acinput-impedance
I want to calculate the input impedance in the following circuit:
simulate this circuit – Schematic created using CircuitLab
From the leftmost loop I have that
$$U_{in}+I_{in}R=V_1$$
so it seems like I need to express V1 in terms of Uin, but I am unable to do that. How should I approach this problem?
Can I replace the driven current source with an equivalent voltage source in series with the resistor? And then add the two voltage sources which would then be in series?
From what I see your analysis is correct but for some reason you have an offset in the inductor current (it goes from 0A to 20A, when it should oscillate between -10A and 10A, just like the cap). I would guess that is the issue, maybe some simulation setup. I don't know where it is coming from, but I was able to simulate the circuit in LTSpice and I get the results you are expecting. 
This circuit model is a logical impossibility.
The control current of the CCCS is the same as its output current. If the gain of the CCCS were anything other than 1, it would be obvious why this is a problem: a current can't be \$Z\$ times itself for any value of \$Z\$ other than 1. (except of course if the current happens to be 0).
Setting the gain to 1 makes this logical contradiction seem to disappear, but the circuit model is still a poor one --- in any real circuit there'd be uncertainty about what that gain is exactly.
So, there's no more sense to trying to reason about this circuit than there is to contemplating the equation
$$3 = 2\times{}5$$
Edit
The question has been clarified to indicate this is meant to be a simplification of the hybrid pi model of a BJT. Here's the usual hybrid pi model:
(source: Wikimedia)
The issue with your model is resolved by including the \$r_o\$ element.
Best Answer
Your equation has a sign error.
Applying KCL at the \$U_{in}\$ node, see that the current through \$R\$ is from right to left which implies
$$U_{in} - I_{in} R = V_1$$
In fact, and more generally, recognize that the parallel resistor and controlled current source can be replaced with a resistor of resistance \$-R\$.
To see this, look at a similar pair in isolation driven by a current test source:
simulate this circuit – Schematic created using CircuitLab
The test source "sees", in this case, a \$-1\Omega\$ resistance.
To find the input impedance, use a similar approach to find the equivalent impedance of the circuit to the right of the negative resistor and then you will have reduced the circuit to a simple circuit with two series impedances. The input impedance will then be obvious.