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Question on Morera's theorem proof
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I am trying to understand this proof and I am having trouble understanding a couple things.
2) How do we know that F is analytic?
3) Is it necessary to say the things about primitives?
3) How do we know that $F'=f$? Maybe I am remembering incorrectly, but for some reason I thought the Fundamental Theorem of Calculus didn't apply for complex analysis...
Thanks!
complex-analysis
asked Jul 25 at 5:00
MathIsHard
1,122415
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up vote
0
down vote
favorite
I am trying to understand this proof and I am having trouble understanding a couple things.
2) How do we know that F is analytic?
3) Is it necessary to say the things about primitives?
3) How do we know that $F'=f$? Maybe I am remembering incorrectly, but for some reason I thought the Fundamental Theorem of Calculus didn't apply for complex analysis...
Thanks!
complex-analysis
asked Jul 25 at 5:00
MathIsHard
1,122415
1
By definition, a primitive of $f$ is an analytic function $F$ such that $F'(z) = f(z)$. So (2) yes. (1) and (3) you have to prove that, hopefully it's proved in one of the links.
– D_S
Jul 25 at 5:21
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to understand this proof and I am having trouble understanding a couple things.
2) How do we know that F is analytic?
3) Is it necessary to say the things about primitives?
3) How do we know that $F'=f$? Maybe I am remembering incorrectly, but for some reason I thought the Fundamental Theorem of Calculus didn't apply for complex analysis...
Thanks!
complex-analysis
asked Jul 25 at 5:00
MathIsHard
1,122415
I am trying to understand this proof and I am having trouble understanding a couple things.
2) How do we know that F is analytic?
3) Is it necessary to say the things about primitives?
3) How do we know that $F'=f$? Maybe I am remembering incorrectly, but for some reason I thought the Fundamental Theorem of Calculus didn't apply for complex analysis...
Thanks!
complex-analysis
asked Jul 25 at 5:00
MathIsHard
1,122415
asked Jul 25 at 5:00
MathIsHard
1,122415
asked Jul 25 at 5:00
MathIsHard
1,122415
asked Jul 25 at 5:00
MathIsHard
1,122415
1,122415
1
By definition, a primitive of $f$ is an analytic function $F$ such that $F'(z) = f(z)$. So (2) yes. (1) and (3) you have to prove that, hopefully it's proved in one of the links.
– D_S
Jul 25 at 5:21
add a comment |Â
1
By definition, a primitive of $f$ is an analytic function $F$ such that $F'(z) = f(z)$. So (2) yes. (1) and (3) you have to prove that, hopefully it's proved in one of the links.
– D_S
Jul 25 at 5:21
1
1
By definition, a primitive of $f$ is an analytic function $F$ such that $F'(z) = f(z)$. So (2) yes. (1) and (3) you have to prove that, hopefully it's proved in one of the links.
– D_S
Jul 25 at 5:21
By definition, a primitive of $f$ is an analytic function $F$ such that $F'(z) = f(z)$. So (2) yes. (1) and (3) you have to prove that, hopefully it's proved in one of the links.
– D_S
Jul 25 at 5:21
add a comment |Â
1 Answer
1
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$$frac F(z+h)-F(z) h = frac int_[z,z+h] f(w) , dw h =frac int_[z,z+h] [f(w)-f(z)] , dw h +f(z) to f(z)$$ as $ h to 0$ by continuity of $f$. [ The first equation follows by considering three paths: one from $z_0$ to $z+h$, one from $z_0$ to $z+h$ and the line segment from $z$ to $z+h$. Drawing a picture helps]. It follows that $f$ is differentiable at all points of $D$ with $F'(z)=f(z)$. The derivative of any analytic functions is also analytic. [This can be proved using power series expansions]. Hence $f$ is analytic.
answered Jul 25 at 5:33
Kavi Rama Murthy
20.1k2829
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$$frac F(z+h)-F(z) h = frac int_[z,z+h] f(w) , dw h =frac int_[z,z+h] [f(w)-f(z)] , dw h +f(z) to f(z)$$ as $ h to 0$ by continuity of $f$. [ The first equation follows by considering three paths: one from $z_0$ to $z+h$, one from $z_0$ to $z+h$ and the line segment from $z$ to $z+h$. Drawing a picture helps]. It follows that $f$ is differentiable at all points of $D$ with $F'(z)=f(z)$. The derivative of any analytic functions is also analytic. [This can be proved using power series expansions]. Hence $f$ is analytic.
answered Jul 25 at 5:33
Kavi Rama Murthy
20.1k2829
add a comment |Â
up vote
4
down vote
accepted
$$frac F(z+h)-F(z) h = frac int_[z,z+h] f(w) , dw h =frac int_[z,z+h] [f(w)-f(z)] , dw h +f(z) to f(z)$$ as $ h to 0$ by continuity of $f$. [ The first equation follows by considering three paths: one from $z_0$ to $z+h$, one from $z_0$ to $z+h$ and the line segment from $z$ to $z+h$. Drawing a picture helps]. It follows that $f$ is differentiable at all points of $D$ with $F'(z)=f(z)$. The derivative of any analytic functions is also analytic. [This can be proved using power series expansions]. Hence $f$ is analytic.
answered Jul 25 at 5:33
Kavi Rama Murthy
20.1k2829
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$$frac F(z+h)-F(z) h = frac int_[z,z+h] f(w) , dw h =frac int_[z,z+h] [f(w)-f(z)] , dw h +f(z) to f(z)$$ as $ h to 0$ by continuity of $f$. [ The first equation follows by considering three paths: one from $z_0$ to $z+h$, one from $z_0$ to $z+h$ and the line segment from $z$ to $z+h$. Drawing a picture helps]. It follows that $f$ is differentiable at all points of $D$ with $F'(z)=f(z)$. The derivative of any analytic functions is also analytic. [This can be proved using power series expansions]. Hence $f$ is analytic.
answered Jul 25 at 5:33
Kavi Rama Murthy
20.1k2829
$$frac F(z+h)-F(z) h = frac int_[z,z+h] f(w) , dw h =frac int_[z,z+h] [f(w)-f(z)] , dw h +f(z) to f(z)$$ as $ h to 0$ by continuity of $f$. [ The first equation follows by considering three paths: one from $z_0$ to $z+h$, one from $z_0$ to $z+h$ and the line segment from $z$ to $z+h$. Drawing a picture helps]. It follows that $f$ is differentiable at all points of $D$ with $F'(z)=f(z)$. The derivative of any analytic functions is also analytic. [This can be proved using power series expansions]. Hence $f$ is analytic.
answered Jul 25 at 5:33
Kavi Rama Murthy
20.1k2829
answered Jul 25 at 5:33
Kavi Rama Murthy
20.1k2829
answered Jul 25 at 5:33
Kavi Rama Murthy
20.1k2829
answered Jul 25 at 5:33
Kavi Rama Murthy
20.1k2829
20.1k2829
add a comment |Â
add a comment |Â
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1
By definition, a primitive of $f$ is an analytic function $F$ such that $F'(z) = f(z)$. So (2) yes. (1) and (3) you have to prove that, hopefully it's proved in one of the links.
– D_S
Jul 25 at 5:21