Complex Analytic Function's Power Series
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Studying complex function, I'm facing some fundamental question:
Given open and bounded set $Omega in mathbb{C}$ and $f:Omega rightarrow Omega$ holomorphic on $Omega$ which $f(z_{0})=z_{0}$. $f$ can be represented as power series around $z_{0}$ as $f(z)=sum_{n=1}^{infty}a_{n}(z-z_{0})^{n}$ so within the radius of convergence around $z_{0}$ the equality holds. therefore when I plug in $z_{0}$ I get $f(z_{0})= sum _{n=1} ^{infty}a_{n}(z_{0}-z_{0})^{n}=0$ so $z_{0}=0$. On the other hand, I can create the function $g(z)=z$ which is also holomorphic on $Omega$ but for $0neq z_{1}in Omega$ I also get that $z_{1}=0$ which contradicts with the definition of $f$. What am I missing?
complex-analysis power-series holomorphic-functions
edited Nov 26 at 10:06
caverac
12.4k21027
asked Nov 26 at 9:45
Yaron Scherf
1187
add a comment |
up vote
1
down vote
favorite
Studying complex function, I'm facing some fundamental question:
Given open and bounded set $Omega in mathbb{C}$ and $f:Omega rightarrow Omega$ holomorphic on $Omega$ which $f(z_{0})=z_{0}$. $f$ can be represented as power series around $z_{0}$ as $f(z)=sum_{n=1}^{infty}a_{n}(z-z_{0})^{n}$ so within the radius of convergence around $z_{0}$ the equality holds. therefore when I plug in $z_{0}$ I get $f(z_{0})= sum _{n=1} ^{infty}a_{n}(z_{0}-z_{0})^{n}=0$ so $z_{0}=0$. On the other hand, I can create the function $g(z)=z$ which is also holomorphic on $Omega$ but for $0neq z_{1}in Omega$ I also get that $z_{1}=0$ which contradicts with the definition of $f$. What am I missing?
complex-analysis power-series holomorphic-functions
edited Nov 26 at 10:06
caverac
12.4k21027
asked Nov 26 at 9:45
Yaron Scherf
1187
If $f(z_0) = z_0$ then $f(z) = z_0 + sum_{i=1}^{infty}a_{n}(z-z_{0})^{n}$ and plugging in $z_0$ gives just $z_0 = z_0$. Your conclusion $z_0 = 0$ seems to be wrong.
– Martin R
Nov 26 at 9:50
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Studying complex function, I'm facing some fundamental question:
Given open and bounded set $Omega in mathbb{C}$ and $f:Omega rightarrow Omega$ holomorphic on $Omega$ which $f(z_{0})=z_{0}$. $f$ can be represented as power series around $z_{0}$ as $f(z)=sum_{n=1}^{infty}a_{n}(z-z_{0})^{n}$ so within the radius of convergence around $z_{0}$ the equality holds. therefore when I plug in $z_{0}$ I get $f(z_{0})= sum _{n=1} ^{infty}a_{n}(z_{0}-z_{0})^{n}=0$ so $z_{0}=0$. On the other hand, I can create the function $g(z)=z$ which is also holomorphic on $Omega$ but for $0neq z_{1}in Omega$ I also get that $z_{1}=0$ which contradicts with the definition of $f$. What am I missing?
complex-analysis power-series holomorphic-functions
edited Nov 26 at 10:06
caverac
12.4k21027
asked Nov 26 at 9:45
Yaron Scherf
1187
Studying complex function, I'm facing some fundamental question:
Given open and bounded set $Omega in mathbb{C}$ and $f:Omega rightarrow Omega$ holomorphic on $Omega$ which $f(z_{0})=z_{0}$. $f$ can be represented as power series around $z_{0}$ as $f(z)=sum_{n=1}^{infty}a_{n}(z-z_{0})^{n}$ so within the radius of convergence around $z_{0}$ the equality holds. therefore when I plug in $z_{0}$ I get $f(z_{0})= sum _{n=1} ^{infty}a_{n}(z_{0}-z_{0})^{n}=0$ so $z_{0}=0$. On the other hand, I can create the function $g(z)=z$ which is also holomorphic on $Omega$ but for $0neq z_{1}in Omega$ I also get that $z_{1}=0$ which contradicts with the definition of $f$. What am I missing?
