(3i^30-i^19)/(2i-1) ··· Eq.1
Here, i^2=-1, i^3=(i^2-i)xi=-i, i^4=(i^2)x(i^2)=1, so i^30=(i^2)^15=-1, i^19=i^(16+3)={(i^4)^4}xi^3=-i
Plug i^30=-1, and i^19=-i into Eq.1. Eq.1 restated as follows:
(3i^30-i^19)/(2i-1)={3(-1)+i}/(2i-1)=(i-3)/(2i-1) ··· Eq.2
To rationalize the denominator of Eq.2, multiply both numerator and denominator by the complex conjugate (2i+1). Eq.2 restated as follows:
(i-3)/(2i-1)=(i-3)(2i+1)/(2i-1)(2i+1)={2(i^2)-5i-3}/{4(i^2)-1}={2(-1)-5i-3}/{4(-1)-1}={-5(i+1)}/-5 =i+1
The answer is: (3i^30-i^19)/(2i-1)=i+1