Why is the argument principle called the argument principle?
up vote
2
down vote
favorite
The Argument principle:
If $f$ is meromorphic in an open connected set $Omega$, with zeros $a_j$ and poles $b_k$ then $$frac{1}{2 pi i}int_{gamma}frac{f'(z)}{f(z)} dz = sum_j n(gamma , a_j) - sum_k n(gamma , b_k)$$
Where the sums include multiplicities and $gamma$ is a cycle homologous to zero in $Omega$ and does not pass through any of the poles and zeros.
Here is am quite confused by the naming of the theorem. The proof does not seem to give me light on the naming either.
Thank you for the insight!
complex-analysis self-learning intuition
asked Nov 16 at 22:07
rannoudanames
531716
add a comment |
up vote
2
down vote
favorite
The Argument principle:
If $f$ is meromorphic in an open connected set $Omega$, with zeros $a_j$ and poles $b_k$ then $$frac{1}{2 pi i}int_{gamma}frac{f'(z)}{f(z)} dz = sum_j n(gamma , a_j) - sum_k n(gamma , b_k)$$
Where the sums include multiplicities and $gamma$ is a cycle homologous to zero in $Omega$ and does not pass through any of the poles and zeros.
Here is am quite confused by the naming of the theorem. The proof does not seem to give me light on the naming either.
Thank you for the insight!
complex-analysis self-learning intuition
asked Nov 16 at 22:07
rannoudanames
531716
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The Argument principle:
If $f$ is meromorphic in an open connected set $Omega$, with zeros $a_j$ and poles $b_k$ then $$frac{1}{2 pi i}int_{gamma}frac{f'(z)}{f(z)} dz = sum_j n(gamma , a_j) - sum_k n(gamma , b_k)$$
Where the sums include multiplicities and $gamma$ is a cycle homologous to zero in $Omega$ and does not pass through any of the poles and zeros.
Here is am quite confused by the naming of the theorem. The proof does not seem to give me light on the naming either.
Thank you for the insight!
complex-analysis self-learning intuition
asked Nov 16 at 22:07
rannoudanames
531716
The Argument principle:
If $f$ is meromorphic in an open connected set $Omega$, with zeros $a_j$ and poles $b_k$ then $$frac{1}{2 pi i}int_{gamma}frac{f'(z)}{f(z)} dz = sum_j n(gamma , a_j) - sum_k n(gamma , b_k)$$
Where the sums include multiplicities and $gamma$ is a cycle homologous to zero in $Omega$ and does not pass through any of the poles and zeros.
Here is am quite confused by the naming of the theorem. The proof does not seem to give me light on the naming either.
Thank you for the insight!
complex-analysis self-learning intuition
complex-analysis self-learning intuition
asked Nov 16 at 22:07
rannoudanames
531716
asked Nov 16 at 22:07
rannoudanames
531716
asked Nov 16 at 22:07
rannoudanames
531716
asked Nov 16 at 22:07
rannoudanames
531716
asked Nov 16 at 22:07
rannoudanames
531716
531716
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
"Formally"
$$int_gamma frac{f'(z)}{f(z)},dz=iint_gamma d(arg f(z))
=iint_{f(gamma)}d(arg w)
$$
is the overall change in the argument of $f(z)$ as $z$ moves around $gamma$
(that is $2pi$ times the winding number of the image contour $f(gamma)$ about
zero).
answered Nov 16 at 22:13
Lord Shark the Unknown
97.8k958129
ohhhh i get it, thanks!
– rannoudanames
Nov 16 at 22:15
ill aprove as soon as they let me
– rannoudanames
Nov 16 at 22:17
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
"Formally"
$$int_gamma frac{f'(z)}{f(z)},dz=iint_gamma d(arg f(z))
=iint_{f(gamma)}d(arg w)
$$
is the overall change in the argument of $f(z)$ as $z$ moves around $gamma$
(that is $2pi$ times the winding number of the image contour $f(gamma)$ about
zero).
answered Nov 16 at 22:13
Lord Shark the Unknown
97.8k958129
ohhhh i get it, thanks!
– rannoudanames
Nov 16 at 22:15
ill aprove as soon as they let me
– rannoudanames
Nov 16 at 22:17
add a comment |
up vote
6
down vote
accepted
"Formally"
$$int_gamma frac{f'(z)}{f(z)},dz=iint_gamma d(arg f(z))
=iint_{f(gamma)}d(arg w)
$$
is the overall change in the argument of $f(z)$ as $z$ moves around $gamma$
(that is $2pi$ times the winding number of the image contour $f(gamma)$ about
zero).
answered Nov 16 at 22:13
Lord Shark the Unknown
97.8k958129
ohhhh i get it, thanks!
– rannoudanames
Nov 16 at 22:15
ill aprove as soon as they let me
– rannoudanames
Nov 16 at 22:17
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
"Formally"
$$int_gamma frac{f'(z)}{f(z)},dz=iint_gamma d(arg f(z))
=iint_{f(gamma)}d(arg w)
$$
is the overall change in the argument of $f(z)$ as $z$ moves around $gamma$
(that is $2pi$ times the winding number of the image contour $f(gamma)$ about
zero).
answered Nov 16 at 22:13
Lord Shark the Unknown
97.8k958129
"Formally"
$$int_gamma frac{f'(z)}{f(z)},dz=iint_gamma d(arg f(z))
=iint_{f(gamma)}d(arg w)
$$
is the overall change in the argument of $f(z)$ as $z$ moves around $gamma$
(that is $2pi$ times the winding number of the image contour $f(gamma)$ about
zero).
answered Nov 16 at 22:13
Lord Shark the Unknown
97.8k958129
answered Nov 16 at 22:13
Lord Shark the Unknown
97.8k958129
answered Nov 16 at 22:13
Lord Shark the Unknown
97.8k958129
answered Nov 16 at 22:13
Lord Shark the Unknown
97.8k958129
97.8k958129
ohhhh i get it, thanks!
– rannoudanames
Nov 16 at 22:15
ill aprove as soon as they let me
– rannoudanames
Nov 16 at 22:17
add a comment |
ohhhh i get it, thanks!
– rannoudanames
Nov 16 at 22:15
ill aprove as soon as they let me
– rannoudanames
Nov 16 at 22:17
ohhhh i get it, thanks!
– rannoudanames
Nov 16 at 22:15
ohhhh i get it, thanks!
– rannoudanames
Nov 16 at 22:15
ill aprove as soon as they let me
– rannoudanames
Nov 16 at 22:17
ill aprove as soon as they let me
– rannoudanames
Nov 16 at 22:17
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001692%2fwhy-is-the-argument-principle-called-the-argument-principle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
