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Why is the proof that contour integrals of triangles on holomorphic domains not trivial?

2

$begingroup$

My professor proved to us the following:

If γ is a closed [piecewise-smooth] curve in an open set Ω whose interior is also contained in Ω, f is continuous and has a primitive in Ω, then

$oint_gammaf(z)dz=0$.

He then went on to prove this:

Let Ω ⊂ C be an open set and T ⊂ Ω be a triangle [piecewise-smooth curve] whose interior is also contained in Ω, then whenever f is holomorphic in Ω

$oint_Tf(z)dz=0$.

The second proof was very long. He used it to justify that the contour integral of any polygon is zero, since any polygon is the union of triangles.

My question is, why was the second proof necessary? Surely it follows immediately from the first result? Apart from the square brackets, everything I've written is directly copy-pasted from his lecture notes. I don't see why the first result doesn't just make the second trivial.

complex-analysis analysis

share|cite|improve this question

asked Mar 26 at 13:19

otah007otah007

1568

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  The entire point of the second statement (Goursat’s theorem) is that you’re no longer assuming that $f$ has a primitive—this is a key step towards eventually proving that holomorphic functions are analytic.
  $endgroup$
  – Branimir Ćaćić
  Mar 26 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BranimirĆaćić Add that as an answer and I can accept it.
  $endgroup$
  – otah007
  Mar 26 at 21:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can take a look at thisone for Goursat not being such a big deal. Goursat says for $f$ holomorphic then $F(z)=int_a^z f(s)ds$ is well-defined (it doesn't depend on the path $a to z$) so $F$ is a primitive of $f$.
  $endgroup$
  – reuns
  Mar 27 at 0:32
  
  

  
  
  
  

add a comment | 

2

$begingroup$

My professor proved to us the following:

If γ is a closed [piecewise-smooth] curve in an open set Ω whose interior is also contained in Ω, f is continuous and has a primitive in Ω, then

$oint_gammaf(z)dz=0$.

He then went on to prove this:

Let Ω ⊂ C be an open set and T ⊂ Ω be a triangle [piecewise-smooth curve] whose interior is also contained in Ω, then whenever f is holomorphic in Ω

$oint_Tf(z)dz=0$.

The second proof was very long. He used it to justify that the contour integral of any polygon is zero, since any polygon is the union of triangles.

My question is, why was the second proof necessary? Surely it follows immediately from the first result? Apart from the square brackets, everything I've written is directly copy-pasted from his lecture notes. I don't see why the first result doesn't just make the second trivial.

complex-analysis analysis

share|cite|improve this question

asked Mar 26 at 13:19

otah007otah007

1568

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  The entire point of the second statement (Goursat’s theorem) is that you’re no longer assuming that $f$ has a primitive—this is a key step towards eventually proving that holomorphic functions are analytic.
  $endgroup$
  – Branimir Ćaćić
  Mar 26 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BranimirĆaćić Add that as an answer and I can accept it.
  $endgroup$
  – otah007
  Mar 26 at 21:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can take a look at thisone for Goursat not being such a big deal. Goursat says for $f$ holomorphic then $F(z)=int_a^z f(s)ds$ is well-defined (it doesn't depend on the path $a to z$) so $F$ is a primitive of $f$.
  $endgroup$
  – reuns
  Mar 27 at 0:32
  
  

  
  
  
  

add a comment | 

$begingroup$

My professor proved to us the following:

If γ is a closed [piecewise-smooth] curve in an open set Ω whose interior is also contained in Ω, f is continuous and has a primitive in Ω, then

$oint_gammaf(z)dz=0$.

He then went on to prove this:

Let Ω ⊂ C be an open set and T ⊂ Ω be a triangle [piecewise-smooth curve] whose interior is also contained in Ω, then whenever f is holomorphic in Ω

$oint_Tf(z)dz=0$.

The second proof was very long. He used it to justify that the contour integral of any polygon is zero, since any polygon is the union of triangles.

