How do I calculate the impedance needed to drop 120V AC?
An ordinary resistor will have virtually no complex component at 50Hz- delays in the optoisolator will be far more significant. 

Your minimum value is set be the dissipation, maximum ambient temperature, maximum current rating, voltage variations and how conservative you want to be.

Your maximum resistor value is determined by the minimum CTR, load resistance, and allowances for aging and voltage variations. 

120VAC is about 170V peak. If you allow something like 40mA peak, a minimum resistor of 4.7K might be appropriate. The resistor power dissipation would be ~\$0.5 \frac{120^2}{R}\$ = 1.5W, so you might use a 2W resistor. The LED power dissipation is about \$0.5\cdot 0.9\cdot V_f\cdot \frac{120}{R}\$ = 14mW, so should be fine even at high Ta. The 0.9 factor comes from the ratio of RMS voltage to average voltage- LED power dissipation is proportional to average current since Vf is fixed to a good approximation. 

Adjust the above as per your requirements and preferences. Then you should calculate maximum allowable resistance depending on your load resistance and required switching threshold. If the maximum resistance is lower than the minimum then you have a problem. If it is much higher consider using a value in between the two. Aging is accentuated by high current and high temperature.