Complex numbers?

Question

i^6+4i^10

and

(2i +5)(3-i)

Answer 1

i^6 = -1
i^10 = -1

-1-4 = -5

2) (2i+5)(3-i) = 6i-2i^2+15-5i = i + 17

Answer 2

In general to find out i^n you'll see
i^1=i
i^2=-1
i^3=-i
i^4=1 and the cycle starts again.So to find out i^n divide n into 4 and take the remainder i raised to that remainder will give you the answer
i^6=i^2 =-1 i^10 = i"=-1 so the first=-5
2)= 17+i

Answer 3

((i)^2)^3=(-1)^3= -1
[(i)^2]^5= -1
-1-4= -5


3(2i+5)-i(2i+5)=6i+15-2i^2-5i
= i+15+2=i+17