Visualizing Cauchy's integral theorem (and complex integration in general)
$begingroup$
(I edited the question due to a hint from Giuseppe Negro who pointed out that I forgot about $dz$.)
Consider Cauchy's integral theorem, i.e.
$$oint_gamma f(z)dz = 0 $$
for holomorphic functions $f: U rightarrow mathbb{C}$ with $U$ a simply connected open subset of $mathbb{C}$ and $gamma$ a closed rectifiable path in $U$.
In visual terms the theorem states that when we draw from each point $z$ along the closed curve $gamma$ an arrow pointing to $z + f(z)Delta z$ then in the limit of ever closer points ($Delta z rightarrow 0$), these arrows add up to $0$ by simple vector addition, i.e. in sum they cancel each other.
I wonder if this visualization does really help to understand Cauchy's theorem better – both its content and its proof.
Here are two examples for $f(z) = z^2$ and some $gamma$:
This is when we draw the arrows not from $z$ to $z + f(z)Delta z$ but from $0$ to $f(z)Delta z$ - which of course doesn't affect the sum, but possibly makes clearer, that the vectors in sum cancel each other:
Note that from this point of view, complex integration works rather similar to real integration: it's an "addition of arrows", only that in the real case they are put next to each other (which works because they point in the same direction, thus giving an area), while in the complex case you cannot put them next to each other (because they point in different directions), so you have to put them along each other, i.e. add them up as vectors. (But you could do the same in the real case: the integral then is the limit area of an ever "higher" rectangle with width $dx rightarrow 0$ – so to speak, with tongue in cheek).
Note, that multiplying by $Delta x$ in the real case gives the arrow a width (and possibly swaps direction), while multiplying by $Delta z$ in the complex case gives the arrow another length and direction.
My question is if this visual point of view on complex integration and Cauchy's
integral theorem has been taken before, esp. where I can find a reference.
And if there are genuine reasons not to adopt this view e.g. because it obscures something.
For the sake of comparison: Here are the original pictures with arrows from $z$ to $z + f(z)$. Interestingly enough, these add up to $0$, too!
For the delight of those who like this kind of pictures: A case for $f(z) = z^3$:
and its $0$ companion:
real-analysis integration complex-analysis contour-integration visualization
asked Jan 17 at 13:23
Hans-Peter StrickerHans-Peter Stricker
6,77243997
$endgroup$
|
show 4 more comments
$begingroup$
(I edited the question due to a hint from Giuseppe Negro who pointed out that I forgot about $dz$.)
Consider Cauchy's integral theorem, i.e.
$$oint_gamma f(z)dz = 0 $$
for holomorphic functions $f: U rightarrow mathbb{C}$ with $U$ a simply connected open subset of $mathbb{C}$ and $gamma$ a closed rectifiable path in $U$.
In visual terms the theorem states that when we draw from each point $z$ along the closed curve $gamma$ an arrow pointing to $z + f(z)Delta z$ then in the limit of ever closer points ($Delta z rightarrow 0$), these arrows add up to $0$ by simple vector addition, i.e. in sum they cancel each other.
I wonder if this visualization does really help to understand Cauchy's theorem better – both its content and its proof.
Here are two examples for $f(z) = z^2$ and some $gamma$:
This is when we draw the arrows not from $z$ to $z + f(z)Delta z$ but from $0$ to $f(z)Delta z$ - which of course doesn't affect the sum, but possibly makes clearer, that the vectors in sum cancel each other:
Note that from this point of view, complex integration works rather similar to real integration: it's an "addition of arrows", only that in the real case they are put next to each other (which works because they point in the same direction, thus giving an area), while in the complex case you cannot put them next to each other (because they point in different directions), so you have to put them along each other, i.e. add them up as vectors. (But you could do the same in the real case: the integral then is the limit area of an ever "higher" rectangle with width $dx rightarrow 0$ – so to speak, with tongue in cheek).
Note, that multiplying by $Delta x$ in the real case gives the arrow a width (and possibly swaps direction), while multiplying by $Delta z$ in the complex case gives the arrow another length and direction.
My question is if this visual point of view on complex integration and Cauchy's
integral theorem has been taken before, esp. where I can find a reference.
And if there are genuine reasons not to adopt this view e.g. because it obscures something.
For the sake of comparison: Here are the original pictures with arrows from $z$ to $z + f(z)$. Interestingly enough, these add up to $0$, too!
For the delight of those who like this kind of pictures: A case for $f(z) = z^3$:
and its $0$ companion:
real-analysis integration complex-analysis contour-integration visualization
asked Jan 17 at 13:23
Hans-Peter StrickerHans-Peter Stricker
6,77243997
$endgroup$
$begingroup$
Aren't you forgetting the dz? It amounts to multiplying $f(z) $ by the tangent vector to the curve, I think.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:21
$begingroup$
@GiuseppeNegro: Yes, you are right, I forgot $dz$ and became aware of this by myself just now. But I am already thinking about it - and how to make the picture more consistent and correct.
