Complex Analysis

  • Thread starter Ted123
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Homework Statement



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Homework Equations





The Attempt at a Solution



How do I go about Q1 and showing the coefficients are unique and then Q2?
 

Answers and Replies

  • #2
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What is your attempt at the solution?
 
  • #3
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What is your attempt at the solution?
I was trying some sort of induction argument.

[itex]P(0)=Q(0) \implies a_0 = b_0[/itex] so the result is true for [itex]z=0[/itex]

Now suppose for induction that [itex]P(k)=Q(k)[/itex] ...
 
  • #4
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OK, so now take the derivative of the polynomials, and compare [tex]P^\prime(0)[/tex] with [tex]Q^\prime(0)[/tex].
 
  • #5
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OK, so now take the derivative of the polynomials, and compare [tex]P^\prime(0)[/tex] with [tex]Q^\prime(0)[/tex].
Bearing in mind z is complex does normal differentiation still hold?

[itex]P'(0)=Q'(0) \implies a_1=b_1[/itex]
 
Last edited:
  • #6
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Scrap that method.

Could I say without loss of generality that [itex]n \geq m[/itex] .

Then if [itex]P(z)=Q(z)[/itex] then [itex]P(z)-Q(z)=0[/itex]

So collecting up like terms we see that

[itex]P(z) - Q(z) = (a_0 - b_0) + (a_1 - b_1)z + (a_2 - b_2)z^2 + ... + (a_m - b_m)z^m + a_{m+1}z^{m+1} + ... + a_nz^n[/itex]

This implies that [itex]a_j=b_j[/itex] for all [itex] 0 \leq j \leq m[/itex] (by equating coefficients).

So The first [itex]m[/itex] terms are all 0.

So we have [itex]a_{m+1}z^{m+1} + ... + a_nz^n=0[/itex] .

Can you help me finish?
 
  • #7
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On the other hand saying you can compare coefficients here might be a bit close to begging the question.

Instead, if [itex]P(z) = Q(z)[/itex] for all z and [itex]F(z) = P(z)-Q(z)[/itex], then [itex]1, 2, 3, ..., m+n+1[/itex] are all roots of [itex]F(z)[/itex], and so [itex](z-1)(z-2)...(z-(m+n+1))[/itex] divides [itex]F(z)[/itex] (factor theorem). Then what does this implies about the degree of F?
 

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