Polynomials Problem Help: How to Solve P(z)Q(z)=0 Using Coefficient Equations

[Ted123]

Homework Statement

[PLAIN]http://img443.imageshack.us/img443/3096/questiond.jpg

The Attempt at a Solution

If P(z)Q(z)=0 then

\displaystyle a_0b_0 + (a_0b_1 + a_1 b_0)z + ... + \left( \sum_{i=0}^k a_i b_{k-i} \right) z^k + ... + a_n b_m z^{n+m} =0

Now what? Equate coefficients?

a_0 b_0 =0 \Rightarrow a_0 = 0 \; \text{or}\; b_0 = 0

a_0 = 0 \Rightarrow a_1 b_0 =0 \Rightarrow a_1 = 0 \; \text{or}\; b_0 =0

b_0 =0 \Rightarrow a_0 b_1 = 0 \Rightarrow b_1 = 0 \; \text{or}\; a_0 = 0

...

 

[tiny-tim]

Hi Ted123!

Hint: start at z^n+m.

 

[Ted123]

Hi Ted123!

Hint: start at z^n+m.

Doing this gives

a_n = 0\;\text{or}\; b_n = 0 but what next?

Would it be better to use the fact that:

If P isn't identically zero, it has at most n roots.
If Q isn't identically zero, it has at most m roots.

So, if neither P and Q are identically zero, PQ has at most m+n roots. How could I use this?

 

[tiny-tim]

… but what next?

think!

(how are a_n and b_m defined?)

 

[Ted123]

think!

(how are a_n and b_m defined?)

Well in the question before this it states that a_n \neq 0 and b_m \neq 0 and it says in this question let P and Q be complex polynomials as in the previous question (I missed these conditions out when I copied the polynomials from the previous question).

So if a_n b_m =0 but a_n , b_m \neq 0 what do we have?

 

[tiny-tim]

a contradiction!

soooo … ? ​

 

[Ted123]

a contradiction!

soooo … ? ​

P(z)Q(z) \neq 0

 

[tiny-tim]

Yes … you've proved that if P and Q are not zero, then PQ is not zero.

 

[Ted123]

Yes … you've proved that if P and Q are not zero, then PQ is not zero.

So does this prove the converse (which I need to prove) that if PQ=0 then P or Q are 0? Oh yes - I see it does now