Greetings,
I am trying to understand some points about the demonstration of the FTA, and there is something confusing.
In order to prove that every polynomial [MATH]P(X) = \Sigma_{i=0}^{n} a_i. X^i [/MATH] has zeros in [MATH]\mathbb{C}[/MATH], the demonstration I am studying goes in two stages :
First, proving that [MATH]\phi(z_0) = |z_0|[/MATH] is bounded below with [MATH] \phi(z_0) = \underset{z \in \mathbb{C}}{Inf} (\phi(z)) [/MATH]. That is the easy part : [MATH]\phi[/MATH] is continuous in the disk [MATH]\lbrace z \in \mathbb{C}, |z| \leq B \rbrace[/MATH]Second, to create a contradiction with the assumption [MATH]P(z) \neq 0[/MATH], we write the Taylor formula of [MATH]P(z_0 + h)[/MATH]. Assuming that there is at least a natural k from whitch [MATH]P^{(k)}(z_0) [/MATH] is not necessarily 0, we can reach to this formula :
[MATH]\forall h \in \mathbb{C}, \frac{P(z_0+h)}{P(z_0)} = 1+h^k .\frac{P^{(k)}(z_0+h)}{k!.P(z_0)}+\dots+h^n .\frac{P^{(n)}(z_0+h)}{n!.P(z_0)} [/MATH]
The problem starts with a variable change : [MATH]h = t/\omega[/MATH] (in order to write [MATH]P(z_0+t/\omega)/P(z_0) = 1-t^k+ o(t^k)[/MATH]) , where the demonstration affirms that there is complex roots ([MATH]\omega^k[/MATH]) so that :
[MATH]\omega^k = -\frac{P^{(k)}(z_0+h)}{k!.P(z_0)}[/MATH]
I can't understand how this is possible, complex roots have all the same measure and passing from [MATH]\omega^k[/MATH] to [MATH]\omega^{k+1}[/MATH] is only a rotation of a fix angle [MATH]2\pi/n[/MATH], which is not the case here.
Can someone help me get around this ?
(P.S. English is not my native language, so sorry if I do not use idiomatic terms or if some grammatical errors occure in some places)
Thank you.
I am trying to understand some points about the demonstration of the FTA, and there is something confusing.
In order to prove that every polynomial [MATH]P(X) = \Sigma_{i=0}^{n} a_i. X^i [/MATH] has zeros in [MATH]\mathbb{C}[/MATH], the demonstration I am studying goes in two stages :
First, proving that [MATH]\phi(z_0) = |z_0|[/MATH] is bounded below with [MATH] \phi(z_0) = \underset{z \in \mathbb{C}}{Inf} (\phi(z)) [/MATH]. That is the easy part : [MATH]\phi[/MATH] is continuous in the disk [MATH]\lbrace z \in \mathbb{C}, |z| \leq B \rbrace[/MATH]Second, to create a contradiction with the assumption [MATH]P(z) \neq 0[/MATH], we write the Taylor formula of [MATH]P(z_0 + h)[/MATH]. Assuming that there is at least a natural k from whitch [MATH]P^{(k)}(z_0) [/MATH] is not necessarily 0, we can reach to this formula :
[MATH]\forall h \in \mathbb{C}, \frac{P(z_0+h)}{P(z_0)} = 1+h^k .\frac{P^{(k)}(z_0+h)}{k!.P(z_0)}+\dots+h^n .\frac{P^{(n)}(z_0+h)}{n!.P(z_0)} [/MATH]
The problem starts with a variable change : [MATH]h = t/\omega[/MATH] (in order to write [MATH]P(z_0+t/\omega)/P(z_0) = 1-t^k+ o(t^k)[/MATH]) , where the demonstration affirms that there is complex roots ([MATH]\omega^k[/MATH]) so that :
[MATH]\omega^k = -\frac{P^{(k)}(z_0+h)}{k!.P(z_0)}[/MATH]
I can't understand how this is possible, complex roots have all the same measure and passing from [MATH]\omega^k[/MATH] to [MATH]\omega^{k+1}[/MATH] is only a rotation of a fix angle [MATH]2\pi/n[/MATH], which is not the case here.
Can someone help me get around this ?
(P.S. English is not my native language, so sorry if I do not use idiomatic terms or if some grammatical errors occure in some places)
Thank you.
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