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Cauchy integral formula help [closed]

0

$begingroup$

Please help me with this problem

$$int_{|z| = 2} frac{e^{z}}{z(z - 3)} , mathrm{d}z$$

$z-3$ lies outside the circle but $z$ is inside, how should you use Cauchy's Theorem or Cauchy's Integral Formula

integration complex-analysis cauchy-integral-formula

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edited Mar 26 at 1:44

Brian

1,495416

asked Mar 26 at 1:43

user657840

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closed as off-topic by Saad, Leucippus, mrtaurho, Cesareo, YiFan Mar 26 at 8:19

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Leucippus, mrtaurho, Cesareo, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Which part is confusing you, exactly?
  $endgroup$
  – Ryan Goulden
  Mar 26 at 1:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know about the residue theorem?
  $endgroup$
  – Ryan Goulden
  Mar 26 at 1:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
  $endgroup$
  – robjohn♦
  Mar 26 at 1:57
  

  
  
  
  

add a comment | 

0

$begingroup$

Please help me with this problem

$$int_{|z| = 2} frac{e^{z}}{z(z - 3)} , mathrm{d}z$$

$z-3$ lies outside the circle but $z$ is inside, how should you use Cauchy's Theorem or Cauchy's Integral Formula

integration complex-analysis cauchy-integral-formula

share|cite|improve this question

edited Mar 26 at 1:44

Brian

1,495416

asked Mar 26 at 1:43

user657840

$endgroup$

closed as off-topic by Saad, Leucippus, mrtaurho, Cesareo, YiFan Mar 26 at 8:19

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Leucippus, mrtaurho, Cesareo, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Which part is confusing you, exactly?
  $endgroup$
  – Ryan Goulden
  Mar 26 at 1:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know about the residue theorem?
  $endgroup$
  – Ryan Goulden
  Mar 26 at 1:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
  $endgroup$
  – robjohn♦
  Mar 26 at 1:57
  

  
  
  
  

add a comment | 

$begingroup$

Please help me with this problem

$$int_{|z| = 2} frac{e^{z}}{z(z - 3)} , mathrm{d}z$$

$z-3$ lies outside the circle but $z$ is inside, how should you use Cauchy's Theorem or Cauchy's Integral Formula

integration complex-analysis cauchy-integral-formula

share|cite|improve this question

edited Mar 26 at 1:44

Brian

1,495416

asked Mar 26 at 1:43

user657840

$endgroup$

share|cite|improve this question

edited Mar 26 at 1:44

Brian

1,495416

asked Mar 26 at 1:43

user657840

share|cite|improve this question

edited Mar 26 at 1:44

Brian

1,495416

asked Mar 26 at 1:43

user657840

edited Mar 26 at 1:44

Brian

1,495416

edited Mar 26 at 1:44

Brian

1,495416

3 Answers
3

active

oldest

votes

2

$begingroup$

The singularities of the integrand are $0$ and $3$. But only $0$ lies inside the circle $|z|=2$. Thus $$oint_{|z|=2}frac{mathrm{e}^z}{z(z-3)}mathrm{d}z=2pi i,mathrm{Res}left(frac{mathrm{e}^z}{z(z-3)},0right).$$The computation of this residue is straightforward by writing the correct Laurent series.

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answered Mar 26 at 1:53

Ivo TerekIvo Terek

46.8k954146

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add a comment | 

1

$begingroup$

Let $f (z)=dfrac {e^z}{z-3} $. Note that $f (z) $ is analytic inside and on the curve $|z|=2$. Then a direct application of Cauchy's Integral formula gives
$$oint_{|z| = 2} frac{e^{z}}{z(z - 3)} , mathrm{d}z=2pi i frac{e^{0}}{0- 3}=frac {-2pi i}{3}. $$

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answered Mar 26 at 2:01

Thomas ShelbyThomas Shelby

4,8082727

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0

$begingroup$

Define $f(z)=dfrac{e^z}{z(z-3)}$. Clearly the only singularities are $0$ and $3$ with ${0}in left| zright| =2$. So by the Residue Theorem we have: $$int_{left|zright|=1}dfrac{e^z}{z(z-3)}mathrm dz=2pi icdottext{Res }_{z=0}$$

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answered Mar 26 at 3:59

Paras KhoslaParas Khosla

3,3121627

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add a comment | 

2

$begingroup$

The singularities of the integrand are $0$ and $3$. But only $0$ lies inside the circle $|z|=2$. Thus $$oint_{|z|=2}frac{mathrm{e}^z}{z(z-3)}mathrm{d}z=2pi i,mathrm{Res}left(frac{mathrm{e}^z}{z(z-3)},0right).$$The computation of this residue is straightforward by writing the correct Laurent series.

