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# the multiples of i

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the multiples of i

i^9  i^100  i^2017

Guest Apr 15, 2017
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#1
+7203
+2

the multiples of i

i^9  i^100  i^2017

$$i=\sqrt{-1}$$

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
edited by asinus  Apr 15, 2017
#2
+6351
+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

hectictar  Apr 16, 2017
#3
+82863
+1

Notice something else:

If we use modular arithmetic.....it's easy to evaluate i to any positive integer power

Exponent mod 4  = 1    →   i

Exponent mod 4  = 2    → -1

Exponent mod 4  = 3   →   -i

Exponent mod 4  = 0   → 1

So

9 mod 4  = 1   =  i

100 mod 4  = 0  = 1

2017 mod 4  = 1  = i

CPhill  Apr 16, 2017

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