Complex functions average [closed]
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If $f(z)$ is analitic inside and over C, a circle centered in $a$. Show that the average of $f(z)$ in C is $f(a)$.
$barf(z)=fracint_C f(z) dzint_C dz$
But if I use Cauchy-integral-formula $int_C f(z) dz=0$, so I do not know what I have to do.
complex-analysis complex-integration cauchy-integral-formula
closed as off-topic by Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus Aug 29 at 0:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus
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up vote
-2
down vote
favorite
If $f(z)$ is analitic inside and over C, a circle centered in $a$. Show that the average of $f(z)$ in C is $f(a)$.
$barf(z)=fracint_C f(z) dzint_C dz$
But if I use Cauchy-integral-formula $int_C f(z) dz=0$, so I do not know what I have to do.
complex-analysis complex-integration cauchy-integral-formula
closed as off-topic by Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus Aug 29 at 0:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus
You should be able to find this in basically any complex analysis textbook.
â Sobi
Aug 28 at 17:18
What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
â Greg Martin
Aug 28 at 17:20
You should show your steps for arriving at $int_C f(z)dz=0$.
â Dave
Aug 28 at 17:35
Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
â user557651
Aug 28 at 17:40
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
If $f(z)$ is analitic inside and over C, a circle centered in $a$. Show that the average of $f(z)$ in C is $f(a)$.
$barf(z)=fracint_C f(z) dzint_C dz$
But if I use Cauchy-integral-formula $int_C f(z) dz=0$, so I do not know what I have to do.
complex-analysis complex-integration cauchy-integral-formula
If $f(z)$ is analitic inside and over C, a circle centered in $a$. Show that the average of $f(z)$ in C is $f(a)$.
$barf(z)=fracint_C f(z) dzint_C dz$
But if I use Cauchy-integral-formula $int_C f(z) dz=0$, so I do not know what I have to do.
complex-analysis complex-integration cauchy-integral-formula
edited Aug 28 at 17:32
asked Aug 28 at 17:10
user557651
373
373
closed as off-topic by Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus Aug 29 at 0:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus
closed as off-topic by Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus Aug 29 at 0:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Greg Martin, Matthew Leingang, José Carlos Santos, T. Bongers, Leucippus
You should be able to find this in basically any complex analysis textbook.
â Sobi
Aug 28 at 17:18
What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
â Greg Martin
Aug 28 at 17:20
You should show your steps for arriving at $int_C f(z)dz=0$.
â Dave
Aug 28 at 17:35
Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
â user557651
Aug 28 at 17:40
add a comment |Â
You should be able to find this in basically any complex analysis textbook.
â Sobi
Aug 28 at 17:18
What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
â Greg Martin
Aug 28 at 17:20
You should show your steps for arriving at $int_C f(z)dz=0$.
â Dave
Aug 28 at 17:35
Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
â user557651
Aug 28 at 17:40
You should be able to find this in basically any complex analysis textbook.
â Sobi
Aug 28 at 17:18
You should be able to find this in basically any complex analysis textbook.
â Sobi
Aug 28 at 17:18
What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
â Greg Martin
Aug 28 at 17:20
What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
â Greg Martin
Aug 28 at 17:20
You should show your steps for arriving at $int_C f(z)dz=0$.
â Dave
Aug 28 at 17:35
You should show your steps for arriving at $int_C f(z)dz=0$.
â Dave
Aug 28 at 17:35
Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
â user557651
Aug 28 at 17:40
Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
â user557651
Aug 28 at 17:40
add a comment |Â
1 Answer
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Hint: Cauchy integral theorem.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: Cauchy integral theorem.
add a comment |Â
up vote
1
down vote
Hint: Cauchy integral theorem.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: Cauchy integral theorem.
Hint: Cauchy integral theorem.
answered Aug 28 at 17:17
Robert Israel
306k22201443
306k22201443
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add a comment |Â
You should be able to find this in basically any complex analysis textbook.
â Sobi
Aug 28 at 17:18
What are your thoughts? What have you tried? Where are you stuck? You need to provide context for your question. Right now, it just looks like you want somebody to do your homework for you; that's not what this site is for. If you add some appropriate context, we will be happy to help.
â Greg Martin
Aug 28 at 17:20
You should show your steps for arriving at $int_C f(z)dz=0$.
â Dave
Aug 28 at 17:35
Because in Cauchy integral formula, the closed integral of an analitic function in this region is zero.
â user557651
Aug 28 at 17:40