Fundamental theorem of calculus for analytic functions












2












$begingroup$


I am confused about the proof in my literature of the fundamental theorem of calculus for analytic functions.



The theorem is that if $f(z)$ is continuous on some domain $D$, and $F(z)$ is a primitive for $f(z)$, then $$int_A^B f(z) dz = F(B) - F(A), $$ over any path in $D$ from $A$ to $B$.



In my literature, they go on to say that this follows from the "regular" fundamental theorem of calculus. Then they say the following:



Since $$F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y},$$ we have that $$F(B)-F(A) = int_A^B dF = int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y}dy $$
$$= int_A^B F'(z)(dx+idy) = int_A^B f(z) dz. $$



I don't really understand what they just showed there. I don't understand if they start by saying that it follows from the "regular" version of the theorem, and that the other part shows something else (perhaps relating to how this holds independently of which path we've chosen) or if they have shown HOW it follows from the "regular" version. And either way, I don't see it. Since I am teaching myself, I am hoping that someone can help fill in the blanks.



(Sorry that this is such a badly formulated question, but it is because that is how confused I am, that I can't even ask a good question about it.)










share|cite|improve this question









$endgroup$












  • $begingroup$
    The second part is a proof of the first part. What is the first step of the equation $F(B)-F(A)=dots$ that you don't understand?
    $endgroup$
    – Eric Wofsey
    Jan 13 at 20:35










  • $begingroup$
    Let me see if I can get this. :p So is it true that the first equality is what follows from the "regular" version of the theorem, and that the rest is simply manipulation of the intragrand? Because then I think I understand it. (Since I didn't know what they were trying to show, I didn't know what I was supposed to convince myself of, so to speak.)
    $endgroup$
    – j.eee
    Jan 13 at 20:42
















2












$begingroup$


I am confused about the proof in my literature of the fundamental theorem of calculus for analytic functions.



The theorem is that if $f(z)$ is continuous on some domain $D$, and $F(z)$ is a primitive for $f(z)$, then $$int_A^B f(z) dz = F(B) - F(A), $$ over any path in $D$ from $A$ to $B$.



In my literature, they go on to say that this follows from the "regular" fundamental theorem of calculus. Then they say the following:



Since $$F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y},$$ we have that $$F(B)-F(A) = int_A^B dF = int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y}dy $$
$$= int_A^B F'(z)(dx+idy) = int_A^B f(z) dz. $$



I don't really understand what they just showed there. I don't understand if they start by saying that it follows from the "regular" version of the theorem, and that the other part shows something else (perhaps relating to how this holds independently of which path we've chosen) or if they have shown HOW it follows from the "regular" version. And either way, I don't see it. Since I am teaching myself, I am hoping that someone can help fill in the blanks.



(Sorry that this is such a badly formulated question, but it is because that is how confused I am, that I can't even ask a good question about it.)










share|cite|improve this question









$endgroup$












  • $begingroup$
    The second part is a proof of the first part. What is the first step of the equation $F(B)-F(A)=dots$ that you don't understand?
    $endgroup$
    – Eric Wofsey
    Jan 13 at 20:35










  • $begingroup$
    Let me see if I can get this. :p So is it true that the first equality is what follows from the "regular" version of the theorem, and that the rest is simply manipulation of the intragrand? Because then I think I understand it. (Since I didn't know what they were trying to show, I didn't know what I was supposed to convince myself of, so to speak.)
    $endgroup$
    – j.eee
    Jan 13 at 20:42














2












2








2





$begingroup$


I am confused about the proof in my literature of the fundamental theorem of calculus for analytic functions.



The theorem is that if $f(z)$ is continuous on some domain $D$, and $F(z)$ is a primitive for $f(z)$, then $$int_A^B f(z) dz = F(B) - F(A), $$ over any path in $D$ from $A$ to $B$.



