How do you simplify i^20?

2 Answers
Surya K.
Mar 20, 2018

1

Explanation:

We have:

i^1=i

i^2=-1

i^3=-i

i^4=1

i^5=i

i^6=-1

i^7=-i

i^8=1

You see a pattern? Let's try to define it mathematically.

Let's define a function. For any division function, our function will return its remainder. We call our function zeta(u/v), where v!=0. For example,

zeta(8/2)=0, or:

zeta(15/4)=3

Note: For these imaginary numbers sequences, ninNN.

Anyways, we find zeta(n/4) for any i^n. If:

  • zeta(n/4)=0, we know that i^n=1

  • zeta(n/4)=1, we know that i^n=i

  • zeta(n/4)=2, we know that i^n=-1

  • zeta(n/4)=3, we know that i^n=-i

Lets confirm this. Take i^6. We know that i^6=-1. According to our "remainder function,"

zeta(6/4)=2, and we confirm that i^6=-1.

Let's tackle our problem. Here, n=20, so we have:

zeta(20/4)=0. Looking at our key, we find that when zeta(n/4)=0, i^n=1.

So i^20=1.

Alan N.
Mar 20, 2018

1

Explanation:

Given z in CC = i^n : n in NN

z=+1 :forall n mod_4 =0

z=+i : forall n mod_4 =1

z=-1 :forall n mod_4 =2

z=-i : forall n mod_4 =3

In this case n=20

20 mod_4 =0 -> z =1

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