Question

I may need help again finding i^31 and i^413, anyone know how to do this?

Answer 1

\[i^1 = \sqrt{-1}\]
\[i^2 = -1\]
\[i^3 = -i\]
\[i^4 = 1\]
\[i^5 = i = \sqrt{-1}\]
The pattern repeats after four terms. i^31 has to be one of those four terms, you just have to use the pattern to see which one it is.

Answer 2

\[i ^{31}\]

Answer 3

Use Joe's help, now divide 31 by 4 because you want to know at what stage in this sequence \[i^{31}\] falls

Answer 4

the mixed number is 7and 3/4

Answer 5

yep, it's not exact, so you can look at 31 as x + y*4

Answer 6

you can cycle through this, at i^4 you get 1, at i^8 you get 1, at i^12 you get 1, at i^16 you get 1

Answer 7

at i^32 you 1

Answer 8

i^32 = 1, i ^31 is thus the previous value in the cycle

Answer 9

31 - 7*4 = 31-28 =3

Answer 10

i^3 is equal to i^31

Answer 11

so, how many times can 4 go into 31 without leaving a fractional amount? 31/4 was 7 3/4

Answer 12

so 4 goes into 31, a total of 7 times before giving a fractional amount

Answer 13

x + y*4 = 31, you found y to be 7, so x + (7)*(4) = 31, therefore x = 3

Answer 14

now you find i^3

Answer 15

so let's check out \[i^413\]

Answer 16

\[i^{413}\]

Answer 17

413 = x + y*4, so, 4 goes into 413 how many times before obtaining a fractional amount?

Answer 18

just take the integer from 413/4, then multiply that by 4 and subtract it from 413, then raise i that value you discovered

Answer 19

0

Answer 20

How did you obtain 0?

Answer 21

413/4 x 4 -413

Answer 22

that is true, but not the right steps to take

Answer 23

413/4 = 103 1/4

Answer 24

so take the integer, which is 103

Answer 25

now, you do 413 - (103)*4, the value you find is what you will raise \[i\] to

Answer 26

\[i^{(413-integer from (413/4)}\]