Even holomorphic function on the punctured disk has a primitive












2












$begingroup$


Let $f$ such that $f$ is holomorphic on ${z|0<|z|<1}$, and $f$ is even.



I need to show that f has a primitive. Any ideas?






complex-analysis holomorphic-functions







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$endgroup$












  • $begingroup$
    Now what, odd or even?
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 18:53










  • $begingroup$
    Even, sorry fixed.
    $endgroup$
    – user3708158
    Jan 22 at 18:53


















2












$begingroup$


Let $f$ such that $f$ is holomorphic on ${z|0<|z|<1}$, and $f$ is even.



I need to show that f has a primitive. Any ideas?






complex-analysis holomorphic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Now what, odd or even?
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 18:53










  • $begingroup$
    Even, sorry fixed.
    $endgroup$
    – user3708158
    Jan 22 at 18:53
















2












2








2


1



$begingroup$


Let $f$ such that $f$ is holomorphic on ${z|0<|z|<1}$, and $f$ is even.



I need to show that f has a primitive. Any ideas?






complex-analysis holomorphic-functions







share|cite|improve this question











$endgroup$




Let $f$ such that $f$ is holomorphic on ${z|0<|z|<1}$, and $f$ is even.



I need to show that f has a primitive. Any ideas?






complex-analysis holomorphic-functions




complex-analysis holomorphic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 18:54







user3708158

















asked Jan 22 at 18:52









user3708158user3708158

1307




1307












  • $begingroup$
    Now what, odd or even?
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 18:53










  • $begingroup$
    Even, sorry fixed.
    $endgroup$
    – user3708158
    Jan 22 at 18:53




















  • $begingroup$
    Now what, odd or even?
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 18:53










  • $begingroup$
    Even, sorry fixed.
    $endgroup$
    – user3708158
    Jan 22 at 18:53


















$begingroup$
Now what, odd or even?
$endgroup$
– Hagen von Eitzen
Jan 22 at 18:53




$begingroup$
Now what, odd or even?
$endgroup$
– Hagen von Eitzen
Jan 22 at 18:53












$begingroup$
Even, sorry fixed.
$endgroup$
– user3708158
Jan 22 at 18:53






$begingroup$
Even, sorry fixed.
$endgroup$
– user3708158
Jan 22 at 18:53












3 Answers
3






active

oldest

votes


















2












$begingroup$

Write $f$ in the form of Laurent series around $0$. You get the form $f(z)=sum_{n=-infty}^infty a_nz^n$. Now, if we prove that $a_{-1}=0$ then $F(z)=sum_{n=-infty}^{infty}frac{a_n}{n+1}z^{n+1}$ is a primitive function of $f$. So we just have to prove that $a_{-1}=0$ and we are done. And here we are going to use that $f$ is even. We know that $sum_{n=-infty}^infty a_nz^n=f(z)=f(-z)=sum_{n=-infty}^infty a_n(-z)^n$. So if we define $g(z)=f(z)-f(-z)$ then $g$ is the zero function and its Laurent expansion is $sum_{n=-infty}^infty a_{2n+1}z^{2n+1}$. Because Laurent expansion is unique we conclude that $a_{2n+1}=0$ for all $ninmathbb{Z}$. So $a_{-1}=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can we take Laurent series around 0, if 0 is not in our domain?
    $endgroup$
    – user3708158
    Jan 22 at 19:20










  • $begingroup$
    The origin is a singular point so you can take Laurent series around it.
    $endgroup$
    – Mark
    Jan 22 at 19:23










  • $begingroup$
    Perfect, thank you.
    $endgroup$
    – user3708158
    Jan 22 at 19:26



















1












$begingroup$

Let $F$ be a primitive of $frestriction_{Bbb Dsetminus(-infty,0]}$. Then $zmapsto F(z)+F(-z)$ is constant where defined, i..e, constant on the upper half disk and constant on the lower half disk. Clearly, the constants are negatives of each other. By adding a suitable constant to $F$, we can achieve that $F(z)+F(-z)=0$ thoughout, i.e., $F$ is odd. If we defien $F(z)=-F(-z)$ on $(-1,0)$, we obtain a continuos (and necesarily holomorphic) function on the whole punctured disk.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not sure it is constant on the lower and upper half disc. the opration $z rightarrow -z$ is symetric relatively to the line $y=-x$, right?
    $endgroup$
    – user3708158
    Jan 22 at 19:21










