Question

any idea on how to solve this?

Answer 1

\[(\frac{ i^3 }{ 3 })^3\]

Answer 2

\(\huge (\frac{a}{b})^x=\frac{a^x}{b^x} \\ (a^b)^c=a^{bc}\)

Answer 3

\(\huge (\frac{ i^3 }{ 3 })^3=\frac{(i^3)^3}{3^3}=?\)

Answer 4

that would be i^9/27 ?\[\frac{ i^9 }{ 27 }\]

Answer 5

yup, thats correct :)
now is 'i' just a variable here or square root of -1 ??

Answer 6

its a square root of -1

Answer 7

so u know what is i^2 ?

Answer 8

its -1

Answer 9

and i^3 = ?

Answer 10

and i^4 = ?

Answer 11

1^3 = -i
i^4 = 1

Answer 12

now we write i^9 as i*(i^4)^2
got this ?

Answer 13

so it would be a -i ?

Answer 14

why - *minus* ?

Answer 15

as in -i/27

Answer 16

i^4 =+1
its square = +1
1*i=i

Answer 17

so it would be i/27

Answer 18

ohhh well its because i thought that i to the power of an odd number ( i.e., i^9 ) would equal a negative i

Answer 19

now u got it ?

Answer 20

i mean any more doubts ?

Answer 21

yeah, so is there no such thing as a -i ?

Answer 22

yes it it,
i^-1 = -i
i^3 = -i
i^7 = -i
i^11= -i
i^15 =-i
....
did u get the pattern ?

Answer 23

umm i think. not too sure.

Answer 24

after every 4th power, the value gets repeated,
like i^0 = i^4 = i^8 = i^12 =.... = 1
i^1 = i^ 5 = i^9 =.....=i
and so on.
ok ?

Answer 25

okay yeah thats what i thought, i just didnt want to sound dumb lol thanks!

Answer 26

welcome :)