# Is This True For Complex Numbers?

1. Nov 29, 2004

### aisha

i^57 is simplified to i ?

2. Nov 29, 2004

### Nonok

i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i

57 is divisible by 3. So, if I remember my calc class then it would be...

-i

(Don't be mad if I am completely wrong though, its just what I remember)

3. Nov 29, 2004

### aisha

Who is correct lol? i or -i? which one?

OH NO!! NOW im not sure well I divided the exponent by 4 and got a remainder of 1 which made me think that the answer is simply i
hmmm can someone tell us who is right?

4. Nov 29, 2004

i have to type some stuff to make my message longer

i^57=i

5. Nov 29, 2004

### Gokul43201

Staff Emeritus
This is correct.
$$i^{57} = i^{(56+1)} = i^{56}*i = (i^4)^{14}*i = 1^{14}*i = 1*i = i$$

Last edited: Nov 29, 2004
6. Nov 29, 2004

### Nonok

Oh, so there has to be a remainder of 1, guess I forgot that.

Sorry.

7. Nov 30, 2004

### aisha

WOW GOKU ur answer is COMPLEX!! lol
holy made me think! A simple question but a long way of simplifying it. Thanks soooo much yayay I got it right. Thanks everyone else for ur help! :tongue2: