Write in the form of a + bi the expression i^(3217) - i^(427) + i^(18)?

Question

anyone please help

Answer 1

i^3217= i^(3216+1) = i^3216*i=1*i = i

i^427 = i^(424+3) = i^424*i^3= 1*(-i) = -i

i^18 = i^(16+2) = i^16*i^2 = 1*(-1)= -1

i-(-i)-1 = 2i-1

Answer 2

it is important to consider that there are 4 cases of i.
i=i
i^2=-1
i^3=-i
i^4=1

Therefore, for i raised to any power, you can reduced it based on the fact that i^5=i and i^6=i^2 (it repeats the pattern) Notice that if you divide the exponent by 4, the equivalent expression is equal to i raised to the remainder. For your problem, i^(3217)-i^(427)+i^(18).
3217/4= 806 with remainder 1
427/4= 106 with remainder 3
18/4= 4 with remainder 2

therefore, the original expression is equivalent to i^1-i^3+i^2. Using the information above, we see that i^1=i, i^3=-i, and i^2=-1. So we get i-(-i)-1 or 2i-1 as the final answer.

Answer 3

look at the following table
i^0=1
i^1=i
i^2=-1
i^3=-i
i^4=1 :So we have a cycle of 4
Divide each exponent by four and take take the remainder
3217=804*4+1 so i^3217 = i^1=i
427=104*4+1 so i^427 =i^1=i
1804*4+2 so i18 = i^2=-1
so you get -1+2i

Answer 4

i^3217 - i^426 + i^18
i^(3216+1)-i^(424+3)+i^(16+2)
i^3216*i^1-i^424*i^2+i^16*i^2
i^4n=1
therefore,
[i^(4*804)*i]-[i^(4*106)*i^3]+[i^4*i^2]
i-i^3+i^2
i-(-i)-1
-1+2i