Roots of UnityDate: 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 pi/2x are radian values 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. Thanks in advance, 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/