Show closure of image equals $mathbb{C}$ (I.e. is dense in $mathbb{C})$
Let $f(z)$ be a nonconstant analytic function on $mathbb{C}backslash S$, where $S$ is a finite subset of $mathbb{C}$. Show that $overline{f(mathbb{C}backslash S)}=mathbb{C}$
In preparation for a final examination, I am reviewing various questions for complex analysis. In particular, I am not too sure how to go about solving this question. I was thinking to prove it by contradiction, and suppose that the image is not dense in $mathbb{C}$, but am not too sure.
My apologies for the lack of work and would appreciate any help. Thanks in advance!
complex-analysis
asked Dec 13 '18 at 8:35
The math godThe math god
1947
add a comment |
Let $f(z)$ be a nonconstant analytic function on $mathbb{C}backslash S$, where $S$ is a finite subset of $mathbb{C}$. Show that $overline{f(mathbb{C}backslash S)}=mathbb{C}$
In preparation for a final examination, I am reviewing various questions for complex analysis. In particular, I am not too sure how to go about solving this question. I was thinking to prove it by contradiction, and suppose that the image is not dense in $mathbb{C}$, but am not too sure.
My apologies for the lack of work and would appreciate any help. Thanks in advance!
complex-analysis
asked Dec 13 '18 at 8:35
The math godThe math god
1947
add a comment |
Let $f(z)$ be a nonconstant analytic function on $mathbb{C}backslash S$, where $S$ is a finite subset of $mathbb{C}$. Show that $overline{f(mathbb{C}backslash S)}=mathbb{C}$
In preparation for a final examination, I am reviewing various questions for complex analysis. In particular, I am not too sure how to go about solving this question. I was thinking to prove it by contradiction, and suppose that the image is not dense in $mathbb{C}$, but am not too sure.
My apologies for the lack of work and would appreciate any help. Thanks in advance!
complex-analysis
asked Dec 13 '18 at 8:35
The math godThe math god
1947
Let $f(z)$ be a nonconstant analytic function on $mathbb{C}backslash S$, where $S$ is a finite subset of $mathbb{C}$. Show that $overline{f(mathbb{C}backslash S)}=mathbb{C}$
In preparation for a final examination, I am reviewing various questions for complex analysis. In particular, I am not too sure how to go about solving this question. I was thinking to prove it by contradiction, and suppose that the image is not dense in $mathbb{C}$, but am not too sure.
My apologies for the lack of work and would appreciate any help. Thanks in advance!
complex-analysis
complex-analysis
asked Dec 13 '18 at 8:35
The math godThe math god
1947
asked Dec 13 '18 at 8:35
The math godThe math god
1947
asked Dec 13 '18 at 8:35
The math godThe math god
1947
asked Dec 13 '18 at 8:35
The math godThe math god
1947
asked Dec 13 '18 at 8:35
The math godThe math god
1947
1947
add a comment |
add a comment |
1 Answer
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Suppose it is not dense. Then there exists an open disk $B(c,r)$ disjoint from $f(mathbb Csetminus S)$. Let $g(z)=frac 1 {f(z)-c}$ for $z in mathbb Csetminus S$ . Then $g$ is analytic on $mathbb Csetminus S$ and it is bounded there (because $|f(x)-c| geq r$). It follows that at the points of $S$ it has a removable singularity. Hence it extends to a bounded entire function. This makes $g$ constant and hence $f$ also constant.
answered Dec 13 '18 at 8:40
Kavi Rama MurthyKavi Rama Murthy
52.4k32055
@S Sorry, there was a typo.
– Kavi Rama Murthy
Dec 13 '18 at 8:47
Haha, I suspected this was similar to the proof showing the closure of $f(mathbb{C})=mathbb{C}$. I just realized that each point in $S$ has a ball sufficiently small around it, not including that point in $S$. Thus we can extend each point to a bounded entire function. Thank you so much!
– The math god
Dec 13 '18 at 8:47
add a comment |
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1 Answer
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1 Answer
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votes
Suppose it is not dense. Then there exists an open disk $B(c,r)$ disjoint from $f(mathbb Csetminus S)$. Let $g(z)=frac 1 {f(z)-c}$ for $z in mathbb Csetminus S$ . Then $g$ is analytic on $mathbb Csetminus S$ and it is bounded there (because $|f(x)-c| geq r$). It follows that at the points of $S$ it has a removable singularity. Hence it extends to a bounded entire function. This makes $g$ constant and hence $f$ also constant.
answered Dec 13 '18 at 8:40
Kavi Rama MurthyKavi Rama Murthy
52.4k32055
@S Sorry, there was a typo.