share|cite|improve this answer

answered Mar 26 at 1:53

Ivo TerekIvo Terek

46.8k954146

$endgroup$

add a comment | 

2

$begingroup$

The singularities of the integrand are $0$ and $3$. But only $0$ lies inside the circle $|z|=2$. Thus $$oint_{|z|=2}frac{mathrm{e}^z}{z(z-3)}mathrm{d}z=2pi i,mathrm{Res}left(frac{mathrm{e}^z}{z(z-3)},0right).$$The computation of this residue is straightforward by writing the correct Laurent series.

share|cite|improve this answer

answered Mar 26 at 1:53

Ivo TerekIvo Terek

46.8k954146

$endgroup$

add a comment | 

$begingroup$

The singularities of the integrand are $0$ and $3$. But only $0$ lies inside the circle $|z|=2$. Thus $$oint_{|z|=2}frac{mathrm{e}^z}{z(z-3)}mathrm{d}z=2pi i,mathrm{Res}left(frac{mathrm{e}^z}{z(z-3)},0right).$$The computation of this residue is straightforward by writing the correct Laurent series.

share|cite|improve this answer

answered Mar 26 at 1:53

Ivo TerekIvo Terek

46.8k954146

$endgroup$

share|cite|improve this answer

answered Mar 26 at 1:53

Ivo TerekIvo Terek

46.8k954146

answered Mar 26 at 1:53

Ivo TerekIvo Terek

46.8k954146

answered Mar 26 at 1:53

Ivo TerekIvo Terek

46.8k954146

1

$begingroup$

Let $f (z)=dfrac {e^z}{z-3} $. Note that $f (z) $ is analytic inside and on the curve $|z|=2$. Then a direct application of Cauchy's Integral formula gives
$$oint_{|z| = 2} frac{e^{z}}{z(z - 3)} , mathrm{d}z=2pi i frac{e^{0}}{0- 3}=frac {-2pi i}{3}. $$

share|cite|improve this answer

answered Mar 26 at 2:01

Thomas ShelbyThomas Shelby

4,8082727

$endgroup$

add a comment | 

1

$begingroup$

Let $f (z)=dfrac {e^z}{z-3} $. Note that $f (z) $ is analytic inside and on the curve $|z|=2$. Then a direct application of Cauchy's Integral formula gives
$$oint_{|z| = 2} frac{e^{z}}{z(z - 3)} , mathrm{d}z=2pi i frac{e^{0}}{0- 3}=frac {-2pi i}{3}. $$

share|cite|improve this answer

answered Mar 26 at 2:01

Thomas ShelbyThomas Shelby

4,8082727

$endgroup$

add a comment | 

$begingroup$

Let $f (z)=dfrac {e^z}{z-3} $. Note that $f (z) $ is analytic inside and on the curve $|z|=2$. Then a direct application of Cauchy's Integral formula gives
$$oint_{|z| = 2} frac{e^{z}}{z(z - 3)} , mathrm{d}z=2pi i frac{e^{0}}{0- 3}=frac {-2pi i}{3}. $$

share|cite|improve this answer

answered Mar 26 at 2:01

Thomas ShelbyThomas Shelby

4,8082727

$endgroup$

share|cite|improve this answer

answered Mar 26 at 2:01

Thomas ShelbyThomas Shelby

4,8082727

answered Mar 26 at 2:01

Thomas ShelbyThomas Shelby

4,8082727

answered Mar 26 at 2:01

Thomas ShelbyThomas Shelby

4,8082727

0

$begingroup$

Define $f(z)=dfrac{e^z}{z(z-3)}$. Clearly the only singularities are $0$ and $3$ with ${0}in left| zright| =2$. So by the Residue Theorem we have: $$int_{left|zright|=1}dfrac{e^z}{z(z-3)}mathrm dz=2pi icdottext{Res }_{z=0}$$

share|cite|improve this answer

answered Mar 26 at 3:59

Paras KhoslaParas Khosla

3,3121627

$endgroup$

add a comment | 

0

$begingroup$

Define $f(z)=dfrac{e^z}{z(z-3)}$. Clearly the only singularities are $0$ and $3$ with ${0}in left| zright| =2$. So by the Residue Theorem we have: $$int_{left|zright|=1}dfrac{e^z}{z(z-3)}mathrm dz=2pi icdottext{Res }_{z=0}$$

share|cite|improve this answer

answered Mar 26 at 3:59

Paras KhoslaParas Khosla

3,3121627

$endgroup$

add a comment | 

$begingroup$

Define $f(z)=dfrac{e^z}{z(z-3)}$. Clearly the only singularities are $0$ and $3$ with ${0}in left| zright| =2$. So by the Residue Theorem we have: $$int_{left|zright|=1}dfrac{e^z}{z(z-3)}mathrm dz=2pi icdottext{Res }_{z=0}$$

share|cite|improve this answer

answered Mar 26 at 3:59

Paras KhoslaParas Khosla

3,3121627

$endgroup$

share|cite|improve this answer

answered Mar 26 at 3:59

Paras KhoslaParas Khosla

3,3121627

answered Mar 26 at 3:59

Paras KhoslaParas Khosla

3,3121627

answered Mar 26 at 3:59

Paras KhoslaParas Khosla

3,3121627