My question is, why was the second proof necessary? Surely it follows immediately from the first result? Apart from the square brackets, everything I've written is directly copy-pasted from his lecture notes. I don't see why the first result doesn't just make the second trivial.

complex-analysis analysis

share|cite|improve this question

asked Mar 26 at 13:19

otah007otah007

1568

$endgroup$

share|cite|improve this question

asked Mar 26 at 13:19

otah007otah007

1568

share|cite|improve this question

asked Mar 26 at 13:19

otah007otah007

1568

asked Mar 26 at 13:19

otah007otah007

1568

asked Mar 26 at 13:19

otah007otah007

1568

1 Answer
1

active

oldest

votes

0

$begingroup$

Branimir Ćaćić already explained why the first result doesn't imply the second result. But there's something else: The first result is stated in a strange way. There is no need to assume the interior of $gamma$ is contained in $Omega.$ If $f$ is continuous on an open set $Omegasubset mathbb C$ and $f$ has a primitive in $Omega,$ then $int_gamma f(z),dz=0$ for any closed contour $gamma$ in $Omega.$ This follows from the fundamental theorem of calculus. Yes, the FTC you learned in ordinary calculus. There's nothing of any depth (beyond the depth of Calc. 1) happening here. It's good to understand this.

share|cite|improve this answer

edited Mar 27 at 19:01

answered Mar 26 at 21:40

zhw.zhw.

74.8k43175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know the second result doesn't follow from the first, I'm asking why it doesn't. Branimir Ćaćić's comment is correct - it is to do with the assumption of the existence of a primitive. Also, it does depend on the interior of γ being contained in Ω, since if it wasn't then there might be a pole inside γ and the integral would be non-zero. Restating the result and closing with 'I don't know' isn't a very helpful answer...
  $endgroup$
  – otah007
  Mar 26 at 21:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @zhw: Nothing weird is happening; as far as I know this is a perfectly normal sequence of results in a complex analysis course. Perhaps you have misread the question? The OP is asking why the second result cannot be deduced immediately from the first.
  $endgroup$
  – Will R
  Mar 26 at 22:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @otah007 Yes, the comment of Branimir Ćaćić explained why the second result doesn't follow from the first. I felt no need to comment further on it. But I noticed something else, which was the way you stated the first result. That result does not depend on the interior of $gamma$ being contained in $Omega.$ As i said, the proof is elementary; it follows from the FTC. If your professor stated it as you have it, then that is indeed strange.
  $endgroup$
  – zhw.
  Mar 27 at 18:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WillR It's not normal to state the first result as the OP has it. That's what I was addressing.
  $endgroup$
  – zhw.
  Mar 27 at 18:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I edited my answer; hopefully it's clearer now.
  $endgroup$
  – zhw.
  Mar 27 at 19:02
  

  
  
  
  

 | 
show 2 more comments

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0

$begingroup$

Branimir Ćaćić already explained why the first result doesn't imply the second result. But there's something else: The first result is stated in a strange way. There is no need to assume the interior of $gamma$ is contained in $Omega.$ If $f$ is continuous on an open set $Omegasubset mathbb C$ and $f$ has a primitive in $Omega,$ then $int_gamma f(z),dz=0$ for any closed contour $gamma$ in $Omega.$ This follows from the fundamental theorem of calculus. Yes, the FTC you learned in ordinary calculus. There's nothing of any depth (beyond the depth of Calc. 1) happening here. It's good to understand this.

share|cite|improve this answer

edited Mar 27 at 19:01

answered Mar 26 at 21:40

zhw.zhw.

