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?

Answer

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.