Q: Compute e^(1+iπ) and sin(i). Find log(i). e^(1+iπ) = e×e^(iπ) = -e. sin(i) = (e^(-1)-e^(1))/2i = (1/e-e)/2i = i(e-1/e)/2 ≈ 1.175i. (Pure imaginary! Complex trig is unbounded!) log(i) = ln 1 + iπ/2 = iπ/2 (principal!).