Considering the FFT of Real & Complex Signals I've been implementing a website to perform the FFT of various signals, real &amp; complex.

Examining the first example, a real signal $x[n] = 10 cos(2\pi\times4n)$, I got the following FFT: 

Which was exactly what I expected - two nice peaks of half amplitude at $\pm 4$ 

So I then extended to the FFT of the complex signal $x[n] = 10e^{j2\pi\times4n}$ (or equivalently $x[n] = 10cos(2\pi\times4n) + j10sin(2\pi\times4n)$). This is shown in the time domain as;

However, this time I got the following as the FFT, however, it now only has a single peak at $+4$, rather than the $\pm4$ mirrored peak I was expecting, and received in the real signal.


After reading through a wide number of articles relating to the conjugate symmetric property of the FFT with regard to real, even signals that states $x[n] =&gt; F*(\omega) = F(-\omega)$.

However, this doesn't help me understand what the FFT of a complex signal should look like - i.e. if my figure is correct?  How should an FFT for a real/complex/imaginary signal appear in terms of mirroring and symmetry?

NB: My own way of justifying this so far is that for a real signal, $cos(2\pi ft) = \frac{1}{2}(e^{2\pi ft}+e^{-2\pi ft})$ which creates negative frequencies - hence the mirroring, whilst the FFT of $e^{2\pi ft}$ directly does not...however I may be completely wrong!
There's nothing wrong here -  complex sinusoids like your signal really have only one peak in frequency domain! This is the fundamental idea of why we use the Fourier transform for periodic (even complex) signals.

You can think of it this way: 

the cosine has two peaks, one at +f, the other at -f. That's because Euler's formula actually says $\cos x = \frac12\left(e^{ix}+e^{-ix}\right)$. You can see the two complex sinusoids that lead to your two peaks.

Now, doing the same for the sine, we see that $i \cdot\sin x =i\frac1{2i}\left(e^{ix}-e^{-ix}\right)$. That means, the sine kills one of the two peaks of the cosine, if you will so.

Everything's alright :)