  • $begingroup$
    @user3708158 The derivative of $F(z)+F(-z)$ is $f(z)-f(-z)$. As the latter is $=0$, the former is locally constant.
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 21:07



















1












$begingroup$

Let $0 < r < 1$. Then we have
$$int_{partial B_r(0)} f(z), dz =
int_0^{2pi} f(re^(it)ire^{it}, dt
= int_0^{pi} f(re^{it})ire^{it}, dt + int_pi^{2pi} f(re^{it})ire^{it}, dt.$$
For the second integal we have
$$ int_pi^{2pi} f(re^{it})ire^{it}, dt
= int_0^{pi} f(re^{i(t+ pi)})ire^{i(t+ pi) }, dt
= -int_0^{pi} f(-re^{it})ire^{it }, dt=
-int_0^{pi} f(re^{it})ire^{it }, dt,$$
by the evenness of $f$.
By homotopy, every simple loop avoiding 0 integrates to 0. Now pick a any point $z_0$ in the punctured disk. Define $F(z) = int_gamma f(z), dz,$
where $gamma$ is a path connecting $z_0$ to $z$. This is well-defined because of the fact for that every loop in the punctured disk, the integral along the loop integrates to zero.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is $f$ even implies $f(z)dz$ is odd $1$-form : $f(-z)d(-z) = -f(z)dz$
    $endgroup$
    – reuns
    Jan 22 at 21:11












  • $begingroup$
    Can you elaborate on the part from: "By homotopy, every...." ? I don't think that the punctured disk is a simply connected set.
    $endgroup$
    – user3708158
    Jan 22 at 21:38










  • $begingroup$
    It is not. See John Conway's Complex Variables for this connection. It is very cool.
    $endgroup$
    – ncmathsadist
    Jan 23 at 0:33











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Write $f$ in the form of Laurent series around $0$. You get the form $f(z)=sum_{n=-infty}^infty a_nz^n$. Now, if we prove that $a_{-1}=0$ then $F(z)=sum_{n=-infty}^{infty}frac{a_n}{n+1}z^{n+1}$ is a primitive function of $f$. So we just have to prove that $a_{-1}=0$ and we are done. And here we are going to use that $f$ is even. We know that $sum_{n=-infty}^infty a_nz^n=f(z)=f(-z)=sum_{n=-infty}^infty a_n(-z)^n$. So if we define $g(z)=f(z)-f(-z)$ then $g$ is the zero function and its Laurent expansion is $sum_{n=-infty}^infty a_{2n+1}z^{2n+1}$. Because Laurent expansion is unique we conclude that $a_{2n+1}=0$ for all $ninmathbb{Z}$. So $a_{-1}=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can we take Laurent series around 0, if 0 is not in our domain?
    $endgroup$
    – user3708158
    Jan 22 at 19:20










  • $begingroup$
    The origin is a singular point so you can take Laurent series around it.
    $endgroup$
    – Mark
    Jan 22 at 19:23










  • $begingroup$
    Perfect, thank you.
    $endgroup$
    – user3708158
    Jan 22 at 19:26
















2












$begingroup$

Write $f$ in the form of Laurent series around $0$. You get the form $f(z)=sum_{n=-infty}^infty a_nz^n$. Now, if we prove that $a_{-1}=0$ then $F(z)=sum_{n=-infty}^{infty}frac{a_n}{n+1}z^{n+1}$ is a primitive function of $f$. So we just have to prove that $a_{-1}=0$ and we are done. And here we are going to use that $f$ is even. We know that $sum_{n=-infty}^infty a_nz^n=f(z)=f(-z)=sum_{n=-infty}^infty a_n(-z)^n$. So if we define $g(z)=f(z)-f(-z)$ then $g$ is the zero function and its Laurent expansion is $sum_{n=-infty}^infty a_{2n+1}z^{2n+1}$. Because Laurent expansion is unique we conclude that $a_{2n+1}=0$ for all $ninmathbb{Z}$. So $a_{-1}=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can we take Laurent series around 0, if 0 is not in our domain?
    $endgroup$
    – user3708158
    Jan 22 at 19:20










