How can I simplify a prime power of i?

You don't have enough information, actually, because primes can be 1 more or 3 more than a multiple of 4.

Use this cycle:

i^1 = i

i^2 = -1

i^3 = -1 * i = -i

i^4 = -i * i = 1

i^5 = i^4 * i = i

i^6 = i^4 * i^2 = i^2 = -1

i^7 = i^4 * i^3 = i^3 = -i

i^8 = i^4 * i^4 = 1

So the powers cycle through {i, -1, -i, 1}. Basically, break down the exponent by dividing it by 4 with a remainder.

So, i^53 = i^52 * i, since 52 is the largest multiple of 4 that's under 53.

i^53 = i^52 * i = (i^4)^13 * i = 1 * i = i

You can't do anything with the fact that the exponent is prime, but you can rewrite the exponent as either:

4k

4k+1

4k+2

4k+3

Basically what you need to do when simplifying a power of i (doesn't have to be a prime number as long as it's an integer) is to find the remainder when that power is divided by 4.

Powers of i work like this:

i^1 = i

i^2 = -1 (by definition)

i^3 = (i^2)i = (-1)i = -i

i^4 = (i^2)(i^2) = (-1)(-1) = 1

i^5 = (i^4) i = 1 i = i

i^6 = (i^4)(i^2) = (1)(-1) = -1

i^7 = (i^4)(i^2)i = (1)(-1)i = -i

i^8 = (i^4)(i^2)(i^2) = (1)(-1)(-1) = 1.

That sequence continues for ever. If the power that i is raised to is a multiple of 4, it simplifies to 1.

If the power that i is raised to is one more than a multiple of 4, it is i. If it two more than a multiple of 4, it is -1 and if it is three more than a multiple of 4 it is -i.

Now when 53 is divided by 4, has a remainder of 1. So i^53 = i^1 = i.

i = √(-1) = i

i² = -1

i³ = -1*√(-1) = -√(-1) = -i

i⁴ = √(-1)*√(-1)*√(-1)*√(-1) = -1 * -1 = 1

and it keeps on going in that pattern as you go up

i, -1, -i, 1, i, -1, -i, 1, i, -1 . . . . . forever

so to find out which one 53 is, you divide 53 by 4

53/4 = 13.25

from this you know it reduces to i

how?

you look only at the decimal portion

and compare it to the decimal portions of these

1/4 = .25

2/4 = .5

3/4 = .75

4/4 = a whole number, so nothing after decimal place

and since it had a decimal of .25, then it has the same value as i¹

see, the 1 from the 1/4 came from the i¹

hope this helps

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