Simplifying i^57 with complex numbers

[aisha]

i^57 is simplified to i ?

 

[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)

 

[aisha]

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

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)

OH NO! NOW I am 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?

 

[vladimir69]

i have to type some stuff to make my message longer
answer is:

i^57=i

 

[Gokul43201]

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?

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

 

[Nonok]

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

Sorry.

 

[aisha]

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

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!