# Fourier transform of even/odd parts of a complex signal

Why does Oppenheim state the following properties:

\begin{align} \mathcal F\big\{x_e (t) \big\} &= \Re\big\{ X(j\omega) \big\}\\ \mathcal F\big\{x_o (t) \big\} &= j \Im\big\{ X(j\omega) \big\} \end{align}

where $x_e (t)$ and $x_o (t)$ denote the even and odd parts of a signal $x(t)$, respectively, only for a real signal $x(t)$? I don't understand how those two properties could break down if $x(t)$ were complex in general.

• Oppenheim and Schaefer's book is OK for learning about Fourier transforms, but Ronald Bracewell's book is much better for the subject matter. www.amazon.com/gp/aw/d/0073039381/ref=dp_ob_neva_mobile Feb 23, 2018 at 2:29

Defining $$x_e(t) = \frac{x(t) + x(-t)}{2}$$ $$x_o(t) = \frac{x(t) - x(-t)}{2}$$ We can take the Fourier transform of each: $$FT\{x_e(t)\} = \int_{-\infty}^{+\infty} x_e(t) (\cos(2\pi\omega t) - j \sin(2\pi\omega t)) dt$$ $$FT\{x_o(t)\} = \int_{-\infty}^{+\infty} x_o(t) (\cos(2\pi\omega t) - j \sin(2\pi\omega t)) dt$$
So if $x_e(t) : \mathbb{R} \mapsto \mathbb{R}$ then the $\sin$ term disappears and we are left with just a real-valued function.
So if $x_o(t) : \mathbb{R} \mapsto \mathbb{R}$ then the $\cos$ term disappears and we are left with a purely imaginary-valued function.
Now, if instead $x_e(t), x_o(t) : \mathbb{R} \mapsto \mathbb{C}$ then the term cancellations still happen, but we are still left with a complex value for each integral because both functions are now complex-valued.