I assume you want to simplify each one
the imaginary number i has very unusual properties
watch this cyclic pattern:
i
i^2 = -1
i^3 = i(i^2) = -i
i^4 = i^2 i^2 = (-1)(-1) = +1
i^5 = i i^4 = i(1) = i , and here we go round again
notice every even power of i is either +1 or -1
if the even power is divisible by 4 we have +1
if the even power is NOT divisible by 4, e.g. 6 , we get -1
if dividing the exponent by 4 leaves a remainder of 1 we get +i
if dividing the exponent by 4 leaves a remainder of 3, we get -i
so i^42 ???
42 รท 4 is not exact, but 42 is even , so
i^42 = -1
of course 2 i^7 will give us 2(-i) = - 2i
So what do you think about i^100 ?
2i^7
i^42
i^100
3 answers
so on 2^7i list i as -1 seven times equals out -1 or -i then what
i still cant solve i^100
i still cant solve i^100
were did you get 2^7i from ???
I thought we did 2 i^7
and I thought I did that for you.
i^100
= (i^4)^25
= (+1)^25
= 1
I strongly suggest your write down the first 4 or 5 terms of the pattern I gave you and I strongly suggest you read my explanation of that pattern.
I thought we did 2 i^7
and I thought I did that for you.
i^100
= (i^4)^25
= (+1)^25
= 1
I strongly suggest your write down the first 4 or 5 terms of the pattern I gave you and I strongly suggest you read my explanation of that pattern.