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:26
$begingroup$
If this interpretation leads somewhere, then it must be in the book of Needham.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:27
$begingroup$
@GiuseppeNegro: OK, I have Needham at hand, I'll go through it. Can you suggest in which chapter(s) to start searching (my copy has >600 pages). What to do, if I won't make a find?
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:35
1
$begingroup$
@GiuseppeNegro: Because you liked the pictures: have a look at the last one.
$endgroup$
– Hans-Peter Stricker
Jan 18 at 16:53
|
show 4 more comments
$begingroup$
(I edited the question due to a hint from Giuseppe Negro who pointed out that I forgot about $dz$.)
Consider Cauchy's integral theorem, i.e.
$$oint_gamma f(z)dz = 0 $$
for holomorphic functions $f: U rightarrow mathbb{C}$ with $U$ a simply connected open subset of $mathbb{C}$ and $gamma$ a closed rectifiable path in $U$.
In visual terms the theorem states that when we draw from each point $z$ along the closed curve $gamma$ an arrow pointing to $z + f(z)Delta z$ then in the limit of ever closer points ($Delta z rightarrow 0$), these arrows add up to $0$ by simple vector addition, i.e. in sum they cancel each other.
I wonder if this visualization does really help to understand Cauchy's theorem better – both its content and its proof.
Here are two examples for $f(z) = z^2$ and some $gamma$:
This is when we draw the arrows not from $z$ to $z + f(z)Delta z$ but from $0$ to $f(z)Delta z$ - which of course doesn't affect the sum, but possibly makes clearer, that the vectors in sum cancel each other:
Note that from this point of view, complex integration works rather similar to real integration: it's an "addition of arrows", only that in the real case they are put next to each other (which works because they point in the same direction, thus giving an area), while in the complex case you cannot put them next to each other (because they point in different directions), so you have to put them along each other, i.e. add them up as vectors. (But you could do the same in the real case: the integral then is the limit area of an ever "higher" rectangle with width $dx rightarrow 0$ – so to speak, with tongue in cheek).
Note, that multiplying by $Delta x$ in the real case gives the arrow a width (and possibly swaps direction), while multiplying by $Delta z$ in the complex case gives the arrow another length and direction.
My question is if this visual point of view on complex integration and Cauchy's
integral theorem has been taken before, esp. where I can find a reference.
And if there are genuine reasons not to adopt this view e.g. because it obscures something.
For the sake of comparison: Here are the original pictures with arrows from $z$ to $z + f(z)$. Interestingly enough, these add up to $0$, too!
For the delight of those who like this kind of pictures: A case for $f(z) = z^3$:
and its $0$ companion:
real-analysis integration complex-analysis contour-integration visualization
asked Jan 17 at 13:23
Hans-Peter StrickerHans-Peter Stricker
6,77243997
$endgroup$
(I edited the question due to a hint from Giuseppe Negro who pointed out that I forgot about $dz$.)
Consider Cauchy's integral theorem, i.e.
$$oint_gamma f(z)dz = 0 $$
for holomorphic functions $f: U rightarrow mathbb{C}$ with $U$ a simply connected open subset of $mathbb{C}$ and $gamma$ a closed rectifiable path in $U$.
In visual terms the theorem states that when we draw from each point $z$ along the closed curve $gamma$ an arrow pointing to $z + f(z)Delta z$ then in the limit of ever closer points ($Delta z rightarrow 0$), these arrows add up to $0$ by simple vector addition, i.e. in sum they cancel each other.
I wonder if this visualization does really help to understand Cauchy's theorem better – both its content and its proof.
Here are two examples for $f(z) = z^2$ and some $gamma$:
This is when we draw the arrows not from $z$ to $z + f(z)Delta z$ but from $0$ to $f(z)Delta z$ - which of course doesn't affect the sum, but possibly makes clearer, that the vectors in sum cancel each other:
Note that from this point of view, complex integration works rather similar to real integration: it's an "addition of arrows", only that in the real case they are put next to each other (which works because they point in the same direction, thus giving an area), while in the complex case you cannot put them next to each other (because they point in different directions), so you have to put them along each other, i.e. add them up as vectors. (But you could do the same in the real case: the integral then is the limit area of an ever "higher" rectangle with width $dx rightarrow 0$ – so to speak, with tongue in cheek).
Note, that multiplying by $Delta x$ in the real case gives the arrow a width (and possibly swaps direction), while multiplying by $Delta z$ in the complex case gives the arrow another length and direction.
My question is if this visual point of view on complex integration and Cauchy's
integral theorem has been taken before, esp. where I can find a reference.
And if there are genuine reasons not to adopt this view e.g. because it obscures something.
For the sake of comparison: Here are the original pictures with arrows from $z$ to $z + f(z)$. Interestingly enough, these add up to $0$, too!