complex-analysis power-series holomorphic-functions
complex-analysis power-series holomorphic-functions
edited Nov 26 at 10:06
caverac
12.4k21027
asked Nov 26 at 9:45
Yaron Scherf
1187
edited Nov 26 at 10:06
caverac
12.4k21027
asked Nov 26 at 9:45
Yaron Scherf
1187
edited Nov 26 at 10:06
caverac
12.4k21027
edited Nov 26 at 10:06
caverac
12.4k21027
edited Nov 26 at 10:06
caverac
12.4k21027
12.4k21027
asked Nov 26 at 9:45
Yaron Scherf
1187
asked Nov 26 at 9:45
Yaron Scherf
1187
asked Nov 26 at 9:45
Yaron Scherf
1187
1187
If $f(z_0) = z_0$ then $f(z) = z_0 + sum_{i=1}^{infty}a_{n}(z-z_{0})^{n}$ and plugging in $z_0$ gives just $z_0 = z_0$. Your conclusion $z_0 = 0$ seems to be wrong.
– Martin R
Nov 26 at 9:50
add a comment |
If $f(z_0) = z_0$ then $f(z) = z_0 + sum_{i=1}^{infty}a_{n}(z-z_{0})^{n}$ and plugging in $z_0$ gives just $z_0 = z_0$. Your conclusion $z_0 = 0$ seems to be wrong.
– Martin R
Nov 26 at 9:50
If $f(z_0) = z_0$ then $f(z) = z_0 + sum_{i=1}^{infty}a_{n}(z-z_{0})^{n}$ and plugging in $z_0$ gives just $z_0 = z_0$. Your conclusion $z_0 = 0$ seems to be wrong.
– Martin R
Nov 26 at 9:50
If $f(z_0) = z_0$ then $f(z) = z_0 + sum_{i=1}^{infty}a_{n}(z-z_{0})^{n}$ and plugging in $z_0$ gives just $z_0 = z_0$. Your conclusion $z_0 = 0$ seems to be wrong.
– Martin R
Nov 26 at 9:50
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You're missing the first term in the sum
$$
f(z) = sum_{color{red}{n = 0}}^{+infty}a_n(z - z_0)^n
$$
with this in mind
$$
f(z_0) = a_0 + a_1(z_0 - z_0) + cdots = a_0
$$
In you case case $a_0 = z_0$
answered Nov 26 at 9:51
caverac
12.4k21027
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You're missing the first term in the sum
$$
f(z) = sum_{color{red}{n = 0}}^{+infty}a_n(z - z_0)^n
$$
with this in mind
$$
f(z_0) = a_0 + a_1(z_0 - z_0) + cdots = a_0
$$
In you case case $a_0 = z_0$
answered Nov 26 at 9:51
caverac
12.4k21027
add a comment |
up vote
1
down vote
accepted
You're missing the first term in the sum
$$
f(z) = sum_{color{red}{n = 0}}^{+infty}a_n(z - z_0)^n
$$
with this in mind
$$
f(z_0) = a_0 + a_1(z_0 - z_0) + cdots = a_0
$$
In you case case $a_0 = z_0$
answered Nov 26 at 9:51
caverac
12.4k21027
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You're missing the first term in the sum
$$
f(z) = sum_{color{red}{n = 0}}^{+infty}a_n(z - z_0)^n
$$
with this in mind
$$
f(z_0) = a_0 + a_1(z_0 - z_0) + cdots = a_0
$$
In you case case $a_0 = z_0$
answered Nov 26 at 9:51
caverac
12.4k21027
You're missing the first term in the sum
$$
f(z) = sum_{color{red}{n = 0}}^{+infty}a_n(z - z_0)^n
$$
with this in mind
$$
f(z_0) = a_0 + a_1(z_0 - z_0) + cdots = a_0
$$
In you case case $a_0 = z_0$
answered Nov 26 at 9:51
caverac
12.4k21027
answered Nov 26 at 9:51
caverac
12.4k21027
answered Nov 26 at 9:51
caverac
12.4k21027
answered Nov 26 at 9:51
caverac
12.4k21027
12.4k21027
add a comment |
add a comment |
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If $f(z_0) = z_0$ then $f(z) = z_0 + sum_{i=1}^{infty}a_{n}(z-z_{0})^{n}$ and plugging in $z_0$ gives just $z_0 = z_0$. Your conclusion $z_0 = 0$ seems to be wrong.
– Martin R
Nov 26 at 9:50