In my literature, they go on to say that this follows from the "regular" fundamental theorem of calculus. Then they say the following:



Since $$F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y},$$ we have that $$F(B)-F(A) = int_A^B dF = int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y}dy $$
$$= int_A^B F'(z)(dx+idy) = int_A^B f(z) dz. $$



I don't really understand what they just showed there. I don't understand if they start by saying that it follows from the "regular" version of the theorem, and that the other part shows something else (perhaps relating to how this holds independently of which path we've chosen) or if they have shown HOW it follows from the "regular" version. And either way, I don't see it. Since I am teaching myself, I am hoping that someone can help fill in the blanks.



(Sorry that this is such a badly formulated question, but it is because that is how confused I am, that I can't even ask a good question about it.)










share|cite|improve this question









$endgroup$




I am confused about the proof in my literature of the fundamental theorem of calculus for analytic functions.



The theorem is that if $f(z)$ is continuous on some domain $D$, and $F(z)$ is a primitive for $f(z)$, then $$int_A^B f(z) dz = F(B) - F(A), $$ over any path in $D$ from $A$ to $B$.



In my literature, they go on to say that this follows from the "regular" fundamental theorem of calculus. Then they say the following:



Since $$F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y},$$ we have that $$F(B)-F(A) = int_A^B dF = int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y}dy $$
$$= int_A^B F'(z)(dx+idy) = int_A^B f(z) dz. $$



I don't really understand what they just showed there. I don't understand if they start by saying that it follows from the "regular" version of the theorem, and that the other part shows something else (perhaps relating to how this holds independently of which path we've chosen) or if they have shown HOW it follows from the "regular" version. And either way, I don't see it. Since I am teaching myself, I am hoping that someone can help fill in the blanks.



(Sorry that this is such a badly formulated question, but it is because that is how confused I am, that I can't even ask a good question about it.)







complex-analysis






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asked Jan 13 at 20:26









j.eeej.eee

787




787












  • $begingroup$
    The second part is a proof of the first part. What is the first step of the equation $F(B)-F(A)=dots$ that you don't understand?
    $endgroup$
    – Eric Wofsey
    Jan 13 at 20:35










  • $begingroup$
    Let me see if I can get this. :p So is it true that the first equality is what follows from the "regular" version of the theorem, and that the rest is simply manipulation of the intragrand? Because then I think I understand it. (Since I didn't know what they were trying to show, I didn't know what I was supposed to convince myself of, so to speak.)
    $endgroup$
    – j.eee
    Jan 13 at 20:42


















  • $begingroup$
    The second part is a proof of the first part. What is the first step of the equation $F(B)-F(A)=dots$ that you don't understand?
    $endgroup$
    – Eric Wofsey
    Jan 13 at 20:35










  • $begingroup$
    Let me see if I can get this. :p So is it true that the first equality is what follows from the "regular" version of the theorem, and that the rest is simply manipulation of the intragrand? Because then I think I understand it. (Since I didn't know what they were trying to show, I didn't know what I was supposed to convince myself of, so to speak.)
    $endgroup$
    – j.eee
    Jan 13 at 20:42
















$begingroup$
The second part is a proof of the first part. What is the first step of the equation $F(B)-F(A)=dots$ that you don't understand?
$endgroup$
– Eric Wofsey
Jan 13 at 20:35




$begingroup$
The second part is a proof of the first part. What is the first step of the equation $F(B)-F(A)=dots$ that you don't understand?
$endgroup$
– Eric Wofsey
Jan 13 at 20:35












$begingroup$
Let me see if I can get this. :p So is it true that the first equality is what follows from the "regular" version of the theorem, and that the rest is simply manipulation of the intragrand? Because then I think I understand it. (Since I didn't know what they were trying to show, I didn't know what I was supposed to convince myself of, so to speak.)
$endgroup$
– j.eee
Jan 13 at 20:42




$begingroup$
Let me see if I can get this. :p So is it true that the first equality is what follows from the "regular" version of the theorem, and that the rest is simply manipulation of the intragrand? Because then I think I understand it. (Since I didn't know what they were trying to show, I didn't know what I was supposed to convince myself of, so to speak.)
$endgroup$
– j.eee
Jan 13 at 20:42










