Question

simplify i^23.

Answer 1

Remember how i works.
\[i = \sqrt{-1}\]\[i^2 = -1\]\[i^3 = -\sqrt{-1}\]\[i^4 = 1\]Knowing this, divide the exponent by 4 and keep the remainder. Use the remainder to figure out what you get :)

Answer 2

it doenst equal out to an even number .

Answer 3

\[\large{i^{23}=(i^3*i^3*i^3*i^3*i^3*i^3*i^3*i^2)}\]
\[\large{i^{23}=(i^3)^7*i^2=-(i^3)^7}\]

Answer 4

what remainder do you get after dividing 23 by 4?

Answer 5

5.75

Answer 6

so 23 = 5*4 + 3
correct?

Answer 7

yes

Answer 8

and as @Calcmathlete has showed you above, \(i^4=1\) which is why you can ignore all multiples of 4.

Answer 9

which means your final answer should just be i raised to the power of whatever remainder you got after dividing by 4.

Answer 10

\[\large{-(i^3)^7=(\sqrt{-1})^7}\]
\[\large{\sqrt{-1}*\sqrt{-1}*\sqrt{-1}*\sqrt{-1}*\sqrt{-1}*\sqrt{-1}*\sqrt{-1}}\]
\[\large{i}\]

Answer 11

y would i raise it to a decimal ?

Answer 12

not a decimal - the remainder which is 3 in this case.

Answer 13

So as @asnaseer said, when you divide 23 by 4 and get the remainder, this is really what you're doing.
\[i^{23} = i^4 \times i^4 \times i^4 \times i^4 \times i^4 \times i^3 = 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times i^3 \implies i^3\]What is \(i^3\) in my chart above?

Answer 14

because 23 = 5*4 + 3

Answer 15

so 3 is the remainder here.

Answer 16

@Calcmathlete −1√-1 ?

Answer 17

Yup. Which is also known as -i because \(i^3 = -\sqrt{-1} = -i\)

Answer 18

you have the right answer although you would usually leave \(\sqrt{-1}\) in the form of \(i\) as Caclmathlete has shown ^^^

Answer 19

okay , i gotcha ! thanks everyone (:

Answer 20

yw :)

Answer 21

np :)