How can a complex exponential represent a real world quantity? Equations containing complex exponentials are mysterious. The complex exponential merely embodies a complex number but in a more compact form where doing maths is easier. Right? If this complex exponential represents a sinusoid then why can't we just write it as a $A\sin(\omega t+\varphi)$ rather that a weird exponential with $j$ in the power?
OK, lets take example of Fourier analysis and other fields of electronics where this is used. I am confused what complex exponentials actually mean and why they show up in mathematical equations in engineering.
Somebody may answer and say that Oh, the complex exponential can be visualized as a helix. Well, sure, but what does it mean when it is used in an equation containing physical quantities as it is used in engineering for example?
 A: The word "real", in "real number", is a misnomer!!  Don't take it literally.  So is "imaginary".  Imaginary numbers are just as real as real numbers are.
Conventionally the way people seem to be taught complex numbers is that they are told that $i^2=-1$, and then go on from there by using algebra.  That's not how I first learned it.  I was surprised when I found out it could be done that way.  I was taught initially that multiplication by $i$ is $90^\circ$ counterclockwise rotation. Look at $\{i^n : n=\ldots,-3,-2,-1,0,1,2,3,\ldots\}$.  You see circular motion.  In an exponential function of $n$.  That's the beginning of understanding of $x\mapsto e^{ix}$ as circular motion coming from an exponential function.  That's a crucial basic idea from which Fourier analysis flows.  And Fourier analysis is relied on heavily not only be electrical engineering (which you mention) but number theory, wave motion, heat flow, isoperimetric problems, probability, statistics, quantum physics, number theory, cryptography, and many things.
Circular motion is real.
A: Euler's Formula says
$$
e^{ix}=\cos(x)+i\sin(x)\tag{1}
$$
Indeed this is a compact way to write the point on a unit circle at angle $x$. However, a lot of the importance follows from the other laws governing exponentials that extend here. For example
$$
e^{ix}e^{iy}=e^{i(x+y)}\tag{2}
$$
and
$$
\frac{\mathrm{d}}{\mathrm{d}x}e^{ix}=i\,e^{ix}\tag{3}
$$
For starters, equation $(2)$ can be used to derive the formula for the sine and cosine of a sum, and equation $(3)$ yield the derivatives of sine and cosine.
Furthermore, the Fourier Transform of a periodic function can be written as
$$
\hat{f}(k)=\int_0^1f(x)\,e^{-2\pi ikx}\,\mathrm{d}x\tag{4}
$$
which is simpler than the two separate formulas using sine and cosine.
It is probably true that without using complex exponentials, we might be able to derive the formulas derived using them; however, the process would almost certainly be more complicated and the resultant formulas would not be as simple.
A: Complex exponentials provide a convenient way to combine sine and cosine terms with the same frequency. For example, if not both $A$ and $B$ are $0$,
$$
       A\cos(kt)+B\sin(kt)= \sqrt{A^2+B^2}\left[\frac{A}{A^2+B^2}\cos(kt)+\frac{B}{\sqrt{A^2+B^2}}\sin(kt)\right].
$$
Because the sum of squares of the coefficients of $\cos(kt)$ and $\sin(kt)$ is $1$, then there exists $\phi$ such that
$$
       \frac{A}{A^2+B^2}=\cos(\phi),\;\;-\frac{B}{A^{2}+B^{2}}=\sin(\phi)
$$
Then
$$
   A\sin(kt)+B\cos(kt) = \sqrt{A^{2}+B^{2}}\cos(kt+\phi),
$$
which can be compactly written as a real part
$$
               \Re (\sqrt{A^{2}+B^{2}}e^{i(kt+\phi)}) = \Re (Ce^{ikt}),
$$
where $C=\sqrt{A^{2}+B^{2}}e^{i\phi}$ is a complex constant. This allows the combinations of sin and cosine terms to be greatly simplified by "lifting" to the comlex plane, performing calculations, and then taking the real part when you're one. This works very well when dealing with differential equations of Engineering where the coefficients of the equations are real. It's a way of simplifying a general linear combination of sin and cosine of the same frequency as the real part of a simple exponential with a complex coefficient. You have a choice of dealing with
$$
            f \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}a_n\cos nt+b_n\sin nt,
$$
where
$$
         a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(y)\cos(ny)dy,\;\;\;
         b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(y)\sin(nx)dy,
$$
or dealing with
$$
                        \sum_{n=-\infty}^{\infty}c_n e^{int}
$$
where
$$
                      c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(y)e^{-iny}dy.
$$
You can always get back to the real by taking the real part. The sums are simpler and there are no special cases for the exponential coefficients. This is basically how the Math evolved to what it is: everything is simplified by using complex exponentials. The only confusing part is the insistent desire to interpret the complex bits in a physical way, instead of recognizing that complex formulations are great because they provided an enormous simplification in the computations required to solve problems, which should be reason enough. The sin and cosine terms are eigenfunctions of second differentiation, whereas the complex exponentials are eigenfunctions of differentiation. By complementing the real part with a corresponding imaginary component, the Math is easier, and you can recover at the end by taking the real part.
A: Many trigonometric problems may be easily solved if you replaced the trig with exponentials.
For example, $\sum_{n=0}^x\cos(n)$ can be changed into the "real" part of $\sum_{n=0}^xe^{ni}$, which is then found to be a very simple geometric sum.
We can do the exact same thing with sine.  Also, many calculus related trig problems can be easily solved as exponentials.
So in engineering, a problem may be rewritten as an exponential to make the solving step much easier, followed by taking out the part that isn't wanted.
A: Adding to Michael Hardy's answer, multiplying by $i$ only means you're trying to rotate something by $90^{\circ}$. It doesn't mean that the quantity doesn't exist, it just means you're achieving a rotational motion mathematically. A really simple (and visual) explanation can be found here.
Now substituting values for $\theta$ in $e^{i\theta}=\cos{\theta}+i\sin{\theta}$ results in a unit circle plotted by points that are rotated in the counter-clockwise direction! $e^{-i\theta}$ just means you're achieving clockwise rotation.