1 Answer
1






active

oldest

votes


















1












$begingroup$

The second part is the proof of the theorem. Here what they refer to as the "regular" Fundamental Theorem of Calculus is a version of the Fundamental Theorem of Calculus for integrals of one-forms on parametrized curves. Namely, if $gamma:[a,b]to D$ is a curve and $F:Dtomathbb{C}$ is a function, then $$int_gamma dF=F(gamma(b))-F(gamma(a)).$$ (Here you need $gamma$ and $F$ to be sufficiently nice, say $C^1$, for this to make sense.) If you unravel the definition of $int_gamma dF$, you find that it is equal to just $int_a^b g'(t),dt$ where $g(t)=F(gamma(t))$, so this parametrized Fundamental Theorem of Calculus follows from the ordinary Fundamental Theorem of Calculus on $mathbb{R}$.



In particular, then, if $gamma(b)=B$ and $gamma(a)=A$ and we denote the integral along $gamma$ by $int_A^B$, we get $$F(B)-F(A)=int_A^B dF.$$ The rest of the derivation is then just an evaluation of the $1$-form $dF$ to show that it is equal to the $1$-form $f(z),dz$, so that $$F(B)-F(A)=int_A^B dF=int_A^Bf(z),dz.$$ Note that the equation $$F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$$ is wrong and should instead be $$F'(z) = frac{partial F}{partial x} color{red}= frac{1}{i}frac{partial F}{partial y}$$ which may be the source of some of your confusion.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much. I am a little uncertain of the equality $int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y} dy = int_A^B F'(z)(dx+idy). $ I'm thinking that going the other way, we'd have $int_A^B F'(z)(dx+idy) = int_A^B left( frac{partial F}{partial x} dx + frac{partial F}{partial x}idy + frac{1}{i}frac{partial F}{partial y}dx + frac{partial F}{partial y}dy right) $, where the two terms in the middle are equal to zero. Is that correct or am I misunderstanding something?
    $endgroup$
    – j.eee
    Jan 13 at 20:52












  • $begingroup$
    Oh, there was a typo (not sure if it's yours or from your book). The equation $F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$ should instead be $F'(z) = frac{partial F}{partial x} = frac{1}{i}frac{partial F}{partial y}$.
    $endgroup$
    – Eric Wofsey
    Jan 13 at 20:54










  • $begingroup$
    Oh. It wasn't a typo. It was my mind trying to autocorrect something that didn't need correcting, apparently. Now I'm confused again. So I suppose that means that $frac{partial F}{partial x}$ is the derivative taken in the $x$-direction and the other in the $y$ direction? But then and I am definitely very unsure of the equality I mentioned in my above comment...
    $endgroup$
    – j.eee
    Jan 13 at 21:00










  • $begingroup$
    It's trivial: just substitute $F'(z)$ for $frac{partial F}{partial x}$ and substitute $iF'(z)$ for $frac{partial F}{partial y}$.
    $endgroup$
    – Eric Wofsey
    Jan 13 at 21:01










  • $begingroup$
    No! Then I'm not confused. Yes, I get it, haha. :D That makes total sense. Thanks a bunch.
    $endgroup$
    – j.eee
    Jan 13 at 21:02













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1 Answer
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1 Answer
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active

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oldest

votes









1












$begingroup$

The second part is the proof of the theorem. Here what they refer to as the "regular" Fundamental Theorem of Calculus is a version of the Fundamental Theorem of Calculus for integrals of one-forms on parametrized curves. Namely, if $gamma:[a,b]to D$ is a curve and $F:Dtomathbb{C}$ is a function, then $$int_gamma dF=F(gamma(b))-F(gamma(a)).$$ (Here you need $gamma$ and $F$ to be sufficiently nice, say $C^1$, for this to make sense.) If you unravel the definition of $int_gamma dF$, you find that it is equal to just $int_a^b g'(t),dt$ where $g(t)=F(gamma(t))$, so this parametrized Fundamental Theorem of Calculus follows from the ordinary Fundamental Theorem of Calculus on $mathbb{R}$.



