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- July 13th 2006, 09:37 AMharoldDeMoivre's Theorem problem
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- July 13th 2006, 10:44 AMThePerfectHackerQuote:

Originally Posted by**harold**

- July 13th 2006, 12:58 PMThePerfectHackerQuote:

Originally Posted by**harold**

In polar complex form,

- July 13th 2006, 07:11 PMThePerfectHackerQuote:

Originally Posted by**harold**

As you understand (I hope),

In order to use de Moiver's theorem you write,

Now you use de Moiver's theorem (actually it is not cuz his theorem if for integral exponents not rational).

Thus,

And each solution is for - July 15th 2006, 08:01 PMThePerfectHackerQuote:

Originally Posted by**harold**

(Basically you use the factor theorem is zero of if and only if divides ) - July 15th 2006, 09:01 PMThePerfectHacker
Ignore what I said before I made a mistake.

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This is the factorization

- July 16th 2006, 07:07 AMThePerfectHackerQuote:

Originally Posted by**harold**

The first pair of conjugates are,

The polynomial they produce is,

The other pairs of conjugates are,

The polynomial they produce is,

So the factorization of,

in is,