Is This True For Complex Numbers?

AI Thread Summary
The discussion centers on simplifying the expression i^57. Participants clarify that i^57 simplifies to i, using the properties of complex numbers where i^2 = -1 and the cyclical nature of powers of i. One contributor explains the calculation by dividing the exponent by 4, leading to a remainder of 1, confirming the result is i. The conversation highlights the importance of understanding the cyclical pattern in powers of i. Ultimately, the correct answer is established as i.
aisha
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i^57 is simplified to i ?
 
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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)
 
Who is correct lol? i or -i? which one?

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

i^57=i
 
aisha said:
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
 
Last edited:
Oh, so there has to be a remainder of 1, guess I forgot that.

Sorry.
 
Gokul43201 said:
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! :-p
 

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