When working with complex numbers, how can you solve for $x$ when it's inside $Re()$? I'm trying to figure out the impedance of a capacitor. My textbook tells me the answer is $\frac{-i}{\omega C}$ and plugging that into the equation does work but I wanted to come up with that answer myself. So I wrote out the equation with what I know:
$$-V_0\omega C\sin\omega t = Re\left( \frac{V_0(\cos\omega t + i\sin\omega t)}{x} \right)$$
This is where I get stuck. I don't know how to isolate $x$ given that it is inside the $Re()$ function. Trying to get somewhere, I tried this:
$$x = \frac{V_0(\cos\omega t + i\sin\omega t)}{-V_0\omega C\sin\omega t} = \frac{\cos\omega t}{-\omega C\sin\omega t} - \frac{i}{\omega C}$$
Seeing $-\frac{i}{\omega C}$ makes me feel like I'm on the right track. Now I just need to figure out how to get rid of the first part of that answer. And I'm guessing that if I knew how to isolate $x$ from the first equation, that would do the trick. So how can I isolate $x$ when it is included in the $Re()$ function?
 A: Re() is a projective map; Re(a+bi) = a. Thus Re(z) = z-iIm(z). So given RHS = Re(z), we have that RHS+bi = z for some real b. Note that Re(z) = a does not yield a single value of z as a solution, but instead gives a vertical line in the complex plane. Each point on that line will give a different value for x.
$$-V_0\omega C\sin\omega t + bi = \frac{V_0(\cos\omega t + i\sin\omega t)}{x} $$
In terms of b, x will be:
$$\frac{V_0(\cos\omega t + i\sin\omega t)}{-V_0\omega C\sin\omega t + bi} $$
Assuming that $\omega$, $V_0$, and C are real numbers, they can be "absorbed" into b; b is an arbitrary real number, so dividing by a real number just gives another arbitrary real number. So the above can be rewritten as
$$\frac{(\cos\omega t + i\sin\omega t)}{-\omega C(\sin\omega t + bi)} $$
Factoring an i out of the numerator, we get
$$\frac{i(\sin\omega t-i\cos\omega t )}{-\omega C(\sin\omega t + ib)} $$
Again, this describes a solution set, not a particular x. But if you take $b = -\cos\omega t$, then you recover the given expression. Any motivation for that choice will have to come from further facts about the capacitance rather than mathematical properties.
A: For a capacitor, there is the relation:
$$\text{I}_\text{C}\left(t\right)=\text{C}\cdot\frac{\text{d}\text{V}_\text{C}\left(t\right)}{\text{d}t}\tag1$$
Considering the voltage signal to be:
$$\text{V}_\text{C}\left(t\right)=\text{V}_\text{p}\sin\left(\omega t\right)\tag2$$
It follows that:
$$\frac{\text{d}\text{V}_\text{C}\left(t\right)}{\text{d}t}=\omega\text{V}_\text{p}\cos\left(\omega t\right)\tag3$$
And thus:
$$\frac{\text{V}_\text{C}\left(t\right)}{\text{I}_\text{C}\left(t\right)}=\frac{\text{V}_\text{p}\sin\left(\omega t\right)}{\omega\text{V}_\text{p}\cos\left(\omega t\right)}=\frac{\sin\left(\omega t\right)}{\omega\text{C}\sin\left(\omega t+\frac{\pi}{2}\right)}\tag4$$
This says that the ratio of AC voltage amplitude to AC current amplitude across a capacitor is $\frac{1}{\omega\text{C}}$, and that the AC voltage lags the AC current across a capacitor by $90$ degrees (or the AC current leads the AC voltage across a capacitor by $90$ degrees).
This result is commonly expressed in polar form as:
$$\text{Z}_\text{c}=\frac{1}{\omega\text{C}}\cdot e^{-\frac{\pi}{2}\cdot\text{j}}\tag5$$
Or, by applying Euler's formula, as:
$$\text{Z}_\text{C}=-\text{j}\cdot\frac{1}{\omega\text{C}}=\frac{1}{\text{j}\omega\text{C}}\tag6$$
Now for $\text{X}_\text{C}$:
$$\text{X}_\text{C}=\left|-\text{j}\cdot\frac{1}{\omega\text{C}}\right|=\frac{1}{\omega\text{C}}\tag7$$
Where $\omega=2\pi\text{f}$