  • $begingroup$
    The origin is a singular point so you can take Laurent series around it.
    $endgroup$
    – Mark
    Jan 22 at 19:23










  • $begingroup$
    Perfect, thank you.
    $endgroup$
    – user3708158
    Jan 22 at 19:26














2












2








2





$begingroup$

Write $f$ in the form of Laurent series around $0$. You get the form $f(z)=sum_{n=-infty}^infty a_nz^n$. Now, if we prove that $a_{-1}=0$ then $F(z)=sum_{n=-infty}^{infty}frac{a_n}{n+1}z^{n+1}$ is a primitive function of $f$. So we just have to prove that $a_{-1}=0$ and we are done. And here we are going to use that $f$ is even. We know that $sum_{n=-infty}^infty a_nz^n=f(z)=f(-z)=sum_{n=-infty}^infty a_n(-z)^n$. So if we define $g(z)=f(z)-f(-z)$ then $g$ is the zero function and its Laurent expansion is $sum_{n=-infty}^infty a_{2n+1}z^{2n+1}$. Because Laurent expansion is unique we conclude that $a_{2n+1}=0$ for all $ninmathbb{Z}$. So $a_{-1}=0$.






share|cite|improve this answer









$endgroup$



Write $f$ in the form of Laurent series around $0$. You get the form $f(z)=sum_{n=-infty}^infty a_nz^n$. Now, if we prove that $a_{-1}=0$ then $F(z)=sum_{n=-infty}^{infty}frac{a_n}{n+1}z^{n+1}$ is a primitive function of $f$. So we just have to prove that $a_{-1}=0$ and we are done. And here we are going to use that $f$ is even. We know that $sum_{n=-infty}^infty a_nz^n=f(z)=f(-z)=sum_{n=-infty}^infty a_n(-z)^n$. So if we define $g(z)=f(z)-f(-z)$ then $g$ is the zero function and its Laurent expansion is $sum_{n=-infty}^infty a_{2n+1}z^{2n+1}$. Because Laurent expansion is unique we conclude that $a_{2n+1}=0$ for all $ninmathbb{Z}$. So $a_{-1}=0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 22 at 19:13









MarkMark

9,459622




9,459622












  • $begingroup$
    How can we take Laurent series around 0, if 0 is not in our domain?
    $endgroup$
    – user3708158
    Jan 22 at 19:20










  • $begingroup$
    The origin is a singular point so you can take Laurent series around it.
    $endgroup$
    – Mark
    Jan 22 at 19:23










  • $begingroup$
    Perfect, thank you.
    $endgroup$
    – user3708158
    Jan 22 at 19:26


















  • $begingroup$
    How can we take Laurent series around 0, if 0 is not in our domain?
    $endgroup$
    – user3708158
    Jan 22 at 19:20










  • $begingroup$
    The origin is a singular point so you can take Laurent series around it.
    $endgroup$
    – Mark
    Jan 22 at 19:23










  • $begingroup$
    Perfect, thank you.
    $endgroup$
    – user3708158
    Jan 22 at 19:26
















$begingroup$
How can we take Laurent series around 0, if 0 is not in our domain?
$endgroup$
– user3708158
Jan 22 at 19:20




$begingroup$
How can we take Laurent series around 0, if 0 is not in our domain?
$endgroup$
– user3708158
Jan 22 at 19:20












$begingroup$
The origin is a singular point so you can take Laurent series around it.
$endgroup$
– Mark
Jan 22 at 19:23




$begingroup$
The origin is a singular point so you can take Laurent series around it.
$endgroup$
– Mark
Jan 22 at 19:23












