calculate the following

Question

calculate the following

i²=

i³=

i⁴=

i⁵=

i⁷⁰=

i⁰=

i^-1=

i^-2=

i^-3=

i^-48=

Answer 1

calculate the following:

i^2 = (√(-1))^2 = -1 b/c (√(a))^2 = a

 

i^3 = i(i^2) = i(-1) = -i

 

i^4 = i(i^3) = i(-i) = -(i^2) = -(-1) = 1

 

i^5 = i(i^4) = i(1) = i

 

i^70 = i^(4*17+2) = ((i^4)^17)(i^2) = ((1)^17)(i^2)= (1)(-1) = -1

 

i^0 = i^(1-1) = i^1/(i^1) = 1

 

i^(-1) = i^(1-2) = i^1/i^2 = i/(-1) = -i

 

i^(-2) = i^(1-3) = i^1/i^3 = i/(-i) = -1

 

i^(-3) = i^(1-4) = i^1/i^4 = i/1 = i

 

i^(-48) = i^(1-49) = i/i^49 = i/i^(4*12+1) = i/i = 1

Answer 2

Dalia,

        

i² =√-1 x √-1 =-1

i³ = i x i² = -i

i⁴ = i² x i² = 1

i⁵ = i x i⁴ = i

see the pattern?

i^{0 = 1}

i^-1 =1/i multiply numerator and denominator by (-i)=1(-i)/((i)(-i))= -i

i^-2 =-1

Regards

Jim

 

 

Answer 3

i^2 = -1

 

i^3 = i^2 . i = -i

 

i ^4 =( i^2 ) ^2 = (-1) ^2 = 1

 

i ^5 = (i ^4 ) i = i

 

i ^ 0 = 1

 

i ^ -1 = 1/i =( 1 . i )/ i . i = -i

 

 i ^ - 2 = 1/i^2 = -1

 

 i ^ -3 = (1/i ) ^3 = 1/ -i = 1. i / - i . i = i/ 1 = i

 

 i ^ -48 = 1/ i ^ 48 = 1/( i ^4 ) ^ 12 = 1