
Re: The cube roots of 8
Posted:
Oct 7, 1999 6:13 PM


G'day G'day Folks,
By way of appreciation to all those people who showed me ways to find the complex roots of 8 I' d like to share this application.
One of the features of three phase electricity is that it delivers constant total power to a load.
The three phase can be named red, yellow and blue.
Power = ((v_r)^2 + (v_y)^2 + (v_b)^2 ) /R for resistive loads.
The trick is to prove (v_r)^2 + (v_y)^2 + (v_b)^2 is a constant.
Let one of the voltages be the principle root of 8 ie 2.
Letting j = sqrt(1) The complex roots become 1 + j*sqrt(3) and 1  j*sqrt(3) And v_r)^2 + (v_y)^2 + (v_b)^2 becomes 6 which is constant.
In general v_r)^2 + (v_y)^2 + (v_b)^2 = 1.5 * (V_peak)^2
It makes a nice change from the trig method.
Thanks,
On Tue, 28 Sep 1999 07:39:37 +1200, Quentin Grady <quentin@inhb.co.nz> wrote:
>G'day G'day Folks, > > I grew up happy in the knowledge that the cube root of 8 was 2. >Life was simple then. Calculators gave 2 as the answer and it was the >only answer I thought of looking for. > >Then I read a statement that astounded me, > "Every number has n nth roots." > >Is there a simple proof of the statement? > >If for the moment we assume that the statement is true then certain >conclusions follow in an obvious fashion. Since 8 has one real cube >root, it must have two complex number cube roots. It is easy enough >to prove the complex cube roots by multiplication once they are found, > > >My question is, >"How does one systematically find complex roots of positive integers?" > >Thanks,

