# Complex numbers, All possible values of i^n

• Aug 12th 2014, 10:39 PM
andy000
Complex numbers, All possible values of i^n
Hi, Sorry if this is not in the right section - I did look but couldn't be sure.

I am trying to find all possible values of i^n

I know that the angle of i is pi/2 therefore there will be four possible values, being i, -1, -i and 1 as the number is rotated around, however I don't know how to really show this mathematically.

Any advice would be appreciated, thanks.
• Aug 13th 2014, 03:40 AM
dennydengler
Re: Complex numbers, All possible values of i^n
I think the key is cos(θ)+i sin(θ). Set that equal to i, -1, -i and 1 to figure out the angles.
• Aug 13th 2014, 03:49 AM
Prove It
Re: Complex numbers, All possible values of i^n
Do you have any restrictions on n? For example, is n an integer?
• Aug 13th 2014, 04:05 AM
dennydengler
Re: Complex numbers, All possible values of i^n
Thing is, i = cos(π)+ i sin(π), and De Movrie's Theorem says that i^n = cos(πn)+i*sin(πn). It should be good as long as n is a real number.
• Aug 13th 2014, 05:31 AM
HallsofIvy
Re: Complex numbers, All possible values of i^n
Since andy000 said "there will be four possible values, being i, -1, -i and 1" it is clear that he intended n to be an integer.

Frankly, I wouldn't use D'Moivre. I would just note that $i^0= 1$, $i^1= i$, $i^3= -i$, $i^4= 1$. Since we are at "1" again and each step just multiplies by i again that will repeat. $i^{4n}= 1$, $i^{4n+1}= i$, $i^{4n+2}= -1$, and $i^{4n+3}= -i$ for any integer n. If you want a more "formal" way of showing that, note that $i^{4n+k}= (i^{4n})(i^k)= (1)(i^k)$ so you really only need the first four values.
• Aug 13th 2014, 08:05 AM
Deveno
Re: Complex numbers, All possible values of i^n
Quote:

Originally Posted by andy000
Hi, Sorry if this is not in the right section - I did look but couldn't be sure.

I am trying to find all possible values of i^n

I know that the angle of i is pi/2 therefore there will be four possible values, being i, -1, -i and 1 as the number is rotated around, however I don't know how to really show this mathematically.

Any advice would be appreciated, thanks.

Your description lacks formality, but that is really all there is to it.

Let's clean this up a bit:

Suppose n = 4k + m, where m = 0,1,2 or 3. Every integer n falls into exactly one of these four categories.

Now $i^0 = 1$, by definition (we define $z^0 = 1$ for any non-zero complex number $z$, and $i$ is certainly not 0).

Also, $i^1 = i$. This is clear.

From the definition of $i$ (as a square root of -1), we have: $i^2 = -1$. So far, no heavy lifting at all.

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

$i^4 = (i^2)(i^2) = (-1)(-1) = 1$ <---this is what we will use from here on out.

******************

So, for ANY integer $k$, we have:

$i^{4k} = (i^4)^k = 1^k = 1$

$i^{4k+1} = (i^{4k})(i) = (1)(i) = i$ (see above)

$i^{4k+2} = (i^{4k})(i^2) = (1)(-1) = -1$

$i^{4k+3} = (i^{4k})(i^3) = (1)(-i) = -i$

....and that is all.

*******************

Your intuition is correct, multiplying by $i$ corresponds to rotating (counter-clockwise) by $\dfrac{\pi}{2}$, which is a "quarter-turn". Four quarter-turns makes a whole turn, and we are back where we started, and repeat.

I did this for integers, rather than natural numbers, because:

$i^{-1} = \dfrac{1}{i} = \dfrac{1}{i}\cdot \dfrac{i}{i} = \dfrac{i}{i^2} = \dfrac{i}{-1} = -i$.

So $i^{-1} = i^3$ (note that -1 and 3 are "four apart"). This corresponds to our natural intuition that "a quarter turn backwards = three quarters of a turn forwards"

Notice the prominent role the number 4 plays, here. As we have seen, $i^4 = 1$, which means $i$ is a solution to $x^4 - 1 = 0$.

This is not surprising, because $x^4 - 1 = (x^2 + 1)(x^2 - 1)$ and $i$ is a root of $x^2 + 1$ (if $x^2 + 1 = 0$ then $x^2 = -1$, so $x$ is a square root of -1).

We have also seen $(-i)^4 = (-i)^2(-i)^2 = (-i)(-i(-i)(-i) = (-1)(i)(-1)(i)(-1)(i)(-1)(i) = (-1)^4i^4 = (1)(1) = 1$. So $-i$ is also a root of $x^4 - 1$.

This is ALSO not surprising as $x^2 + 1 = (x - i)(x + i) = (x - i)(x - (-i))$

Since -1 and 1 are the two roots of $x^2 - 1 = (x - 1)(x + 1)$, we conclude that $\{1,-1,i,-i\}$ are indeed ALL FOUR fourth roots of one (the four roots of $x^4 - 1$).

Note the various ways 4 figures in this:

$x^4 - 1$ <--a polynomial of degree 4
$\frac{\pi}{2}$ <--1/4th of a whole turn
$\{1,-1,i,-i\}$ <--a set of 4 things

This is no coincidence. There's a pattern, here.
• Aug 14th 2014, 03:41 AM
andy000
Re: Complex numbers, All possible values of i^n
thanks all!