Question

What is the simplified version of.....

(see problem below)

Answer 1

yes another root , exponent prob maybe!?

Answer 2

\[i^{17}\sqrt{-400}\]

Answer 3

okay,

Do you remember the property that i^2 = -1

Answer 4

yes

Answer 5

if you list them out in order

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

i^5 = i^4*i = i
i^6 = i^4 *i^2 = -1
... and so on

Answer 6

so what is 17 divided by 4?

Answer 7

since the cycle repeats every 4 lines, i, -1, -i, +1 ....

Answer 8

I believe its 4.25

Answer 9

right, so you are left with remainder 1/4 right

Answer 10

what is the first entry in the series

Answer 11

i , -1 , -i , +1

Answer 12

so
i^17 = i

Answer 13

just divide by 4, and you can figure out which of the 4 answers the i^some power will be

Answer 14

ok thank you so much!

Answer 15

if you had remainder 2/4, then i^(power) would be -1

you see what i mean?

Answer 16

yes I do

Answer 17

since the cycle repeats every 4 increases in power

Answer 18

alrighty, so you have i^17 = i, you can rewrite that now

\[i ^{17}\sqrt{-400} = i \sqrt{-400}\]

Answer 19

now for the root...You know there will be an 'i' because it is a negative number under the root

\[\sqrt{-1 * 400} =\sqrt{-1}\sqrt{400} = \sqrt{-1}\sqrt{20^2} = \sqrt{-1}*20\]

Answer 20

and \[i^2 = -1 ~~~~or~~~i=\sqrt{-1}\]

Answer 21

overall you get, i*(20i) = 20i^2

i^2 = -1 so

20i^2 = -20

Answer 22

all you have to remember is
\[i^2 = -1~~~or~~~i=\sqrt{-1}\]

everything else works out from that

Answer 23

If you forget how to reduce i^473 for example,, write out the series first

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

i^473 = i^(4*118 + 1)

remainder of 1, so

i^473 = i

Answer 24

thank you so much!!

Answer 25

welcome, anytime