Question

i^-13

Answer 1

hint:
\[\large i^1 = i\]
\[\large i^2 = -1\]
\[\large i^3 = i^2 \times i - -1 \times i = -i\]
\[\large i^4 = (i^2)^2 = (-1)^2 = 1\]
\[\large i^5 = i\]
so the process repeats

does that help?

Answer 2

it is raised to a negative number though. how does that work?

Answer 3

simple..change it first to \[\huge \frac{1}{i^{13}}\]

Answer 4

I was told to always go with the i^4 so what I have so far is:
1/(i^4)^3i=1/i

Answer 5

that's right

Answer 6

so then how is the answer -i, is it the reciprocal?

Answer 7

maybe you mean \[i^{-1}?\]

Answer 8

oh no I'm confused now. LOL. how did you get i^-1

Answer 9

\[i^{-1} = \frac 1i\]

Answer 10

then how is the answer expected to be -i? I have the answer sheets for my review, I just don't know how she got to -i, when I keep getting 1/i

Answer 11

are you sure the question is \[\huge i^{-13}\] maybe it's \[\huge -i^{13}\]

Answer 12

\[\huge i^{-13} = \frac 1i\]
\[\huge -i^{13} = -i\]

Answer 13

uhmm wait...

Answer 14

I'm waiting. lol.

Answer 15

OHHH i see

Answer 16

\[\huge \frac 1i \implies \frac 1i \times \frac ii \implies \frac{i}{i^2}\implies \frac{i}{-1} \implies -i\]
it's rationalization

Answer 17

how could i forget =))

Answer 18

did you get what i did?

Answer 19

BRILLIANT!!!!!! Thanks a million.

Answer 20

welcome ^_^ just remember to do that everytime there's an i in the denominator