In particular, then, if $gamma(b)=B$ and $gamma(a)=A$ and we denote the integral along $gamma$ by $int_A^B$, we get $$F(B)-F(A)=int_A^B dF.$$ The rest of the derivation is then just an evaluation of the $1$-form $dF$ to show that it is equal to the $1$-form $f(z),dz$, so that $$F(B)-F(A)=int_A^B dF=int_A^Bf(z),dz.$$ Note that the equation $$F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$$ is wrong and should instead be $$F'(z) = frac{partial F}{partial x} color{red}= frac{1}{i}frac{partial F}{partial y}$$ which may be the source of some of your confusion.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much. I am a little uncertain of the equality $int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y} dy = int_A^B F'(z)(dx+idy). $ I'm thinking that going the other way, we'd have $int_A^B F'(z)(dx+idy) = int_A^B left( frac{partial F}{partial x} dx + frac{partial F}{partial x}idy + frac{1}{i}frac{partial F}{partial y}dx + frac{partial F}{partial y}dy right) $, where the two terms in the middle are equal to zero. Is that correct or am I misunderstanding something?
    $endgroup$
    – j.eee
    Jan 13 at 20:52












  • $begingroup$
    Oh, there was a typo (not sure if it's yours or from your book). The equation $F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$ should instead be $F'(z) = frac{partial F}{partial x} = frac{1}{i}frac{partial F}{partial y}$.
    $endgroup$
    – Eric Wofsey
    Jan 13 at 20:54










  • $begingroup$
    Oh. It wasn't a typo. It was my mind trying to autocorrect something that didn't need correcting, apparently. Now I'm confused again. So I suppose that means that $frac{partial F}{partial x}$ is the derivative taken in the $x$-direction and the other in the $y$ direction? But then and I am definitely very unsure of the equality I mentioned in my above comment...
    $endgroup$
    – j.eee
    Jan 13 at 21:00










  • $begingroup$
    It's trivial: just substitute $F'(z)$ for $frac{partial F}{partial x}$ and substitute $iF'(z)$ for $frac{partial F}{partial y}$.
    $endgroup$
    – Eric Wofsey
    Jan 13 at 21:01










  • $begingroup$
    No! Then I'm not confused. Yes, I get it, haha. :D That makes total sense. Thanks a bunch.
    $endgroup$
    – j.eee
    Jan 13 at 21:02


















1












$begingroup$

The second part is the proof of the theorem. Here what they refer to as the "regular" Fundamental Theorem of Calculus is a version of the Fundamental Theorem of Calculus for integrals of one-forms on parametrized curves. Namely, if $gamma:[a,b]to D$ is a curve and $F:Dtomathbb{C}$ is a function, then $$int_gamma dF=F(gamma(b))-F(gamma(a)).$$ (Here you need $gamma$ and $F$ to be sufficiently nice, say $C^1$, for this to make sense.) If you unravel the definition of $int_gamma dF$, you find that it is equal to just $int_a^b g'(t),dt$ where $g(t)=F(gamma(t))$, so this parametrized Fundamental Theorem of Calculus follows from the ordinary Fundamental Theorem of Calculus on $mathbb{R}$.



