Question

Simplify.
3i^23

Answer 1

you know what "i" is ?

Answer 2

no

Answer 3

its the unit imaginary number.
\(\Large i = \sqrt{-1}\)

can you find the value of i^2 now?

Answer 4

um -1?

Answer 5

thats correct! :)

\(i^2 =-1\)
what about
\(\large i^4 =(i^2)^2 =?? \)

Answer 6

umm...-1^4..?

Answer 7

or do i have to simplify more

Answer 8

you mean (-1)^4
isn't that just =1 ? :)
and yes

Answer 9

oh, and it should be (-1)^2
and not (-1)^4

Answer 10

oh i just saw that too lol

Answer 11

right,
so we have, so far
i^2 =-1
i^4 = 1

how about i^3 ?
this is a tricky one !

Answer 12

thennn would the answer be 3?

Answer 13

and no, the answer isn't 3 :)

Answer 14

aww :( i^3= -1?

Answer 15

this is confusing lol

Answer 16

how -1??
\(\Large i^3 = i^2 \times i = ...?\)

Answer 17

ohhhh 1

Answer 18

i^2 is -1

so,
\(\Large i^3 = i^2 \times i = (-1)
\times i = -i\)

makes sense?

Answer 19

oh. wow thats confuusingg. but ok :D

Answer 20

so we have

i^2 = -1
i^3 = -i
i^4 = 1

here's some easy questions
\(i^8 = (i^4)^2 = ..? \\ i^{12} =(i^4)^3 = ... ?\)

Answer 21

oh my goodness...um

Answer 22

just plug in the value of i^4 in them :)

Answer 23

1^2 = ... ?
1^3 = .. .

Answer 24

-1 and -1

Answer 25

-1?

\(1^2 = 1\times 1 =1 \\ 1^3 = 1\times 1\times 1 =1 \)

so both
\(i^8 = 1 \\ i^{12} = 1\)

Answer 26

infact if the 'i' has an exponent which is divisible by 4,
the answer is always 1

Answer 27

\(\large i^{4n} =1\)

Answer 28

ohh wow

Answer 29

ok

Answer 30

so lets break i^23
\(\large i^{23 } = i^{20}\times i^3 =...?\)

we already know i^3
can you guess value of i^20 ? :)