Calculating higher powers of i?

Question

I know there are many ways to calculate higher powers of i. Could anyone please tell, and show, me which one works best for you. Which method do you find the easiest and fastest to use.
For example in calculating

i to the 23rd power
i to the 100th power and so on.

Thank you in advanced. Your help is very much appreciated.

Answer 1

i= sqrt -1
i2= -1
i3=-1 sqrt-1
i4=1
i5=sqrt -1
i6=-1
i7=-1sqrt-1
i8=1
this goes on forever so ever by patterns of 4

Answer 2

It is very simple once you understand. First off, know that i=the square root of -1, i^2=-1, i^3=-1times the square root of -1, and i^4=1. So, no matter what you are raising i by, divide that number by four. the remainder will give you what you are looking for. So the example i^23=i^20 times i^3. i^20=(i^4)^5, which is 1^5, which is 1, so now it it 1 times i^3, which is -1times the square root of -1. i^100=i^4)^25, so the answer is 1.

Answer 3

A trick my teachers used to say was "I won I won" as in i 1 i 1 and just remember the middle two are negative.
With that in mind, take the power that the i is to and divide by four.Then take the remainder and see what it is. That is the power to get your answer from.
I.E. 23/4= r=3 so i^3=-i
so your answer would be -i

Answer 4

Not all equipment runs at full load. For example, a centrifugal pump will run at minimum load with its shutoff valve closed. It will merely spin (and heat) the water inside it. The pump may be required to start "shutoff" to minimize starting current. As the valve is opened the pump does more work (perhaps moving fluid up two levels?) and draws more current and power. Other equipment (fans, heaters and a/c's) may have high, medium and low settings. Computers may have "sleep modes" when not in use. Nameplate data helps prevent overloading a circuit, but you must also know the "duty cycle" for any given piece of equipment.

Answer 5

i repeats itself in powers of four:
i=i
i^2=-1
i^3=-i
i^4=1
i^5=i
i^6=-1
i^7=-i
i^8=1
And so on...
Notice that the pattern goes by 4s. So divide the power by 4, and if the remainder is 0, the answer is 1. If the remainder is 1 (0.25), the answer is i. If the remainder is 2 (0.5), the answer is -1. If the remainder is 3(0.75), the answer is -i.
For example, i^23. 23/4 is 5.75, so the answer is -i. 100/4 is just 25, so the answer is 1.

Answer 6

Hi,
I sometime use the "i" function of a graphing calculator, but frankly, I usually just divide the exponent by 4 using any calculator. I interpret the decimal point in the following way.

Decimal .....Result of power.
0....................1
.25..................i
.50..................-1
.75..................-i

For example, i^23:
23/4=5.75. So, i^23 = -i.
You don't need to remember the decimals, it's easy to see that those decimals correspond to remainders of 0, 1,2,and 3 from the division.

FE

Answer 7

well i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
therfore i ^ 5 = i ^ 4 * i = i
this patter repeats.. ie i -1 -i 1 i -1 -i 1 etc.

so the 0th, 4th 8th power is 1 etc
1st 5th 9th etc is i
2nd 6th 10th etc is -1
3rd 7th 11th etc is - i

hope this helps

Answer 8

i^2 = -1

i^3 = -i

i^4 = 1

i^5 = i

for even numbers

if n is even and divisible by 4 ( i)^n = +1

if n is even but not divisible by 4 (i)^n = -1

if n is odd write n = 2m+ 1

(i)^(2m+1)

if m is odd , (i)^2m+1 = -i

if m is even , (i)^2m+ 1 = +i

examples for odd numbers

(i)^23 => i^(2*11+1) = -i (since 11 is odd)

(i)^25 => i^(2*12 + 1) = + i ( since 12 is even)

examples for even numbers

(i)^100 = + 1 ( since 100 is divisible by 4)

(i)^50 = -1 ( since 50 is even but not divisible by 4)

Answer 9

exp(i*pi) = -1,

exp(i*pi/2) = i

exp(i*pi*n/2) = cos(n*pi/2) + isin(n*pi/2) = i^n

eg, n=0, 1

n=1, i

n=2, -1

n=3, -i . . .

n=23, -i

n=100, 1

Answer 10

i = e^(i Pi/2)

i ^(n) = {e^(i Pi/2)}^(n) = e^(i nPi/2)

I like this better!

n =23 ; i^23 = e^(i 23Pi/2) = cos (23 Pi/2) +i sin(23Pi/2) = i sin(23 Pi/2) = i sin(12 Pi - Pi/2) = -i sin(Pi/2) = -i

i^100 = cos(100 Pi) + isin(100 Pi) =
cos(100 Pi) = 1