Real discrete Fourier transform
The advantage of the complex DFT or complex Fourier transform or
complex Fourier series is that linear
systems have the nice property that the response to $A\exp(j\omega t)$ is
$H(\omega)A\exp(j\omega t)$.  (Here $A$ can be a complex constant).
So the output is just a scalar multiple of the input.
More importantly, if we have a representation of the input
as a weighted sum of complex exponentials, the output is
just another weighted sum of the same exponentials.  Different
weights, but same set of exponentials. Furthermore, each new weight
is obtained by multiplying the old weight by an appropriate number.

Of course, no physical system has complex-values
signals going in and coming out; at least, not as of today though
one can always hope for better things in the future.  In the mean time,
we take real parts of the complex signals, or get the response to $\cos(\omega t)$ or
$\sin(\omega t)$ via linearity and superposition and liberal use of
$$\begin{align*}
\cos(\omega t) &amp;= \frac{\exp(j\omega t) + \exp(-j\omega t)}{2}\\
\sin(\omega t) &amp;= \frac{\exp(j\omega t) - \exp(-j\omega t)}{2j}
\end{align*}$$

In contrast, the response to $\cos(\omega t)$ is of the form 
$B(\omega)\cos(\omega t) + C(\omega)\sin(\omega t)$.  So,
while linearity and superposition etc all work, the output
might well need the use of different basis functions than the 
input does.  Very closely related, of course, but still possibly
different and maybe more  basis functions might be needed. For
example, input $\cos(\omega t)$ is represented by one basis function,
output $B(\omega)\cos(\omega t) + C(\omega)\sin(\omega t)$ by
two basis functions.  It can be argued that complex functions
require twice as much work as real functions and so any savings
are purely imaginary (pun intended), but complex representations allow
uniform treatment while sin/cos representations do not.  Quick!
Given the response to $\cos(\omega t)$ is 
$B(\omega)\cos(\omega t) + C(\omega)\sin(\omega t)$, 
what is the response to $\sin(\omega t)$?  You have to work at it
a bit, you may need to invoke formulas such as 
$$\cos (\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$$
and so on.  With complex exponentials, life is a lot easier.

But, as in real life, your mileage may vary, and if you feel
that sin/cos representations are the way to go and complex
exponentials should be eschewed, you are free to follow your heart.
If you have difficulty communicating your ideas to colleagues, 
bosses, clients or consultants, that will be their loss, not yours.