e^{i(n-m)x}+1\right)\mathrm dx$$ I cannot see how this was done, please help!

Answer 1

I would recommend using the complex conjugate to calculate the squared absolute value.

$$\int_{-\pi}^\pi \left|{\frac{1}{\sqrt{2\pi}} e^{imx} - \frac{1}{\sqrt{2\pi}} e^{inx}}\right|^2\mathrm dx$$

$$\int_{-\pi}^\pi \left({\frac{1}{\sqrt{2\pi}} e^{imx} - \frac{1}{\sqrt{2\pi}} e^{inx}}\right) \left({\frac{1}{\sqrt{2\pi}} e^{-imx} - \frac{1}{\sqrt{2\pi}} e^{-inx}}\right)\mathrm dx$$

$$\int_{-\pi}^\pi \left({\frac{1}{2\pi} - \frac{1}{2\pi} e^{i(n-m)x}} - {\frac{1}{2\pi} e^{i(m-n)x} + \frac{1}{2\pi}}\right)\mathrm dx$$

$$\int_{-\pi}^{\pi} \frac{1}{2\pi} \left(1-e^{i(m-n)x}-e^{i(n-m)x}+1\right)\mathrm dx$$

Answer 2

$$(e^{imx}-e^{inx})^2=e^{2imx}-2e^{i(m+n)x}+e^{2inx}=e^{i(m+n)x}(e^{i(m-n)x}-2+e^{2i(n-m)x}).$$

Now take the modulus.

The first factor vanishes, and the second is real, negative.