In particular, then, if $gamma(b)=B$ and $gamma(a)=A$ and we denote the integral along $gamma$ by $int_A^B$, we get $$F(B)-F(A)=int_A^B dF.$$ The rest of the derivation is then just an evaluation of the $1$-form $dF$ to show that it is equal to the $1$-form $f(z),dz$, so that $$F(B)-F(A)=int_A^B dF=int_A^Bf(z),dz.$$ Note that the equation $$F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$$ is wrong and should instead be $$F'(z) = frac{partial F}{partial x} color{red}= frac{1}{i}frac{partial F}{partial y}$$ which may be the source of some of your confusion.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much. I am a little uncertain of the equality $int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y} dy = int_A^B F'(z)(dx+idy). $ I'm thinking that going the other way, we'd have $int_A^B F'(z)(dx+idy) = int_A^B left( frac{partial F}{partial x} dx + frac{partial F}{partial x}idy + frac{1}{i}frac{partial F}{partial y}dx + frac{partial F}{partial y}dy right) $, where the two terms in the middle are equal to zero. Is that correct or am I misunderstanding something?
    $endgroup$
    – j.eee
    Jan 13 at 20:52












  • $begingroup$
    Oh, there was a typo (not sure if it's yours or from your book). The equation $F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$ should instead be $F'(z) = frac{partial F}{partial x} = frac{1}{i}frac{partial F}{partial y}$.
    $endgroup$
    – Eric Wofsey
    Jan 13 at 20:54










  • $begingroup$
    Oh. It wasn't a typo. It was my mind trying to autocorrect something that didn't need correcting, apparently. Now I'm confused again. So I suppose that means that $frac{partial F}{partial x}$ is the derivative taken in the $x$-direction and the other in the $y$ direction? But then and I am definitely very unsure of the equality I mentioned in my above comment...
    $endgroup$
    – j.eee
    Jan 13 at 21:00










  • $begingroup$
    It's trivial: just substitute $F'(z)$ for $frac{partial F}{partial x}$ and substitute $iF'(z)$ for $frac{partial F}{partial y}$.
    $endgroup$
    – Eric Wofsey
    Jan 13 at 21:01










  • $begingroup$
    No! Then I'm not confused. Yes, I get it, haha. :D That makes total sense. Thanks a bunch.
    $endgroup$
    – j.eee
    Jan 13 at 21:02
















1












1








1





$begingroup$

The second part is the proof of the theorem. Here what they refer to as the "regular" Fundamental Theorem of Calculus is a version of the Fundamental Theorem of Calculus for integrals of one-forms on parametrized curves. Namely, if $gamma:[a,b]to D$ is a curve and $F:Dtomathbb{C}$ is a function, then $$int_gamma dF=F(gamma(b))-F(gamma(a)).$$ (Here you need $gamma$ and $F$ to be sufficiently nice, say $C^1$, for this to make sense.) If you unravel the definition of $int_gamma dF$, you find that it is equal to just $int_a^b g'(t),dt$ where $g(t)=F(gamma(t))$, so this parametrized Fundamental Theorem of Calculus follows from the ordinary Fundamental Theorem of Calculus on $mathbb{R}$.



In particular, then, if $gamma(b)=B$ and $gamma(a)=A$ and we denote the integral along $gamma$ by $int_A^B$, we get $$F(B)-F(A)=int_A^B dF.$$ The rest of the derivation is then just an evaluation of the $1$-form $dF$ to show that it is equal to the $1$-form $f(z),dz$, so that $$F(B)-F(A)=int_A^B dF=int_A^Bf(z),dz.$$ Note that the equation $$F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$$ is wrong and should instead be $$F'(z) = frac{partial F}{partial x} color{red}= frac{1}{i}frac{partial F}{partial y}$$ which may be the source of some of your confusion.






share|cite|improve this answer











$endgroup$



The second part is the proof of the theorem. Here what they refer to as the "regular" Fundamental Theorem of Calculus is a version of the Fundamental Theorem of Calculus for integrals of one-forms on parametrized curves. Namely, if $gamma:[a,b]to D$ is a curve and $F:Dtomathbb{C}$ is a function, then $$int_gamma dF=F(gamma(b))-F(gamma(a)).$$ (Here you need $gamma$ and $F$ to be sufficiently nice, say $C^1$, for this to make sense.) If you unravel the definition of $int_gamma dF$, you find that it is equal to just $int_a^b g'(t),dt$ where $g(t)=F(gamma(t))$, so this parametrized Fundamental Theorem of Calculus follows from the ordinary Fundamental Theorem of Calculus on $mathbb{R}$.



