Zero-State Solution using Complex Numbers

Recall:
$\tau \frac{di_f(t)}{dt}+i(t)=Kcos(\omega t)$
We can rewrite this as:
$\tau \frac{di_f(t)}{dt}+i(t)=Re\left\{Ke^{j\omega t}\right\}$

We will ignore that we are taking the real part of the polar form of the complex number momentarily.
Instead, we will solve the problem as if we had a complex input voltage.
$\tau \frac{di_f(t)}{dt}+i(t)=Ke^{j\omega t}$

Then, once we find $i_f(t)$ that solves the complex input voltage, we'll remember to take the real part of this answer to find the real zero-state solution for current.

Suppose $i_f(t)=Be^{j\omega t}$
$Bj\omega \tau e{j\omega t} + Be{j\omega t} = Ke{j\omega t}$
Since $e{j\omega t} \neq 0$:
$Bj\omega \tau + B = K$
$B=\frac {K} {1+j\omega \tau}$
$i_f(t)=\frac {K} {1+j\omega \tau}e^{j\omega t}$

Now, we take the real part of this solution to find our real answer:
$i_f(t)=Re\left\{\frac {K} {1+j\omega \tau}e^{j\omega t}\right\}$
$i_f(t)=Re\left\{\frac {v_{in}\omega C} {1+j\omega \tau}e^{j\omega t}\right\}$