Long name, I understand, I love to overname things, bit of a hobby of mine actually. Anyways, to get on to the actual formula(s). The following formula is used to determine the exact value of i^n where i is the imaginary number and n is an integer.
i^n = i^R(n/4) where R = the remainder of ()
or
i^n = i^[d(n/4)]4 where d = the decimal of ()
The first one is good for those who are quick at finding remainders (as I am) and the secondary is for those who are quicker at decimals (long division). Recently in my maths B30 course (Canadian course) we have started to play around with complex numbers, something I've been doing for the past year and a half (about time the curriculum caught up). Anyways, my teacher had a few things written on the board when I walked in, these were i^103 and i^74 if memory serves me right (it's photographic so it should -.-). I automatically sat down and got to work on these questions. Unlike the rest of the class I already knew that:
i = sqrt(-1)
i^2 = -1
i^3 = -i
i^4 = 1
and then the pattern repeated, both forward and into the negative numbers. This lead me to the automatic thought that "Okay, so if I divide 103 by 4 I get 25.75, .75 = 3/4 and i^3 is 3/4 down the list". This allowed me to conclude that i^103=-i, which it does. I did the same with i^75. "Well, we have 74/4, that's .5 meaning that 1/2 down the list we have i^2". My conclusion was obviously that i^74=-1. I was, again, right. The day after this event I decided to make some formulas, so I did just that, as you can see way above.
Let's try with the first formula for i^385.
Okay:
i^n = i^R(n/4)
i^385 = i^R(385/4)
i^385 = i^1
i^385 = i this is right
Let's try the second formula with i^638. (these are just button mashed numbers by the way):
i^n = i^[d(n/4)]4
i^638 = i^[d(638/4)]4
i^638 = i^(.5)4
i^638 = i^2
i^638 = -1 This is, again, right
i^n = i^R(n/4) where R = the remainder of ()
or
i^n = i^[d(n/4)]4 where d = the decimal of ()
The first one is good for those who are quick at finding remainders (as I am) and the secondary is for those who are quicker at decimals (long division). Recently in my maths B30 course (Canadian course) we have started to play around with complex numbers, something I've been doing for the past year and a half (about time the curriculum caught up). Anyways, my teacher had a few things written on the board when I walked in, these were i^103 and i^74 if memory serves me right (it's photographic so it should -.-). I automatically sat down and got to work on these questions. Unlike the rest of the class I already knew that:
i = sqrt(-1)
i^2 = -1
i^3 = -i
i^4 = 1
and then the pattern repeated, both forward and into the negative numbers. This lead me to the automatic thought that "Okay, so if I divide 103 by 4 I get 25.75, .75 = 3/4 and i^3 is 3/4 down the list". This allowed me to conclude that i^103=-i, which it does. I did the same with i^75. "Well, we have 74/4, that's .5 meaning that 1/2 down the list we have i^2". My conclusion was obviously that i^74=-1. I was, again, right. The day after this event I decided to make some formulas, so I did just that, as you can see way above.
Let's try with the first formula for i^385.
Okay:
i^n = i^R(n/4)
i^385 = i^R(385/4)
i^385 = i^1
i^385 = i this is right
Let's try the second formula with i^638. (these are just button mashed numbers by the way):
i^n = i^[d(n/4)]4
i^638 = i^[d(638/4)]4
i^638 = i^(.5)4
i^638 = i^2
i^638 = -1 This is, again, right
.