In particular, then, if $gamma(b)=B$ and $gamma(a)=A$ and we denote the integral along $gamma$ by $int_A^B$, we get $$F(B)-F(A)=int_A^B dF.$$ The rest of the derivation is then just an evaluation of the $1$-form $dF$ to show that it is equal to the $1$-form $f(z),dz$, so that $$F(B)-F(A)=int_A^B dF=int_A^Bf(z),dz.$$ Note that the equation $$F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$$ is wrong and should instead be $$F'(z) = frac{partial F}{partial x} color{red}= frac{1}{i}frac{partial F}{partial y}$$ which may be the source of some of your confusion.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 at 20:55

























answered Jan 13 at 20:45









Eric WofseyEric Wofsey

190k14216348




190k14216348












  • $begingroup$
    Thank you so much. I am a little uncertain of the equality $int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y} dy = int_A^B F'(z)(dx+idy). $ I'm thinking that going the other way, we'd have $int_A^B F'(z)(dx+idy) = int_A^B left( frac{partial F}{partial x} dx + frac{partial F}{partial x}idy + frac{1}{i}frac{partial F}{partial y}dx + frac{partial F}{partial y}dy right) $, where the two terms in the middle are equal to zero. Is that correct or am I misunderstanding something?
    $endgroup$
    – j.eee
    Jan 13 at 20:52












  • $begingroup$
    Oh, there was a typo (not sure if it's yours or from your book). The equation $F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$ should instead be $F'(z) = frac{partial F}{partial x} = frac{1}{i}frac{partial F}{partial y}$.
    $endgroup$
    – Eric Wofsey
    Jan 13 at 20:54










  • $begingroup$
    Oh. It wasn't a typo. It was my mind trying to autocorrect something that didn't need correcting, apparently. Now I'm confused again. So I suppose that means that $frac{partial F}{partial x}$ is the derivative taken in the $x$-direction and the other in the $y$ direction? But then and I am definitely very unsure of the equality I mentioned in my above comment...
    $endgroup$
    – j.eee
    Jan 13 at 21:00










  • $begingroup$
    It's trivial: just substitute $F'(z)$ for $frac{partial F}{partial x}$ and substitute $iF'(z)$ for $frac{partial F}{partial y}$.
    $endgroup$
    – Eric Wofsey
    Jan 13 at 21:01










  • $begingroup$
    No! Then I'm not confused. Yes, I get it, haha. :D That makes total sense. Thanks a bunch.
    $endgroup$
    – j.eee
    Jan 13 at 21:02




















  • $begingroup$
    Thank you so much. I am a little uncertain of the equality $int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y} dy = int_A^B F'(z)(dx+idy). $ I'm thinking that going the other way, we'd have $int_A^B F'(z)(dx+idy) = int_A^B left( frac{partial F}{partial x} dx + frac{partial F}{partial x}idy + frac{1}{i}frac{partial F}{partial y}dx + frac{partial F}{partial y}dy right) $, where the two terms in the middle are equal to zero. Is that correct or am I misunderstanding something?
    $endgroup$
    – j.eee
    Jan 13 at 20:52












  • $begingroup$
    Oh, there was a typo (not sure if it's yours or from your book). The equation $F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$ should instead be $F'(z) = frac{partial F}{partial x} = frac{1}{i}frac{partial F}{partial y}$.
    $endgroup$
    – Eric Wofsey
    Jan 13 at 20:54










  • $begingroup$
    Oh. It wasn't a typo. It was my mind trying to autocorrect something that didn't need correcting, apparently. Now I'm confused again. So I suppose that means that $frac{partial F}{partial x}$ is the derivative taken in the $x$-direction and the other in the $y$ direction? But then and I am definitely very unsure of the equality I mentioned in my above comment...
    $endgroup$
    – j.eee
    Jan 13 at 21:00










