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Computation of integral $int_rho fracdz(z-a)(z-b)$ [duplicate]

0

$begingroup$

This question already has an answer here:

• 
  Complex integral over a circle
  
  3 answers
  
  

Let $a,b$ be complex number and $|a| < r < |b|$, compute.

$int_rho fracdz(z-a)(z-b)$

where $rho$ is the circle with radius $r$ and the usual orientation.

I've tried the common path $int_rho fracdz(z-a)(z-b) = int_0^2pifracrie^i theta dtheta(re^i theta -a)(re^i theta-b)$ but in this point I don't know how use the $r$

complex-analysis complex-numbers contour-integration residue-calculus

share|cite|improve this question

edited Apr 7 at 19:31

Arnaud D.

16.2k52445

asked Apr 7 at 19:29

jfruizc273jfruizc273

126

$endgroup$

marked as duplicate by Arnaud D., Lord Shark the Unknown, user477343, Cesareo, Joel Reyes Noche 2 days ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Complex integral over a circle
  
  3 answers
  
  

Let $a,b$ be complex number and $|a| < r < |b|$, compute.

$int_rho fracdz(z-a)(z-b)$

where $rho$ is the circle with radius $r$ and the usual orientation.

I've tried the common path $int_rho fracdz(z-a)(z-b) = int_0^2pifracrie^i theta dtheta(re^i theta -a)(re^i theta-b)$ but in this point I don't know how use the $r$

complex-analysis complex-numbers contour-integration residue-calculus

share|cite|improve this question

edited Apr 7 at 19:31

Arnaud D.

16.2k52445

asked Apr 7 at 19:29

jfruizc273jfruizc273

126

$endgroup$

marked as duplicate by Arnaud D., Lord Shark the Unknown, user477343, Cesareo, Joel Reyes Noche 2 days ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Complex integral over a circle
  
  3 answers
  
  

Let $a,b$ be complex number and $|a| < r < |b|$, compute.

$int_rho fracdz(z-a)(z-b)$

where $rho$ is the circle with radius $r$ and the usual orientation.

I've tried the common path $int_rho fracdz(z-a)(z-b) = int_0^2pifracrie^i theta dtheta(re^i theta -a)(re^i theta-b)$ but in this point I don't know how use the $r$

complex-analysis complex-numbers contour-integration residue-calculus

share|cite|improve this question

edited Apr 7 at 19:31

Arnaud D.

16.2k52445

asked Apr 7 at 19:29

jfruizc273jfruizc273

126

$endgroup$

share|cite|improve this question

edited Apr 7 at 19:31

Arnaud D.

16.2k52445

asked Apr 7 at 19:29

jfruizc273jfruizc273

126

share|cite|improve this question

edited Apr 7 at 19:31

Arnaud D.

16.2k52445

asked Apr 7 at 19:29

jfruizc273jfruizc273

126

edited Apr 7 at 19:31

Arnaud D.

16.2k52445

edited Apr 7 at 19:31

Arnaud D.

16.2k52445

asked Apr 7 at 19:29

jfruizc273jfruizc273

126

asked Apr 7 at 19:29

jfruizc273jfruizc273

126

1 Answer
1

active

oldest

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0

$begingroup$

By residue theorem the integral is
$$
2pi itext Res_z=afrac1 (z-a)(z-b)=frac 2pi ia-b,
$$
since only the pole $z=a $ is inside the integration contour.

share|cite|improve this answer

answered Apr 7 at 20:19

useruser

6,47811031

$endgroup$

add a comment | 

0

$begingroup$

By residue theorem the integral is
$$
2pi itext Res_z=afrac1 (z-a)(z-b)=frac 2pi ia-b,
$$
since only the pole $z=a $ is inside the integration contour.

share|cite|improve this answer

answered Apr 7 at 20:19

useruser

6,47811031

$endgroup$

add a comment | 

0

$begingroup$

By residue theorem the integral is
$$
2pi itext Res_z=afrac1 (z-a)(z-b)=frac 2pi ia-b,
$$
since only the pole $z=a $ is inside the integration contour.

share|cite|improve this answer

answered Apr 7 at 20:19

useruser

6,47811031

$endgroup$

add a comment | 

$begingroup$

By residue theorem the integral is
$$
2pi itext Res_z=afrac1 (z-a)(z-b)=frac 2pi ia-b,
$$
since only the pole $z=a $ is inside the integration contour.

share|cite|improve this answer

answered Apr 7 at 20:19

useruser

6,47811031

$endgroup$

share|cite|improve this answer

answered Apr 7 at 20:19

useruser

6,47811031

answered Apr 7 at 20:19

useruser

6,47811031

answered Apr 7 at 20:19

useruser

6,47811031