Evaluate a complex function
$begingroup$
Let $varepsilon>0$. Let $f:B_{varepsilon}(0)rightarrowmathbb{C} $ be an analytic function such that $f(0)=0$, $f(a)=a$ for some $ain B_{varepsilon}(0)-{0}$ and ${0,a}$ are the only fixed points of $f$ in $B_{varepsilon}(0)$ (both multiplicity 1). Define,
$$g(z)=frac{f(z)-z}{z(z-a)}$$
It is clear that $0$ and $a$ are removable singularities of $g$. But how can I show that $g(a)neq 0$ and $g(0)neq 0$.
complex-analysis holomorphic-functions singularity
asked Dec 3 '18 at 21:13
Galois BrazilGalois Brazil
538
$endgroup$
add a comment |
$begingroup$
Let $varepsilon>0$. Let $f:B_{varepsilon}(0)rightarrowmathbb{C} $ be an analytic function such that $f(0)=0$, $f(a)=a$ for some $ain B_{varepsilon}(0)-{0}$ and ${0,a}$ are the only fixed points of $f$ in $B_{varepsilon}(0)$ (both multiplicity 1). Define,
$$g(z)=frac{f(z)-z}{z(z-a)}$$
It is clear that $0$ and $a$ are removable singularities of $g$. But how can I show that $g(a)neq 0$ and $g(0)neq 0$.
complex-analysis holomorphic-functions singularity
asked Dec 3 '18 at 21:13
Galois BrazilGalois Brazil
538
$endgroup$
add a comment |
$begingroup$
Let $varepsilon>0$. Let $f:B_{varepsilon}(0)rightarrowmathbb{C} $ be an analytic function such that $f(0)=0$, $f(a)=a$ for some $ain B_{varepsilon}(0)-{0}$ and ${0,a}$ are the only fixed points of $f$ in $B_{varepsilon}(0)$ (both multiplicity 1). Define,
$$g(z)=frac{f(z)-z}{z(z-a)}$$
It is clear that $0$ and $a$ are removable singularities of $g$. But how can I show that $g(a)neq 0$ and $g(0)neq 0$.
complex-analysis holomorphic-functions singularity
asked Dec 3 '18 at 21:13
Galois BrazilGalois Brazil
538
$endgroup$
Let $varepsilon>0$. Let $f:B_{varepsilon}(0)rightarrowmathbb{C} $ be an analytic function such that $f(0)=0$, $f(a)=a$ for some $ain B_{varepsilon}(0)-{0}$ and ${0,a}$ are the only fixed points of $f$ in $B_{varepsilon}(0)$ (both multiplicity 1). Define,
$$g(z)=frac{f(z)-z}{z(z-a)}$$
It is clear that $0$ and $a$ are removable singularities of $g$. But how can I show that $g(a)neq 0$ and $g(0)neq 0$.
complex-analysis holomorphic-functions singularity
complex-analysis holomorphic-functions singularity
asked Dec 3 '18 at 21:13
Galois BrazilGalois Brazil
538
asked Dec 3 '18 at 21:13
Galois BrazilGalois Brazil
538
edited Dec 3 '18 at 21:36
Galois Brazil
asked Dec 3 '18 at 21:13
Galois BrazilGalois Brazil
538
asked Dec 3 '18 at 21:13
Galois BrazilGalois Brazil
538
asked Dec 3 '18 at 21:13
Galois BrazilGalois Brazil
538
538
add a comment |
add a comment |
1 Answer
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$begingroup$
What if we take $f(z) = z + big(z(z-a)big)^2$?
answered Dec 3 '18 at 21:19
metamorphymetamorphy
3,6921621
$endgroup$
$begingroup$
In this edited form, your claim is true (if you mean $0$ and $a$ are zeros of $f(z)-z$ of multiplicity $1$ (i.e. simple zeros) - and then it holds by the definition of multiplicity (or simplicity)).
$endgroup$
– metamorphy
Dec 3 '18 at 21:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
What if we take $f(z) = z + big(z(z-a)big)^2$?
answered Dec 3 '18 at 21:19
metamorphymetamorphy
3,6921621
$endgroup$
$begingroup$
In this edited form, your claim is true (if you mean $0$ and $a$ are zeros of $f(z)-z$ of multiplicity $1$ (i.e. simple zeros) - and then it holds by the definition of multiplicity (or simplicity)).
$endgroup$
– metamorphy
Dec 3 '18 at 21:41
add a comment |
$begingroup$
What if we take $f(z) = z + big(z(z-a)big)^2$?
answered Dec 3 '18 at 21:19
metamorphymetamorphy
3,6921621
$endgroup$
$begingroup$
In this edited form, your claim is true (if you mean $0$ and $a$ are zeros of $f(z)-z$ of multiplicity $1$ (i.e. simple zeros) - and then it holds by the definition of multiplicity (or simplicity)).
$endgroup$
– metamorphy
Dec 3 '18 at 21:41
add a comment |
$begingroup$
What if we take $f(z) = z + big(z(z-a)big)^2$?
answered Dec 3 '18 at 21:19
metamorphymetamorphy
3,6921621
$endgroup$
What if we take $f(z) = z + big(z(z-a)big)^2$?
answered Dec 3 '18 at 21:19
metamorphymetamorphy
3,6921621
answered Dec 3 '18 at 21:19
metamorphymetamorphy
3,6921621
answered Dec 3 '18 at 21:19
metamorphymetamorphy
3,6921621
answered Dec 3 '18 at 21:19
metamorphymetamorphy
3,6921621
3,6921621
$begingroup$
In this edited form, your claim is true (if you mean $0$ and $a$ are zeros of $f(z)-z$ of multiplicity $1$ (i.e. simple zeros) - and then it holds by the definition of multiplicity (or simplicity)).
$endgroup$
– metamorphy
Dec 3 '18 at 21:41
add a comment |
$begingroup$
In this edited form, your claim is true (if you mean $0$ and $a$ are zeros of $f(z)-z$ of multiplicity $1$ (i.e. simple zeros) - and then it holds by the definition of multiplicity (or simplicity)).
$endgroup$
– metamorphy
Dec 3 '18 at 21:41
$begingroup$
In this edited form, your claim is true (if you mean $0$ and $a$ are zeros of $f(z)-z$ of multiplicity $1$ (i.e. simple zeros) - and then it holds by the definition of multiplicity (or simplicity)).
$endgroup$
– metamorphy
Dec 3 '18 at 21:41
$begingroup$
In this edited form, your claim is true (if you mean $0$ and $a$ are zeros of $f(z)-z$ of multiplicity $1$ (i.e. simple zeros) - and then it holds by the definition of multiplicity (or simplicity)).
$endgroup$
– metamorphy
Dec 3 '18 at 21:41
add a comment |
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