Getting the transfer function of a parallel RC circuit. Am I allowed to simplify/remove the component I am trying to solve for?
You are allowed to do whatever artifices, they only say it's not &quot;recommended&quot; (why? I don't know). So, one other way is by Kirchoff: solve for the currents or voltages and derive the transfer function from there. Or try differential equations, they'll be 1st order, so it won't be as complicated. There are other methods, FACTS comes to mind (Verbal Kint might have a word or two about it), but if your teachers advised you not to modify the circuit, it might not please them. I thought I'd throw in some MathJax, but it looks like you are good with the equations, so I'll leave the pleasure to you. But, if you think it's necessary, I'll oblige.

@Thriloknath palla bellow showed something similar to what I was talking about. For example, apply mesh equations, considering two (obvious) meshes, with currents \$i_1\$ and \$i_2\$ running clockwise in each:
$$\left\{
\begin{aligned}
v_i&amp;=R_1i_1+\frac{i_1-i_2}{sC} \\
\frac{i_1-i_2}{sC}&amp;=R_2i_2 \\
\end{aligned}
\right. \\
\Rightarrow \\
\left\{
\begin{aligned}
i_1&amp;=\frac{v_i(sR_2C+1)}{sR_1R_2C+R_1+R_2} \\
i_2&amp;=\frac{v_i}{sR_1R_2C+R_1+R_2} \\
\end{aligned}
\right.$$
Since the voltage across \$R_2\$ is the one that's needed, \$i_2\$ is the solution of interest, and \$\frac{v_i}{i_2}=R_2\$, thus the transfer function becomes:
$$H(s)=\frac{R_2}{sR_1R_2C+R_1+R_2}=\frac{\frac{1}{R_1C}}{s+\frac{R_1+R_2}{R_1R_2C}}$$
\$H(0)=\frac{R_2}{R_1+R_2}\$ gives you the attenuation at DC. Which is a rather more convoluted way of solving the way you did. It might pay off for higher orders, though. BTW, you didn't simplify anything, you simply considered a temporary variable holding the result of \$R_2||C\$, that's all (in case your teacher says anything).