From the equation

Vrms = square root of ( (1/T) * integration(v(t)^2))

Integration boundaries: 0 to T

even if v(t) is under the x-axis due to charging a source v(t) ^ 2 will be above the curve and the integration should always be positive.

Therefore, no imaginary parts are to exist.

(even in case some miraculous happened that the integration is negative should Vrms be just imaginary not cmplex)

I hope someone help!