$begingroup$
Perfect, thank you.
$endgroup$
– user3708158
Jan 22 at 19:26




$begingroup$
Perfect, thank you.
$endgroup$
– user3708158
Jan 22 at 19:26











1












$begingroup$

Let $F$ be a primitive of $frestriction_{Bbb Dsetminus(-infty,0]}$. Then $zmapsto F(z)+F(-z)$ is constant where defined, i..e, constant on the upper half disk and constant on the lower half disk. Clearly, the constants are negatives of each other. By adding a suitable constant to $F$, we can achieve that $F(z)+F(-z)=0$ thoughout, i.e., $F$ is odd. If we defien $F(z)=-F(-z)$ on $(-1,0)$, we obtain a continuos (and necesarily holomorphic) function on the whole punctured disk.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not sure it is constant on the lower and upper half disc. the opration $z rightarrow -z$ is symetric relatively to the line $y=-x$, right?
    $endgroup$
    – user3708158
    Jan 22 at 19:21










  • $begingroup$
    @user3708158 The derivative of $F(z)+F(-z)$ is $f(z)-f(-z)$. As the latter is $=0$, the former is locally constant.
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 21:07
















1












$begingroup$

Let $F$ be a primitive of $frestriction_{Bbb Dsetminus(-infty,0]}$. Then $zmapsto F(z)+F(-z)$ is constant where defined, i..e, constant on the upper half disk and constant on the lower half disk. Clearly, the constants are negatives of each other. By adding a suitable constant to $F$, we can achieve that $F(z)+F(-z)=0$ thoughout, i.e., $F$ is odd. If we defien $F(z)=-F(-z)$ on $(-1,0)$, we obtain a continuos (and necesarily holomorphic) function on the whole punctured disk.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am not sure it is constant on the lower and upper half disc. the opration $z rightarrow -z$ is symetric relatively to the line $y=-x$, right?
    $endgroup$
    – user3708158
    Jan 22 at 19:21










  • $begingroup$
    @user3708158 The derivative of $F(z)+F(-z)$ is $f(z)-f(-z)$. As the latter is $=0$, the former is locally constant.
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 21:07














1












1








1





$begingroup$

Let $F$ be a primitive of $frestriction_{Bbb Dsetminus(-infty,0]}$. Then $zmapsto F(z)+F(-z)$ is constant where defined, i..e, constant on the upper half disk and constant on the lower half disk. Clearly, the constants are negatives of each other. By adding a suitable constant to $F$, we can achieve that $F(z)+F(-z)=0$ thoughout, i.e., $F$ is odd. If we defien $F(z)=-F(-z)$ on $(-1,0)$, we obtain a continuos (and necesarily holomorphic) function on the whole punctured disk.






share|cite|improve this answer









$endgroup$



Let $F$ be a primitive of $frestriction_{Bbb Dsetminus(-infty,0]}$. Then $zmapsto F(z)+F(-z)$ is constant where defined, i..e, constant on the upper half disk and constant on the lower half disk. Clearly, the constants are negatives of each other. By adding a suitable constant to $F$, we can achieve that $F(z)+F(-z)=0$ thoughout, i.e., $F$ is odd. If we defien $F(z)=-F(-z)$ on $(-1,0)$, we obtain a continuos (and necesarily holomorphic) function on the whole punctured disk.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 22 at 19:14









Hagen von EitzenHagen von Eitzen

281k23272505




281k23272505












  • $begingroup$
    I am not sure it is constant on the lower and upper half disc. the opration $z rightarrow -z$ is symetric relatively to the line $y=-x$, right?
    $endgroup$
    – user3708158
    Jan 22 at 19:21










  • $begingroup$
    @user3708158 The derivative of $F(z)+F(-z)$ is $f(z)-f(-z)$. As the latter is $=0$, the former is locally constant.
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 21:07


















  • $begingroup$
    I am not sure it is constant on the lower and upper half disc. the opration $z rightarrow -z$ is symetric relatively to the line $y=-x$, right?
    $endgroup$
    – user3708158
    Jan 22 at 19:21










  • $begingroup$
    @user3708158 The derivative of $F(z)+F(-z)$ is $f(z)-f(-z)$. As the latter is $=0$, the former is locally constant.
    $endgroup$
    – Hagen von Eitzen
    Jan 22 at 21:07
















$begingroup$
I am not sure it is constant on the lower and upper half disc. the opration $z rightarrow -z$ is symetric relatively to the line $y=-x$, right?
$endgroup$
– user3708158
Jan 22 at 19:21




$begingroup$
I am not sure it is constant on the lower and upper half disc. the opration $z rightarrow -z$ is symetric relatively to the line $y=-x$, right?
$endgroup$
– user3708158
Jan 22 at 19:21












$begingroup$
@user3708158 The derivative of $F(z)+F(-z)$ is $f(z)-f(-z)$. As the latter is $=0$, the former is locally constant.
$endgroup$
– Hagen von Eitzen
Jan 22 at 21:07




$begingroup$
@user3708158 The derivative of $F(z)+F(-z)$ is $f(z)-f(-z)$. As the latter is $=0$, the former is locally constant.
$endgroup$
– Hagen von Eitzen
Jan 22 at 21:07











1












$begingroup$

Let $0 < r < 1$. Then we have
$$int_{partial B_r(0)} f(z), dz =
int_0^{2pi} f(re^(it)ire^{it}, dt
= int_0^{pi} f(re^{it})ire^{it}, dt + int_pi^{2pi} f(re^{it})ire^{it}, dt.$$
For the second integal we have
$$ int_pi^{2pi} f(re^{it})ire^{it}, dt
= int_0^{pi} f(re^{i(t+ pi)})ire^{i(t+ pi) }, dt
= -int_0^{pi} f(-re^{it})ire^{it }, dt=
-int_0^{pi} f(re^{it})ire^{it }, dt,$$
by the evenness of $f$.
By homotopy, every simple loop avoiding 0 integrates to 0. Now pick a any point $z_0$ in the punctured disk. Define $F(z) = int_gamma f(z), dz,$
where $gamma$ is a path connecting $z_0$ to $z$. This is well-defined because of the fact for that every loop in the punctured disk, the integral along the loop integrates to zero.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is $f$ even implies $f(z)dz$ is odd $1$-form : $f(-z)d(-z) = -f(z)dz$
    $endgroup$
    – reuns
    Jan 22 at 21:11












  • $begingroup$
    Can you elaborate on the part from: "By homotopy, every...." ? I don't think that the punctured disk is a simply connected set.
    $endgroup$
    – user3708158
    Jan 22 at 21:38










  • $begingroup$
    It is not. See John Conway's Complex Variables for this connection. It is very cool.
    $endgroup$
    – ncmathsadist
    Jan 23 at 0:33
















1












$begingroup$

Let $0 < r < 1$. Then we have
$$int_{partial B_r(0)} f(z), dz =
int_0^{2pi} f(re^(it)ire^{it}, dt
= int_0^{pi} f(re^{it})ire^{it}, dt + int_pi^{2pi} f(re^{it})ire^{it}, dt.$$
For the second integal we have
$$ int_pi^{2pi} f(re^{it})ire^{it}, dt
= int_0^{pi} f(re^{i(t+ pi)})ire^{i(t+ pi) }, dt
= -int_0^{pi} f(-re^{it})ire^{it }, dt=
-int_0^{pi} f(re^{it})ire^{it }, dt,$$
by the evenness of $f$.
By homotopy, every simple loop avoiding 0 integrates to 0. Now pick a any point $z_0$ in the punctured disk. Define $F(z) = int_gamma f(z), dz,$
where $gamma$ is a path connecting $z_0$ to $z$. This is well-defined because of the fact for that every loop in the punctured disk, the integral along the loop integrates to zero.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is $f$ even implies $f(z)dz$ is odd $1$-form : $f(-z)d(-z) = -f(z)dz$
    $endgroup$
    – reuns
    Jan 22 at 21:11












  • $begingroup$
    Can you elaborate on the part from: "By homotopy, every...." ? I don't think that the punctured disk is a simply connected set.
    $endgroup$
    – user3708158
    Jan 22 at 21:38










  • $begingroup$
    It is not. See John Conway's Complex Variables for this connection. It is very cool.
    $endgroup$
    – ncmathsadist
    Jan 23 at 0:33