  • $begingroup$
    It's trivial: just substitute $F'(z)$ for $frac{partial F}{partial x}$ and substitute $iF'(z)$ for $frac{partial F}{partial y}$.
    $endgroup$
    – Eric Wofsey
    Jan 13 at 21:01










  • $begingroup$
    No! Then I'm not confused. Yes, I get it, haha. :D That makes total sense. Thanks a bunch.
    $endgroup$
    – j.eee
    Jan 13 at 21:02


















$begingroup$
Thank you so much. I am a little uncertain of the equality $int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y} dy = int_A^B F'(z)(dx+idy). $ I'm thinking that going the other way, we'd have $int_A^B F'(z)(dx+idy) = int_A^B left( frac{partial F}{partial x} dx + frac{partial F}{partial x}idy + frac{1}{i}frac{partial F}{partial y}dx + frac{partial F}{partial y}dy right) $, where the two terms in the middle are equal to zero. Is that correct or am I misunderstanding something?
$endgroup$
– j.eee
Jan 13 at 20:52






$begingroup$
Thank you so much. I am a little uncertain of the equality $int_A^B frac{partial F}{partial x}dx + frac{partial F}{partial y} dy = int_A^B F'(z)(dx+idy). $ I'm thinking that going the other way, we'd have $int_A^B F'(z)(dx+idy) = int_A^B left( frac{partial F}{partial x} dx + frac{partial F}{partial x}idy + frac{1}{i}frac{partial F}{partial y}dx + frac{partial F}{partial y}dy right) $, where the two terms in the middle are equal to zero. Is that correct or am I misunderstanding something?
$endgroup$
– j.eee
Jan 13 at 20:52














$begingroup$
Oh, there was a typo (not sure if it's yours or from your book). The equation $F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$ should instead be $F'(z) = frac{partial F}{partial x} = frac{1}{i}frac{partial F}{partial y}$.
$endgroup$
– Eric Wofsey
Jan 13 at 20:54




$begingroup$
Oh, there was a typo (not sure if it's yours or from your book). The equation $F'(z) = frac{partial F}{partial x} + frac{1}{i}frac{partial F}{partial y}$ should instead be $F'(z) = frac{partial F}{partial x} = frac{1}{i}frac{partial F}{partial y}$.
$endgroup$
– Eric Wofsey
Jan 13 at 20:54












$begingroup$
Oh. It wasn't a typo. It was my mind trying to autocorrect something that didn't need correcting, apparently. Now I'm confused again. So I suppose that means that $frac{partial F}{partial x}$ is the derivative taken in the $x$-direction and the other in the $y$ direction? But then and I am definitely very unsure of the equality I mentioned in my above comment...
$endgroup$
– j.eee
Jan 13 at 21:00




$begingroup$
Oh. It wasn't a typo. It was my mind trying to autocorrect something that didn't need correcting, apparently. Now I'm confused again. So I suppose that means that $frac{partial F}{partial x}$ is the derivative taken in the $x$-direction and the other in the $y$ direction? But then and I am definitely very unsure of the equality I mentioned in my above comment...
$endgroup$
– j.eee
Jan 13 at 21:00












$begingroup$
It's trivial: just substitute $F'(z)$ for $frac{partial F}{partial x}$ and substitute $iF'(z)$ for $frac{partial F}{partial y}$.
$endgroup$
– Eric Wofsey
Jan 13 at 21:01




$begingroup$
It's trivial: just substitute $F'(z)$ for $frac{partial F}{partial x}$ and substitute $iF'(z)$ for $frac{partial F}{partial y}$.
$endgroup$
– Eric Wofsey
Jan 13 at 21:01












$begingroup$
No! Then I'm not confused. Yes, I get it, haha. :D That makes total sense. Thanks a bunch.
$endgroup$
– j.eee
Jan 13 at 21:02






$begingroup$
No! Then I'm not confused. Yes, I get it, haha. :D That makes total sense. Thanks a bunch.
$endgroup$
– j.eee
Jan 13 at 21:02




















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