Input impedance of a one stage voltage amplifier?
For typical sensible design values, the input impedance is dominated by the transistor base, with some effect from the bias resistor(s). The coupling capacitor is typically chosen to have negligible impedance at the frequencies of interest, so you can usually ignore that. 



If you use the hybrid-\$\pi\$ model for the transistor, the transconductance of the transistor gm is Ic/Vt or about Ic * 40 at room temperature. 

Vt= kT/q where k is the familiar Boltzman constant, T is the temperature (Kelvin) and q is the charge on an electron. It's about 25mV at room temperature. 

The value of r\$\pi\$ is \$\beta_0\$/gm, where beta is the current gain of the transistor at the operating point (it's a parameter of the transistor and can vary over a wide range for a given part number, often more than 2:1). 

You should also take the effect of the bias resistor R2 into account, however it's not as simple as the resistor in parallel because it's affected by the dependent current sink in the transistor, so it's effectively reduced by the voltage gain of the amplifier. More detailed information in general here. Loading the amplifier output will reduce the voltage gain so the effect due to the bias resistor will be less. If you split the bias resistor into two and bypass it to ground you can get a more constant input impedance- just the r\$\pi\$ in parallel with the left part of R2 (with Xc1 in series, but usually that's negligible).  

So to pick some reasonable values, if the collector current (a design parameter) is chosen to be 1.0mA, and beta is 300, then gm = 0.04A/V and the input resistance due to the base is 300/0.04 = 7.5K\$\Omega\$. 

Higher beta (as a darlington would have) proportionally increases the input resistance due to the base. You can also increase the bias resistor value.