On a proof about rational functions and connected components
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I'm trying to follow a demonstration of the following:
Here $Q$ is a complex function and $S^1$ is the unitary circle.
Let $S = Q^{-1}([-1,1])$ where $Q(z)$ is a rational function of degree at most $2n$ such that $Q(S^1) subset mathbb{R}$. If $S cap S^1$ has at least $2n$ connected components then $S$ is the union of $2n$ subintervals of $S^1$.
The autor start saying that $Q$ and $Q'$ has degree at most $2n$ and since $Q(S^1) subset mathbb{R}$ that must exist at least one zero of $Q'(z)$ in each component of $S^1 - S$, then that is no zero of $Q'(z)$ in $S$, which implies that each connected component of $S cap S^1$ is mapped diffeomorphically onto [-1,1].
I'm really not familiar with the tool he used here (I'm not even know what kind of theorem he uses on these affirmations), can someone clarify that for me?
general-topology complex-analysis differential-topology
asked Nov 6 at 23:50
Lucas Resende
1266
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1
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I'm trying to follow a demonstration of the following:
Here $Q$ is a complex function and $S^1$ is the unitary circle.
Let $S = Q^{-1}([-1,1])$ where $Q(z)$ is a rational function of degree at most $2n$ such that $Q(S^1) subset mathbb{R}$. If $S cap S^1$ has at least $2n$ connected components then $S$ is the union of $2n$ subintervals of $S^1$.
The autor start saying that $Q$ and $Q'$ has degree at most $2n$ and since $Q(S^1) subset mathbb{R}$ that must exist at least one zero of $Q'(z)$ in each component of $S^1 - S$, then that is no zero of $Q'(z)$ in $S$, which implies that each connected component of $S cap S^1$ is mapped diffeomorphically onto [-1,1].
I'm really not familiar with the tool he used here (I'm not even know what kind of theorem he uses on these affirmations), can someone clarify that for me?
general-topology complex-analysis differential-topology
asked Nov 6 at 23:50
Lucas Resende
1266
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to follow a demonstration of the following:
Here $Q$ is a complex function and $S^1$ is the unitary circle.
Let $S = Q^{-1}([-1,1])$ where $Q(z)$ is a rational function of degree at most $2n$ such that $Q(S^1) subset mathbb{R}$. If $S cap S^1$ has at least $2n$ connected components then $S$ is the union of $2n$ subintervals of $S^1$.
The autor start saying that $Q$ and $Q'$ has degree at most $2n$ and since $Q(S^1) subset mathbb{R}$ that must exist at least one zero of $Q'(z)$ in each component of $S^1 - S$, then that is no zero of $Q'(z)$ in $S$, which implies that each connected component of $S cap S^1$ is mapped diffeomorphically onto [-1,1].
I'm really not familiar with the tool he used here (I'm not even know what kind of theorem he uses on these affirmations), can someone clarify that for me?
general-topology complex-analysis differential-topology
asked Nov 6 at 23:50
Lucas Resende
1266
I'm trying to follow a demonstration of the following:
Here $Q$ is a complex function and $S^1$ is the unitary circle.
Let $S = Q^{-1}([-1,1])$ where $Q(z)$ is a rational function of degree at most $2n$ such that $Q(S^1) subset mathbb{R}$. If $S cap S^1$ has at least $2n$ connected components then $S$ is the union of $2n$ subintervals of $S^1$.
The autor start saying that $Q$ and $Q'$ has degree at most $2n$ and since $Q(S^1) subset mathbb{R}$ that must exist at least one zero of $Q'(z)$ in each component of $S^1 - S$, then that is no zero of $Q'(z)$ in $S$, which implies that each connected component of $S cap S^1$ is mapped diffeomorphically onto [-1,1].
I'm really not familiar with the tool he used here (I'm not even know what kind of theorem he uses on these affirmations), can someone clarify that for me?
general-topology complex-analysis differential-topology
general-topology complex-analysis differential-topology
asked Nov 6 at 23:50
Lucas Resende
1266
asked Nov 6 at 23:50
Lucas Resende
1266
edited Nov 24 at 1:27
asked Nov 6 at 23:50
Lucas Resende
1266
asked Nov 6 at 23:50
Lucas Resende
1266
asked Nov 6 at 23:50
Lucas Resende
1266
1266
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add a comment |
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