For the delight of those who like this kind of pictures: A case for $f(z) = z^3$:
and its $0$ companion:
real-analysis integration complex-analysis contour-integration visualization
real-analysis integration complex-analysis contour-integration visualization
asked Jan 17 at 13:23
Hans-Peter StrickerHans-Peter Stricker
6,77243997
asked Jan 17 at 13:23
Hans-Peter StrickerHans-Peter Stricker
6,77243997
edited Jan 18 at 16:52
Hans-Peter Stricker
asked Jan 17 at 13:23
Hans-Peter StrickerHans-Peter Stricker
6,77243997
asked Jan 17 at 13:23
Hans-Peter StrickerHans-Peter Stricker
6,77243997
asked Jan 17 at 13:23
Hans-Peter StrickerHans-Peter Stricker
6,77243997
6,77243997
$begingroup$
Aren't you forgetting the dz? It amounts to multiplying $f(z) $ by the tangent vector to the curve, I think.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:21
$begingroup$
@GiuseppeNegro: Yes, you are right, I forgot $dz$ and became aware of this by myself just now. But I am already thinking about it - and how to make the picture more consistent and correct.
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:26
$begingroup$
If this interpretation leads somewhere, then it must be in the book of Needham.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:27
$begingroup$
@GiuseppeNegro: OK, I have Needham at hand, I'll go through it. Can you suggest in which chapter(s) to start searching (my copy has >600 pages). What to do, if I won't make a find?
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:35
1
$begingroup$
@GiuseppeNegro: Because you liked the pictures: have a look at the last one.
$endgroup$
– Hans-Peter Stricker
Jan 18 at 16:53
|
show 4 more comments
$begingroup$
Aren't you forgetting the dz? It amounts to multiplying $f(z) $ by the tangent vector to the curve, I think.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:21
$begingroup$
@GiuseppeNegro: Yes, you are right, I forgot $dz$ and became aware of this by myself just now. But I am already thinking about it - and how to make the picture more consistent and correct.
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:26
$begingroup$
If this interpretation leads somewhere, then it must be in the book of Needham.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:27
$begingroup$
@GiuseppeNegro: OK, I have Needham at hand, I'll go through it. Can you suggest in which chapter(s) to start searching (my copy has >600 pages). What to do, if I won't make a find?
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:35
1
$begingroup$
@GiuseppeNegro: Because you liked the pictures: have a look at the last one.
$endgroup$
– Hans-Peter Stricker
Jan 18 at 16:53
$begingroup$
Aren't you forgetting the dz? It amounts to multiplying $f(z) $ by the tangent vector to the curve, I think.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:21
$begingroup$
Aren't you forgetting the dz? It amounts to multiplying $f(z) $ by the tangent vector to the curve, I think.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:21
$begingroup$
@GiuseppeNegro: Yes, you are right, I forgot $dz$ and became aware of this by myself just now. But I am already thinking about it - and how to make the picture more consistent and correct.
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:26
$begingroup$
@GiuseppeNegro: Yes, you are right, I forgot $dz$ and became aware of this by myself just now. But I am already thinking about it - and how to make the picture more consistent and correct.
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:26
$begingroup$
If this interpretation leads somewhere, then it must be in the book of Needham.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:27
$begingroup$
If this interpretation leads somewhere, then it must be in the book of Needham.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:27
$begingroup$
@GiuseppeNegro: OK, I have Needham at hand, I'll go through it. Can you suggest in which chapter(s) to start searching (my copy has >600 pages). What to do, if I won't make a find?
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:35
$begingroup$
@GiuseppeNegro: OK, I have Needham at hand, I'll go through it. Can you suggest in which chapter(s) to start searching (my copy has >600 pages). What to do, if I won't make a find?
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:35
1
1
$begingroup$
@GiuseppeNegro: Because you liked the pictures: have a look at the last one.
$endgroup$
– Hans-Peter Stricker
Jan 18 at 16:53
$begingroup$
@GiuseppeNegro: Because you liked the pictures: have a look at the last one.
$endgroup$
– Hans-Peter Stricker
Jan 18 at 16:53
|
show 4 more comments
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$begingroup$
Aren't you forgetting the dz? It amounts to multiplying $f(z) $ by the tangent vector to the curve, I think.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:21
$begingroup$
@GiuseppeNegro: Yes, you are right, I forgot $dz$ and became aware of this by myself just now. But I am already thinking about it - and how to make the picture more consistent and correct.
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:26
$begingroup$
If this interpretation leads somewhere, then it must be in the book of Needham.
$endgroup$
– Giuseppe Negro
Jan 17 at 17:27
$begingroup$
@GiuseppeNegro: OK, I have Needham at hand, I'll go through it. Can you suggest in which chapter(s) to start searching (my copy has >600 pages). What to do, if I won't make a find?
$endgroup$
– Hans-Peter Stricker
Jan 17 at 17:35
1
$begingroup$
@GiuseppeNegro: Because you liked the pictures: have a look at the last one.
$endgroup$
– Hans-Peter Stricker
Jan 18 at 16:53