Check your equation. Decreasing the power factor must increase the current, ie. when calculating the current you must divide by the power factor, not multiply by it
Also, the formula with Zbal is a bit abridged, but that has smaller effect on the precision of the outcome
Z = sqrt( X^2 + R^2 ), where X is the inductive impedance, R is the resistance representing the losses (most of it is plain DC resistance of the winding wire, but there is some more to account for the core)
So the ballast itself does have its own angle between Z and X, which depends on the ballast efficiency, and does not have anything to do with the lamp PF or the complete circuit PF
When calculating, the Vdrop of the ballast consists of Vdrop(X) and Vdrop(R), which together form Vdrop(Z). However, Vdrop(R) is the one which is mostly in phase with the Vdrop of the lamp arc
So, the complete Vdrop sum would be something like :
Vline = sqrt( Vdrop(X)^2 + ( Vdrop(R) + Varc )^2 )
