What does it mean to have a complex impedance?
For example, the impedance of a capacitor (in the Laplace domain?) is given by 1/sC (I believe) which equates to \$ \dfrac{1}{j \cdot 2 \pi \cdot f \cdot C}\$ where transients are neglected. What does it mean for the impedance to be imaginary?
I'm currently in my 2nd year of Electrical Engineering at University so, if possible, I'd appreciate a mathematically valid and thorough response if it's not too much trouble, with the reference of study material (web and paper resources) ideal.
Thanks in advance.
Best Answer
TL;DR The imaginary part of the impedence tells you the reactive component of the impedance; this is responsible (among others) for the difference in phase between current and voltage and the reactive power used by the circuit.
The underlying principle is that any periodic signal can be treated as the sum of (sometimes) infinite sinewaves called harmonics, with equally spaced frequencies. Each of them can be treated separately, as a signal of its own.
For these signals you use a representation that is like: $$ v(t) = V_{0} \cos (2 \pi f t + \phi) = \Re \{ V_{0}e^{j 2 \pi f t + \phi} \} $$
And you can see that we already jumped in the domain of complex numbers, because you can use a complex exponential to represent rotation.
So impedance can be active (resistance) or reactive (reactance); while the first one by definition doesn't affect the phase of signals (\$ \phi \$) the reactance does, so using complex numbers is possible to evaluate the variation in the phase that is introduced by the reactance.
So you obtain: $$ V = I \cdot Z = I \cdot |Z| \cdot e^{j \theta} $$
where |Z| is the magnitude of the impedance, given by: $$|Z|=\sqrt{R^2+X^2}$$
and theta is the phase introduced by the impedance, and is given by: $$\theta = \arctan \left( \frac{X}{R} \right) $$
When applied to the previous function, it becomes: $$ v(t) = \Re \{ I_{0}|Z|e^{j 2 \pi f t + \phi + \theta } \} = I_{0} |Z| \cos (2 \pi f t + \phi + \theta ) $$
Let's consider the ideal capacitor: it's impedance will be \$ \frac{1}{j \omega C} = -\frac{j}{\omega C} \$ which is imaginary and negative; if you put it into the trigonometric circumference, you obtain a phase of -90°, which means that with a purely capacitive load the voltage will be 90° behind the current.
So why?
Let's say that you want to sum two impedances, 100 Ohm and 50+i50 Ohm (or, without complex numbers, \$ 70.7 \angle 45 ^\circ \$ ). Then with complex numbers you sum the real and imaginary part and obtain 150+i50 Ohm.
Without using complex numbers, the thing is quite more complicated, as you can either use cosines and sines (but it's the same of using complex numbers then) or get into a mess of magnitudes and phases. It's up to you :).
Theory
Some additional notions, trying to address your questions:
$$ v(t) = \sum_{- \infty}^{+ \infty} c_{n}e^{jnt} , \text{ where } c_{n} = \frac{1}{2 \pi } \int_{-\pi}^{\pi} v(t)e^{-jnt} \, dt $$
$$ cos(x) = \frac{e^{ix}+e^{-ix}}{2} $$