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Exercise [04.06]
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Joined: 11 May 2008, 20:07
Posts: 11
Location: france
Exercise [04.06]
Let's say we have the complex polynomial :

If we name the solutions of , , we can rewrite it as :

With

...

As we can see (if you really pay attention because it is rather small!) every is a "function" of every (except ). So as Mr penrose says in this exercise, if we divide with every will change (except once more ) and the polynomial will have a different fully developed formula.

13 May 2008, 16:25

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Re: Exercise [4.6]
Alternative prooof:

From the discussion in section 4.2, we can infer that P(z) will have at least one complex solution.
Call it
Now we have to prove that is a factor of P(z).
Assume is not a factor of P(z).
From the division algorithm we have,
where R(z) is the remainder which must be non zero for all z because of our assumption.
Setting in this identity

or(is a zero of )which is a contradiction.
Hence,is a factor of P(z) which means

Using the same argument we have,

.
.
.
where is the solution of
.

Now

Now little thought revealsis apolynomial of degree
Therefore is a constant as it has degree n-n=0
On multiplying out the factors the coefficient of is .
Equating this coefficient with the one in original expression for P(z) we have

Hence,

16 May 2008, 12:00
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