1












1








1





$begingroup$

Let $0 < r < 1$. Then we have
$$int_{partial B_r(0)} f(z), dz =
int_0^{2pi} f(re^(it)ire^{it}, dt
= int_0^{pi} f(re^{it})ire^{it}, dt + int_pi^{2pi} f(re^{it})ire^{it}, dt.$$
For the second integal we have
$$ int_pi^{2pi} f(re^{it})ire^{it}, dt
= int_0^{pi} f(re^{i(t+ pi)})ire^{i(t+ pi) }, dt
= -int_0^{pi} f(-re^{it})ire^{it }, dt=
-int_0^{pi} f(re^{it})ire^{it }, dt,$$
by the evenness of $f$.
By homotopy, every simple loop avoiding 0 integrates to 0. Now pick a any point $z_0$ in the punctured disk. Define $F(z) = int_gamma f(z), dz,$
where $gamma$ is a path connecting $z_0$ to $z$. This is well-defined because of the fact for that every loop in the punctured disk, the integral along the loop integrates to zero.






share|cite|improve this answer











$endgroup$



Let $0 < r < 1$. Then we have
$$int_{partial B_r(0)} f(z), dz =
int_0^{2pi} f(re^(it)ire^{it}, dt
= int_0^{pi} f(re^{it})ire^{it}, dt + int_pi^{2pi} f(re^{it})ire^{it}, dt.$$
For the second integal we have
$$ int_pi^{2pi} f(re^{it})ire^{it}, dt
= int_0^{pi} f(re^{i(t+ pi)})ire^{i(t+ pi) }, dt
= -int_0^{pi} f(-re^{it})ire^{it }, dt=
-int_0^{pi} f(re^{it})ire^{it }, dt,$$
by the evenness of $f$.
By homotopy, every simple loop avoiding 0 integrates to 0. Now pick a any point $z_0$ in the punctured disk. Define $F(z) = int_gamma f(z), dz,$
where $gamma$ is a path connecting $z_0$ to $z$. This is well-defined because of the fact for that every loop in the punctured disk, the integral along the loop integrates to zero.







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edited Jan 22 at 21:14

























answered Jan 22 at 21:09









ncmathsadistncmathsadist

42.9k260103




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  • $begingroup$
    That is $f$ even implies $f(z)dz$ is odd $1$-form : $f(-z)d(-z) = -f(z)dz$
    $endgroup$
    – reuns
    Jan 22 at 21:11












  • $begingroup$
    Can you elaborate on the part from: "By homotopy, every...." ? I don't think that the punctured disk is a simply connected set.
    $endgroup$
    – user3708158
    Jan 22 at 21:38










  • $begingroup$
    It is not. See John Conway's Complex Variables for this connection. It is very cool.
    $endgroup$
    – ncmathsadist
    Jan 23 at 0:33


















  • $begingroup$
    That is $f$ even implies $f(z)dz$ is odd $1$-form : $f(-z)d(-z) = -f(z)dz$
    $endgroup$
    – reuns
    Jan 22 at 21:11












  • $begingroup$
    Can you elaborate on the part from: "By homotopy, every...." ? I don't think that the punctured disk is a simply connected set.
    $endgroup$
    – user3708158
    Jan 22 at 21:38










  • $begingroup$
    It is not. See John Conway's Complex Variables for this connection. It is very cool.
    $endgroup$
    – ncmathsadist
    Jan 23 at 0:33
















$begingroup$
That is $f$ even implies $f(z)dz$ is odd $1$-form : $f(-z)d(-z) = -f(z)dz$
$endgroup$
– reuns
Jan 22 at 21:11






$begingroup$
That is $f$ even implies $f(z)dz$ is odd $1$-form : $f(-z)d(-z) = -f(z)dz$
$endgroup$
– reuns
Jan 22 at 21:11














$begingroup$
Can you elaborate on the part from: "By homotopy, every...." ? I don't think that the punctured disk is a simply connected set.
$endgroup$
– user3708158
Jan 22 at 21:38




$begingroup$
Can you elaborate on the part from: "By homotopy, every...." ? I don't think that the punctured disk is a simply connected set.
$endgroup$
– user3708158
Jan 22 at 21:38












$begingroup$
It is not. See John Conway's Complex Variables for this connection. It is very cool.
$endgroup$
– ncmathsadist
Jan 23 at 0:33




$begingroup$
It is not. See John Conway's Complex Variables for this connection. It is very cool.
$endgroup$
– ncmathsadist
Jan 23 at 0:33


















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