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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.

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    $\begingroup$ 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 $\endgroup$ – Andy Walls Feb 23 '18 at 2:29
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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.

As a result, the right-hand sides of your quoted equations do not hold in general.

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To take a single counter example, if x(t) is a constant and strictly imaginary. it is an obvious even function, but it’s FT will be imaginary, thus violating your top most property.

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