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Check my Proof Please (Complex Analysis)
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Question: Let $f$ be an analytic function in an open set $U$. Let $V = z in mathbbC : overlinez in U$. Define $g$ on $V$ by $g(z) = overlinef(overlinez)$. Show that $g$ is analytic on $V$ without using the fact that $g$ is holomorphic.
Proof Attempt: Choose $z_0 in V$. Then $overlinez_0 in U$. Since $f$ is analytic on $U$, there is an $r > 0$ and a sequence of complex numbers $a_n_n = 0^infty$ such that for $overlinez in D(overlinez_0, r)$, we have
$$f(overlinez) = sum_n = 0^infty a_n(overlinez - overlinez_0)^n$$
It follows that
$$g(z) = overlinef(overlinez) = sum_n = 0^infty overlinea_n(z - z_0)^n$$
for all $z in D(z_0, r)$. Thus, $g$ is analytic on $V$.
complex-analysis proof-verification power-series
edited Jul 26 at 20:33
egreg
164k1180187
asked Jul 26 at 19:13
Bobo
244
add a comment |Â
up vote
3
down vote
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Question: Let $f$ be an analytic function in an open set $U$. Let $V = z in mathbbC : overlinez in U$. Define $g$ on $V$ by $g(z) = overlinef(overlinez)$. Show that $g$ is analytic on $V$ without using the fact that $g$ is holomorphic.
Proof Attempt: Choose $z_0 in V$. Then $overlinez_0 in U$. Since $f$ is analytic on $U$, there is an $r > 0$ and a sequence of complex numbers $a_n_n = 0^infty$ such that for $overlinez in D(overlinez_0, r)$, we have
$$f(overlinez) = sum_n = 0^infty a_n(overlinez - overlinez_0)^n$$
It follows that
$$g(z) = overlinef(overlinez) = sum_n = 0^infty overlinea_n(z - z_0)^n$$
for all $z in D(z_0, r)$. Thus, $g$ is analytic on $V$.
complex-analysis proof-verification power-series
edited Jul 26 at 20:33
egreg
164k1180187
asked Jul 26 at 19:13
Bobo
244
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Question: Let $f$ be an analytic function in an open set $U$. Let $V = z in mathbbC : overlinez in U$. Define $g$ on $V$ by $g(z) = overlinef(overlinez)$. Show that $g$ is analytic on $V$ without using the fact that $g$ is holomorphic.
Proof Attempt: Choose $z_0 in V$. Then $overlinez_0 in U$. Since $f$ is analytic on $U$, there is an $r > 0$ and a sequence of complex numbers $a_n_n = 0^infty$ such that for $overlinez in D(overlinez_0, r)$, we have
$$f(overlinez) = sum_n = 0^infty a_n(overlinez - overlinez_0)^n$$
It follows that
$$g(z) = overlinef(overlinez) = sum_n = 0^infty overlinea_n(z - z_0)^n$$
for all $z in D(z_0, r)$. Thus, $g$ is analytic on $V$.
complex-analysis proof-verification power-series
edited Jul 26 at 20:33
egreg
164k1180187
asked Jul 26 at 19:13
Bobo
244
Question: Let $f$ be an analytic function in an open set $U$. Let $V = z in mathbbC : overlinez in U$. Define $g$ on $V$ by $g(z) = overlinef(overlinez)$. Show that $g$ is analytic on $V$ without using the fact that $g$ is holomorphic.
Proof Attempt: Choose $z_0 in V$. Then $overlinez_0 in U$. Since $f$ is analytic on $U$, there is an $r > 0$ and a sequence of complex numbers $a_n_n = 0^infty$ such that for $overlinez in D(overlinez_0, r)$, we have
$$f(overlinez) = sum_n = 0^infty a_n(overlinez - overlinez_0)^n$$
It follows that
$$g(z) = overlinef(overlinez) = sum_n = 0^infty overlinea_n(z - z_0)^n$$
for all $z in D(z_0, r)$. Thus, $g$ is analytic on $V$.
complex-analysis proof-verification power-series
edited Jul 26 at 20:33
egreg
164k1180187
asked Jul 26 at 19:13
Bobo
244
edited Jul 26 at 20:33
egreg
164k1180187
edited Jul 26 at 20:33
egreg
164k1180187
edited Jul 26 at 20:33
egreg
164k1180187
164k1180187
asked Jul 26 at 19:13
Bobo
244
asked Jul 26 at 19:13
Bobo
244
asked Jul 26 at 19:13
Bobo
244
244
add a comment |Â
add a comment |Â
1 Answer
1
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1
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Since conjugation is a homeomorphism, $V$ is open.