– Kavi Rama Murthy
Dec 13 '18 at 8:47
Haha, I suspected this was similar to the proof showing the closure of $f(mathbb{C})=mathbb{C}$. I just realized that each point in $S$ has a ball sufficiently small around it, not including that point in $S$. Thus we can extend each point to a bounded entire function. Thank you so much!
– The math god
Dec 13 '18 at 8:47
add a comment |
Suppose it is not dense. Then there exists an open disk $B(c,r)$ disjoint from $f(mathbb Csetminus S)$. Let $g(z)=frac 1 {f(z)-c}$ for $z in mathbb Csetminus S$ . Then $g$ is analytic on $mathbb Csetminus S$ and it is bounded there (because $|f(x)-c| geq r$). It follows that at the points of $S$ it has a removable singularity. Hence it extends to a bounded entire function. This makes $g$ constant and hence $f$ also constant.
answered Dec 13 '18 at 8:40
Kavi Rama MurthyKavi Rama Murthy
52.4k32055
@S Sorry, there was a typo.
– Kavi Rama Murthy
Dec 13 '18 at 8:47
Haha, I suspected this was similar to the proof showing the closure of $f(mathbb{C})=mathbb{C}$. I just realized that each point in $S$ has a ball sufficiently small around it, not including that point in $S$. Thus we can extend each point to a bounded entire function. Thank you so much!
– The math god
Dec 13 '18 at 8:47
add a comment |
Suppose it is not dense. Then there exists an open disk $B(c,r)$ disjoint from $f(mathbb Csetminus S)$. Let $g(z)=frac 1 {f(z)-c}$ for $z in mathbb Csetminus S$ . Then $g$ is analytic on $mathbb Csetminus S$ and it is bounded there (because $|f(x)-c| geq r$). It follows that at the points of $S$ it has a removable singularity. Hence it extends to a bounded entire function. This makes $g$ constant and hence $f$ also constant.
answered Dec 13 '18 at 8:40
Kavi Rama MurthyKavi Rama Murthy
52.4k32055
Suppose it is not dense. Then there exists an open disk $B(c,r)$ disjoint from $f(mathbb Csetminus S)$. Let $g(z)=frac 1 {f(z)-c}$ for $z in mathbb Csetminus S$ . Then $g$ is analytic on $mathbb Csetminus S$ and it is bounded there (because $|f(x)-c| geq r$). It follows that at the points of $S$ it has a removable singularity. Hence it extends to a bounded entire function. This makes $g$ constant and hence $f$ also constant.
answered Dec 13 '18 at 8:40
Kavi Rama MurthyKavi Rama Murthy
52.4k32055
edited Dec 13 '18 at 8:46
answered Dec 13 '18 at 8:40
Kavi Rama MurthyKavi Rama Murthy
52.4k32055
answered Dec 13 '18 at 8:40
Kavi Rama MurthyKavi Rama Murthy
52.4k32055
answered Dec 13 '18 at 8:40
Kavi Rama MurthyKavi Rama Murthy
52.4k32055
52.4k32055
@S Sorry, there was a typo.
– Kavi Rama Murthy
Dec 13 '18 at 8:47
Haha, I suspected this was similar to the proof showing the closure of $f(mathbb{C})=mathbb{C}$. I just realized that each point in $S$ has a ball sufficiently small around it, not including that point in $S$. Thus we can extend each point to a bounded entire function. Thank you so much!
– The math god
Dec 13 '18 at 8:47
add a comment |
@S Sorry, there was a typo.
– Kavi Rama Murthy
Dec 13 '18 at 8:47
Haha, I suspected this was similar to the proof showing the closure of $f(mathbb{C})=mathbb{C}$. I just realized that each point in $S$ has a ball sufficiently small around it, not including that point in $S$. Thus we can extend each point to a bounded entire function. Thank you so much!
– The math god
Dec 13 '18 at 8:47
@S Sorry, there was a typo.
– Kavi Rama Murthy
Dec 13 '18 at 8:47
@S Sorry, there was a typo.
– Kavi Rama Murthy
Dec 13 '18 at 8:47
Haha, I suspected this was similar to the proof showing the closure of $f(mathbb{C})=mathbb{C}$. I just realized that each point in $S$ has a ball sufficiently small around it, not including that point in $S$. Thus we can extend each point to a bounded entire function. Thank you so much!
– The math god
Dec 13 '18 at 8:47
Haha, I suspected this was similar to the proof showing the closure of $f(mathbb{C})=mathbb{C}$. I just realized that each point in $S$ has a ball sufficiently small around it, not including that point in $S$. Thus we can extend each point to a bounded entire function. Thank you so much!
– The math god
Dec 13 '18 at 8:47
add a comment |
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