Electronic – Impedance across AC circuit

Since I am exciting with voltage and reading the current I assume I have a ratio of current to voltage which is admittance.

More specifically, you can call this a trans-admittance, since the voltage is applied at one port of the filter and the response current is measured at a different port.

If you just talk about an "admittance" people will at least at first assume you are talking about the voltage/current relationship at a single port.

Do I convert to current correctly? Do I find conductance and susceptance correctly? Did I understand correctly that the values I measure is admittance?

I didn't follow through your calculations entirely, but you did not come up with the correct result for this case. Your result may be correct for the normal admittance to impedance conversion for a one-port network.

For converting transadmittance to transimpedance you have to consider both ports of the network.

The admittance representation is like this:

\$\begin{bmatrix}I_1 \\ I_2\end{bmatrix}=\begin{bmatrix}Y_{11} & Y_{12} \\ Y_{21} & Y_{22}\end{bmatrix}\begin{bmatrix}V_1 \\ V_2\end{bmatrix}\$

To get the Z-parameters you need to invert the Y matrix. Rather than calculate the inverse by hand I looked it up on Wikipedia, where I found the result for the Z₂₁ term is

\$Z_{21} = \frac{-Y_{21}}{Y_{11}Y_{22}-Y_{12}Y_{21}}\$

Separating this out into real and imaginary parts is of course another algebraic effort, which I'd much rather do using software (Mathematica or whatever) than sort through by hand.

Assuming this equation is correct: $$Z = \frac{R}{1+i \omega cR} + i \omega L$$

We multiply the first term's nominator and the denominator by the complex conjugate of the denominator: $$.. = \frac{R(1-i\omega c R)}{(1+i \omega cR)(1-i\omega c R)} + i \omega L$$ and thus getting rid of imaginary stuff in the denominator: $$.. = \frac{R(1-i\omega c R)}{(1+ \omega^2 c^2R^2)} + i \omega L$$ And this thing is easily separated to real and imaginary parts: $$.. = \frac{R}{1+ \omega^2 c^2R^2} + i \left(\omega L - \frac{\omega c R^2}{1+ \omega^2 c^2R^2} \right)$$