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Topic: The cube roots of 8
Replies: 45   Last Post: Nov 14, 1999 4:24 PM

 Messages: [ Previous | Next ]
 Quentin Grady Posts: 291 Registered: 12/6/04
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,

Date Subject Author
9/27/99 Doug Norris
9/29/99 Doug Norris
9/27/99 Dann Corbit
9/27/99 Paul Arendt
9/27/99 Lynn Killingbeck
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10/11/99 Mike Mccarty Sr
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