74.8k43175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know the second result doesn't follow from the first, I'm asking why it doesn't. Branimir Ćaćić's comment is correct - it is to do with the assumption of the existence of a primitive. Also, it does depend on the interior of γ being contained in Ω, since if it wasn't then there might be a pole inside γ and the integral would be non-zero. Restating the result and closing with 'I don't know' isn't a very helpful answer...
  $endgroup$
  – otah007
  Mar 26 at 21:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @zhw: Nothing weird is happening; as far as I know this is a perfectly normal sequence of results in a complex analysis course. Perhaps you have misread the question? The OP is asking why the second result cannot be deduced immediately from the first.
  $endgroup$
  – Will R
  Mar 26 at 22:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @otah007 Yes, the comment of Branimir Ćaćić explained why the second result doesn't follow from the first. I felt no need to comment further on it. But I noticed something else, which was the way you stated the first result. That result does not depend on the interior of $gamma$ being contained in $Omega.$ As i said, the proof is elementary; it follows from the FTC. If your professor stated it as you have it, then that is indeed strange.
  $endgroup$
  – zhw.
  Mar 27 at 18:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WillR It's not normal to state the first result as the OP has it. That's what I was addressing.
  $endgroup$
  – zhw.
  Mar 27 at 18:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I edited my answer; hopefully it's clearer now.
  $endgroup$
  – zhw.
  Mar 27 at 19:02
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

Branimir Ćaćić already explained why the first result doesn't imply the second result. But there's something else: The first result is stated in a strange way. There is no need to assume the interior of $gamma$ is contained in $Omega.$ If $f$ is continuous on an open set $Omegasubset mathbb C$ and $f$ has a primitive in $Omega,$ then $int_gamma f(z),dz=0$ for any closed contour $gamma$ in $Omega.$ This follows from the fundamental theorem of calculus. Yes, the FTC you learned in ordinary calculus. There's nothing of any depth (beyond the depth of Calc. 1) happening here. It's good to understand this.

share|cite|improve this answer

edited Mar 27 at 19:01

answered Mar 26 at 21:40

zhw.zhw.

74.8k43175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know the second result doesn't follow from the first, I'm asking why it doesn't. Branimir Ćaćić's comment is correct - it is to do with the assumption of the existence of a primitive. Also, it does depend on the interior of γ being contained in Ω, since if it wasn't then there might be a pole inside γ and the integral would be non-zero. Restating the result and closing with 'I don't know' isn't a very helpful answer...
  $endgroup$
  – otah007
  Mar 26 at 21:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @zhw: Nothing weird is happening; as far as I know this is a perfectly normal sequence of results in a complex analysis course. Perhaps you have misread the question? The OP is asking why the second result cannot be deduced immediately from the first.
  $endgroup$
  – Will R
  Mar 26 at 22:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @otah007 Yes, the comment of Branimir Ćaćić explained why the second result doesn't follow from the first. I felt no need to comment further on it. But I noticed something else, which was the way you stated the first result. That result does not depend on the interior of $gamma$ being contained in $Omega.$ As i said, the proof is elementary; it follows from the FTC. If your professor stated it as you have it, then that is indeed strange.
  $endgroup$
  – zhw.
  Mar 27 at 18:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WillR It's not normal to state the first result as the OP has it. That's what I was addressing.
  $endgroup$
  – zhw.
  Mar 27 at 18:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I edited my answer; hopefully it's clearer now.
  $endgroup$
  – zhw.
  Mar 27 at 19:02
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Branimir Ćaćić already explained why the first result doesn't imply the second result. But there's something else: The first result is stated in a strange way. There is no need to assume the interior of $gamma$ is contained in $Omega.$ If $f$ is continuous on an open set $Omegasubset mathbb C$ and $f$ has a primitive in $Omega,$ then $int_gamma f(z),dz=0$ for any closed contour $gamma$ in $Omega.$ This follows from the fundamental theorem of calculus. Yes, the FTC you learned in ordinary calculus. There's nothing of any depth (beyond the depth of Calc. 1) happening here. It's good to understand this.

share|cite|improve this answer

edited Mar 27 at 19:01

answered Mar 26 at 21:40

zhw.zhw.

74.8k43175

$endgroup$

share|cite|improve this answer

edited Mar 27 at 19:01

answered Mar 26 at 21:40

zhw.zhw.

74.8k43175

answered Mar 26 at 21:40

zhw.zhw.

74.8k43175

answered Mar 26 at 21:40

zhw.zhw.

74.8k43175