Given:
v(t) = |v|cos(ωt)
i(t) = |i|cos(ωt+θ)
where θ = φi - φv
The following appears to be valid:
v(t)/i(t) = z(t) = |v|/|i|*cos(θ)+tan(ωt+θ) ...when θ>0
v(t)/i(t) = z(t) = |v|/|i| ...when θ=0
v(t)/i(t) = z(t) = |v|/|i|*cos(θ)-tan(ωt+θ) ...when θ<0
Ham' might be interested in proofing this claim. I haven't proofed it, but I have plotted it and the 'curves' were identical to plotting v(t)/i(t) for the example I used. Yet...
...even if it is valid, it is not much different than simply writing the equation as:
v(t)/i(t) = z(t) = |v|cos(ωt)/|i|cos(ωt+θ)
btw, I'm not saying the above has any meaningful significance
...though...
Z = |v|/|i|
...is always significant