Let $cin V$; then $barcin U$, so there is an $r>0$ such that $D(barc,r)subseteq U$ and $f$ can be written as
$$
f(z)=sum_n=0^inftya_n(z-barc)^n tag1
$$
for every $zin D(barc,r)$. Therefore
$$
f(barz)=sum_n=0^inftya_n(barz-barc)^n
$$
for every $zinoverlineD(barc,r)=D(c,r)$.
The convergence is absolute on $D(barc,s)$, for every $0<s<r$, so it is not restrictive to assume the series $(1)$ is absolutely convergent on $D(barc,r)$ (by using $r/2$ instead of $r$).
Then the series
$$
sum_n=0^inftyoverlinea_n(z-c)^n
$$
is absolutely convergent on $overlineD(barc,r)=D(c,r)$, so it defines an analytic function over this disk. Since
$$
sum_n=0^inftyoverlinea_n(z-c)^n=
overlinesum_n=0^inftya_n(barz-barc)^n=
overlinef(barz)
$$
we are done.
Your idea is good, but the convergence of the final series should be mentioned.
answered Jul 26 at 22:28
egreg
164k1180187
This might be a bad question, but what allows us to conclude that the power series expansion is still valid outside of the disk of convergence? That is, does the power series expansion for f still hold if /overlinez /notin D(c, r)?
– Bobo
Jul 26 at 23:56
@Bobo The function $g$ admits a power series expansion in a neighborhood of each of its points, which is the same as saying $g$ is analytic. It is generally false that the same power series expansion holds in the whole domain of the function, consider for instance $f(z)=1/(1+z^2)$.
– egreg
Jul 27 at 6:50
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since conjugation is a homeomorphism, $V$ is open.
Let $cin V$; then $barcin U$, so there is an $r>0$ such that $D(barc,r)subseteq U$ and $f$ can be written as
$$
f(z)=sum_n=0^inftya_n(z-barc)^n tag1
$$
for every $zin D(barc,r)$. Therefore
$$
f(barz)=sum_n=0^inftya_n(barz-barc)^n
$$
for every $zinoverlineD(barc,r)=D(c,r)$.
The convergence is absolute on $D(barc,s)$, for every $0<s<r$, so it is not restrictive to assume the series $(1)$ is absolutely convergent on $D(barc,r)$ (by using $r/2$ instead of $r$).
Then the series
$$
sum_n=0^inftyoverlinea_n(z-c)^n
$$
is absolutely convergent on $overlineD(barc,r)=D(c,r)$, so it defines an analytic function over this disk. Since
$$
sum_n=0^inftyoverlinea_n(z-c)^n=
overlinesum_n=0^inftya_n(barz-barc)^n=
overlinef(barz)
$$
we are done.
Your idea is good, but the convergence of the final series should be mentioned.
answered Jul 26 at 22:28
egreg
164k1180187
This might be a bad question, but what allows us to conclude that the power series expansion is still valid outside of the disk of convergence? That is, does the power series expansion for f still hold if /overlinez /notin D(c, r)?
– Bobo
Jul 26 at 23:56
@Bobo The function $g$ admits a power series expansion in a neighborhood of each of its points, which is the same as saying $g$ is analytic. It is generally false that the same power series expansion holds in the whole domain of the function, consider for instance $f(z)=1/(1+z^2)$.
– egreg
Jul 27 at 6:50
add a comment |Â
up vote
1
down vote
accepted
Since conjugation is a homeomorphism, $V$ is open.
Let $cin V$; then $barcin U$, so there is an $r>0$ such that $D(barc,r)subseteq U$ and $f$ can be written as
$$
f(z)=sum_n=0^inftya_n(z-barc)^n tag1
$$
for every $zin D(barc,r)$. Therefore
$$
f(barz)=sum_n=0^inftya_n(barz-barc)^n
$$
for every $zinoverlineD(barc,r)=D(c,r)$.
The convergence is absolute on $D(barc,s)$, for every $0<s<r$, so it is not restrictive to assume the series $(1)$ is absolutely convergent on $D(barc,r)$ (by using $r/2$ instead of $r$).
Then the series
$$
sum_n=0^inftyoverlinea_n(z-c)^n
$$
is absolutely convergent on $overlineD(barc,r)=D(c,r)$, so it defines an analytic function over this disk. Since
$$
sum_n=0^inftyoverlinea_n(z-c)^n=
overlinesum_n=0^inftya_n(barz-barc)^n=
overlinef(barz)
$$
we are done.
Your idea is good, but the convergence of the final series should be mentioned.
answered Jul 26 at 22:28
egreg
164k1180187
This might be a bad question, but what allows us to conclude that the power series expansion is still valid outside of the disk of convergence? That is, does the power series expansion for f still hold if /overlinez /notin D(c, r)?
– Bobo
Jul 26 at 23:56
@Bobo The function $g$ admits a power series expansion in a neighborhood of each of its points, which is the same as saying $g$ is analytic. It is generally false that the same power series expansion holds in the whole domain of the function, consider for instance $f(z)=1/(1+z^2)$.
