#1**+2 **

the multiples of i

i^9 i^100 i^2017

how to answer these questions

\(i=\sqrt{-1}\)

i^1= i

i^2=-1

i^3= -i

i^4= 1

i^5= i

\(i^9 =(i^3)^3=(-i)(-i)(-i)=(-1)(-i)=i\)

\(i^{100}=((i^5)^5)^4=(i^5)^4=i^4=1\)

**\(i^{2017}=(i)(i^{2016})\) ?**

**Excuse me. Unfortunately I do not know.**

!

asinus
Apr 15, 2017

#2**+2 **

This is how I think of it:

For example:

i^7 = i i i i i i i

Then circle each pair of i's and replace each pair with -1.

i^7 = i i i i i i i = (i i)(i i)(i i)i = (-1)(-1)(-1)i

Then circle each pair of -1's and replace each pair with 1.

i^7 = i i i i i i i = (i i)(i i)(i i)i = (-1)(-1)(-1)i = [ (-1)(-1) ] (-1)i = [1](-1)i = -i

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For i^2017 , we could do it the same way,

...but I don't feel like writing out all those i's!

But if I WERE to write them all out, and circle each pair of i's,

how many pairs would I get?

I would get 2017/2 = 1008 pairs of i's with a remainder of 1 i's.

Then replace every pair with a negative one and then __put one i at the end.__

Then if I circled each pair of negative ones, how many pairs would I get?

I would get 1008/2 = exactly 504 pairs of negative ones.

(__There are no negative ones left__ unpaired.)

Then I would replace each pair of negative ones with a positive one.

And then I would "erase" every " 1 " that I wrote because those are not necessary to write. Whatever is left = i^2017

i^2017 = i

I hope this made sense!!

hectictar
Apr 16, 2017