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#

Related questions
See all questions in Powers of Complex Numbers
Impact of this question
12890 views around the world
You can reuse this answer
Creative Commons License