– egreg
Jul 27 at 6:50
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since conjugation is a homeomorphism, $V$ is open.
Let $cin V$; then $barcin U$, so there is an $r>0$ such that $D(barc,r)subseteq U$ and $f$ can be written as
$$
f(z)=sum_n=0^inftya_n(z-barc)^n tag1
$$
for every $zin D(barc,r)$. Therefore
$$
f(barz)=sum_n=0^inftya_n(barz-barc)^n
$$
for every $zinoverlineD(barc,r)=D(c,r)$.
The convergence is absolute on $D(barc,s)$, for every $0<s<r$, so it is not restrictive to assume the series $(1)$ is absolutely convergent on $D(barc,r)$ (by using $r/2$ instead of $r$).
Then the series
$$
sum_n=0^inftyoverlinea_n(z-c)^n
$$
is absolutely convergent on $overlineD(barc,r)=D(c,r)$, so it defines an analytic function over this disk. Since
$$
sum_n=0^inftyoverlinea_n(z-c)^n=
overlinesum_n=0^inftya_n(barz-barc)^n=
overlinef(barz)
$$
we are done.
Your idea is good, but the convergence of the final series should be mentioned.
answered Jul 26 at 22:28
egreg
164k1180187
Since conjugation is a homeomorphism, $V$ is open.
Let $cin V$; then $barcin U$, so there is an $r>0$ such that $D(barc,r)subseteq U$ and $f$ can be written as
$$
f(z)=sum_n=0^inftya_n(z-barc)^n tag1
$$
for every $zin D(barc,r)$. Therefore
$$
f(barz)=sum_n=0^inftya_n(barz-barc)^n
$$
for every $zinoverlineD(barc,r)=D(c,r)$.
The convergence is absolute on $D(barc,s)$, for every $0<s<r$, so it is not restrictive to assume the series $(1)$ is absolutely convergent on $D(barc,r)$ (by using $r/2$ instead of $r$).
Then the series
$$
sum_n=0^inftyoverlinea_n(z-c)^n
$$
is absolutely convergent on $overlineD(barc,r)=D(c,r)$, so it defines an analytic function over this disk. Since
$$
sum_n=0^inftyoverlinea_n(z-c)^n=
overlinesum_n=0^inftya_n(barz-barc)^n=
overlinef(barz)
$$
we are done.
Your idea is good, but the convergence of the final series should be mentioned.
answered Jul 26 at 22:28
egreg
164k1180187
answered Jul 26 at 22:28
egreg
164k1180187
answered Jul 26 at 22:28
egreg
164k1180187
answered Jul 26 at 22:28
egreg
164k1180187
164k1180187
This might be a bad question, but what allows us to conclude that the power series expansion is still valid outside of the disk of convergence? That is, does the power series expansion for f still hold if /overlinez /notin D(c, r)?
– Bobo
Jul 26 at 23:56
@Bobo The function $g$ admits a power series expansion in a neighborhood of each of its points, which is the same as saying $g$ is analytic. It is generally false that the same power series expansion holds in the whole domain of the function, consider for instance $f(z)=1/(1+z^2)$.
– egreg
Jul 27 at 6:50
add a comment |Â
This might be a bad question, but what allows us to conclude that the power series expansion is still valid outside of the disk of convergence? That is, does the power series expansion for f still hold if /overlinez /notin D(c, r)?
– Bobo
Jul 26 at 23:56
@Bobo The function $g$ admits a power series expansion in a neighborhood of each of its points, which is the same as saying $g$ is analytic. It is generally false that the same power series expansion holds in the whole domain of the function, consider for instance $f(z)=1/(1+z^2)$.
– egreg
Jul 27 at 6:50
This might be a bad question, but what allows us to conclude that the power series expansion is still valid outside of the disk of convergence? That is, does the power series expansion for f still hold if /overlinez /notin D(c, r)?
– Bobo
Jul 26 at 23:56
This might be a bad question, but what allows us to conclude that the power series expansion is still valid outside of the disk of convergence? That is, does the power series expansion for f still hold if /overlinez /notin D(c, r)?
– Bobo
Jul 26 at 23:56
@Bobo The function $g$ admits a power series expansion in a neighborhood of each of its points, which is the same as saying $g$ is analytic. It is generally false that the same power series expansion holds in the whole domain of the function, consider for instance $f(z)=1/(1+z^2)$.
– egreg
Jul 27 at 6:50
@Bobo The function $g$ admits a power series expansion in a neighborhood of each of its points, which is the same as saying $g$ is analytic. It is generally false that the same power series expansion holds in the whole domain of the function, consider for instance $f(z)=1/(1+z^2)$.
– egreg
Jul 27 at 6:50
add a comment |Â
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