Question

Evaluate i^12
I want to know how to work this problem out please.

Answer 1

\[i^{12}=(i^2)^6=\dots\]

Answer 2

\[\large i^2 = -1\]

Answer 3

\[i^1 = i\]
\[i^2 = -1\]
\[i^3 = -i\]
\[i^4 = 1\]
does that help?

Answer 4

Thank you. Could you explain if it would be different for i^49 also please?

Answer 5

are you asking what the value of that is?

Answer 6

just divide 49 by 4...then tell me the remainder

Answer 7

It is Intermediate algebra. The directions just state to evaluate.

Answer 8

12 remainder 1

Answer 9

right so 1 is the remainder

Answer 10

now look at the thing i posted earlier which corresponds to i^1

Answer 11

\[i^{49}=(i^4)^{12}\times i\]

Answer 12

ok now that post makes sense thank you both.

Answer 13

Now take out the remainder as the power of \(i\),,

\[\large i^1 = ??\]

Answer 14

i^49 = i correct

Answer 15

Yes...

Answer 16

sorry -i

Answer 17

How ??

Answer 18

i^1 = i

according to that post a while ago

Answer 19

\[i*i ^{48}\]
\[1*i=-i\]

Answer 20

After finding the remainder just make the remainder as the power of \(i\)

Here you got remainder as 1;

\[\large i^1 = ??\]

Answer 21

\[\huge 1 \times i = i \ne -i\]

Answer 22

Not sure where I got the negative from. But your explanation makes more sense.

Answer 23

So it is \(i\) and not \(-i\)..

Answer 24

\[\Large i \times i^{48} \implies i \times (i^2)^{24} \implies i \times (-1)^{24} \implies i \times 1 \implies i\]

Answer 25

does that make sense?

Answer 26

Yes it does.