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### Roots of Unity

```
Date: 04/18/97 at 14:05:30
From: Volante
Subject: Possible new formula?

I was directed here and am hoping you could answer a little question
for me.  I was killing a lot of time today by looking at roots of -1
and eventually devised a formula to help find them:

xth root of i = (cos (pi/2x))(sin (pi/2x) i)

x = real number
pi = pi (3.14159...)
i = square root of -1

I am not in a high enough math class to see if this has already been
found and I really don't know where I'd look now.  I'd like to know if
someone else has found this.  I think it is an accomplishment for a
10th grader to do in the first place.

Jeremy Martz
```

```
Date: 04/18/97 at 15:24:35
From: Doctor Steven
Subject: Re: Possible new formula?

Jeremy,

Basically what you have found is the nth root of i, which is the 2nth
root of unity.  (Is a root of the equation x^(2n) - 1 = 0.)

To look at this we will try to find all roots of x^n = 1. To do this
we first note Euler's formula:

e^(i*Theta) = cos(Theta) + i*sin(Theta).

If you want you can verify this formula by taking the Taylor expansion
of e^(i*Theta) centered at 0.

Now let's create a graph of this function of Theta. First create your
coordinate axes: the horizontal axis will be the real numbers and the
vertical axis will be real numbers times i. We get a coordinate
plane like this:

|
| 2i
|
| i
|
---------------------------
-2    -1    |     1     2
| -i
|
| -2i
|

If you graph the function, you get a unit circle in this plane
centered at the origin.

What happens if we have a value for Theta (which gives us a point on
the circle) and want to know what [e^(i*Theta)]^2 is?  Well,
[e^(i*theta)]^2 = e^(i*2*Theta) = cos(2*Theta) + i*sin(2*Theta).
So basically what we've done is rotated that point we got on the
circle over another Theta degrees.

What does this have to do with roots of unity (i.e., roots of the
equation x^n - 1 = 0)?  Note that:

e^(i*k*2Pi) = cos(k*2Pi) + i*sin(k*2Pi) = 1 + i*0 = 1

To find the nth roots of unity, we're interested in numbers of the
form e^(i*k*2Pi/n). As k goes from 1 to n, we get n different values
for e^(i*k*2Pi/n), each of which raised to the nth power equals 1.  We
have found n roots to the equation x^n - 1 = 0.

As for nth roots of i, we need to find roots of the equation
x^n - i = 0. This gives us the values e^(i*k*(Pi/2)/n), where k goes
from 1 to n. When k = 1, you get the answer you noted above.

Hope this helps.

-Doctor Steven,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 04/18/97 at 15:57:52
From: Doctor Wilkinson
Subject: Re: possible new formula?

Jeremy,

This formula was known to Euler in the 18th century, so I'm afraid it
isn't new.  On the other hand, you're in excellent company!  You're
right: it is an accomplishment and I congratulate you.  Keep at it.

-Doctor Wilkinson,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Imaginary/Complex Numbers

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