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Question about Domain and the Open Disk in Cauchy's Integral Formula

0

$begingroup$

In my class, this is our version of Cauchy's integral formula:

Let $OmegasubseteqmathbbC$ be an open convex set, $Dleft(a,rright)=left<rrightsubseteqOmega$ be the open disk centered at $ainmathbbC$ with radius $r>0$, the closure of $Dleft(a,rright)$ be contained in $Omega$, $bin Dleft(a,rright)$, and $fleft(zright)$ be holomorphic on $Omega$. Then,
$$fleft(bright)=frac12pi iint_gammafracfleft(zright)z-b,$$
where $gamma=lefta+re^ithetainmathbbC:thetainleft[0,2piright]right$.

I've seen some other versions (like on Wikipedia) of this where $Omega$ is only an open set in $mathbbC$. My questions are:

1. What is the difference if $Omega$ is convex or not? Does this change the theorem at all?


2. What happens if $b$ is on the boundary of $Dleft(a,rright)$? Does Cauchy's integral formula still hold? I am actually solving the integral $int_leftlvert zrightrvert=1fraccos zleft(z-iright)left(z+4right)textdz$. I have set $fleft(zright)=fraccos zz+4$ and $Omega=mathbbCsetminusleft-4right$, but I realized that $z=i$ lies on the boundary of $Dleft(0,1right)$ and this made me think about what happens at the boundary.

complex-analysis cauchy-integral-formula

share|cite|improve this question

asked Mar 28 at 1:52

JakeJake

565314

$endgroup$

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  $begingroup$
  "What happens if b is on the boundary" -- Try it with the unit disk, $f(z) = 1$, and, e.g., $b = 1$.
  $endgroup$
  – amsmath
  Mar 28 at 1:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amsmath If I use Cauchy's integral formula, it simply gives us $1$, but actually attempting to evaluate the integral shows that it is not convergent, so it does not necessarily hold for the boundary points?
  $endgroup$
  – Jake
  Mar 28 at 2:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly. The integral does not converge. About $Omega$: The convexity is indeed superfluous.
  $endgroup$
  – amsmath
  Mar 28 at 2:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amsmath, thank you! Greatly appreciate your help.
  $endgroup$
  – Jake
  Mar 28 at 2:14
  

  
  
  
  

add a comment | 

0

$begingroup$

In my class, this is our version of Cauchy's integral formula:

Let $OmegasubseteqmathbbC$ be an open convex set, $Dleft(a,rright)=left<rrightsubseteqOmega$ be the open disk centered at $ainmathbbC$ with radius $r>0$, the closure of $Dleft(a,rright)$ be contained in $Omega$, $bin Dleft(a,rright)$, and $fleft(zright)$ be holomorphic on $Omega$. Then,
$$fleft(bright)=frac12pi iint_gammafracfleft(zright)z-b,$$
where $gamma=lefta+re^ithetainmathbbC:thetainleft[0,2piright]right$.

I've seen some other versions (like on Wikipedia) of this where $Omega$ is only an open set in $mathbbC$. My questions are:

1. What is the difference if $Omega$ is convex or not? Does this change the theorem at all?


2. What happens if $b$ is on the boundary of $Dleft(a,rright)$? Does Cauchy's integral formula still hold? I am actually solving the integral $int_leftlvert zrightrvert=1fraccos zleft(z-iright)left(z+4right)textdz$. I have set $fleft(zright)=fraccos zz+4$ and $Omega=mathbbCsetminusleft-4right$, but I realized that $z=i$ lies on the boundary of $Dleft(0,1right)$ and this made me think about what happens at the boundary.

complex-analysis cauchy-integral-formula

share|cite|improve this question

asked Mar 28 at 1:52

JakeJake

565314

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "What happens if b is on the boundary" -- Try it with the unit disk, $f(z) = 1$, and, e.g., $b = 1$.
  $endgroup$
  – amsmath
  Mar 28 at 1:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amsmath If I use Cauchy's integral formula, it simply gives us $1$, but actually attempting to evaluate the integral shows that it is not convergent, so it does not necessarily hold for the boundary points?
  $endgroup$
  – Jake
  Mar 28 at 2:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Exactly. The integral does not converge. About $Omega$: The convexity is indeed superfluous.
  $endgroup$
  – amsmath
  Mar 28 at 2:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @amsmath, thank you! Greatly appreciate your help.
  $endgroup$
  – Jake
  Mar 28 at 2:14
  

  
  
  
  

add a comment | 

$begingroup$

In my class, this is our version of Cauchy's integral formula:

Let $OmegasubseteqmathbbC$ be an open convex set, $Dleft(a,rright)=left<rrightsubseteqOmega$ be the open disk centered at $ainmathbbC$ with radius $r>0$, the closure of $Dleft(a,rright)$ be contained in $Omega$, $bin Dleft(a,rright)$, and $fleft(zright)$ be holomorphic on $Omega$. Then,
$$fleft(bright)=frac12pi iint_gammafracfleft(zright)z-b,$$
where $gamma=lefta+re^ithetainmathbbC:thetainleft[0,2piright]right$.

I've seen some other versions (like on Wikipedia) of this where $Omega$ is only an open set in $mathbbC$. My questions are:

1. What is the difference if $Omega$ is convex or not? Does this change the theorem at all?


2. What happens if $b$ is on the boundary of $Dleft(a,rright)$? Does Cauchy's integral formula still hold? I am actually solving the integral $int_leftlvert zrightrvert=1fraccos zleft(z-iright)left(z+4right)textdz$. I have set $fleft(zright)=fraccos zz+4$ and $Omega=mathbbCsetminusleft-4right$, but I realized that $z=i$ lies on the boundary of $Dleft(0,1right)$ and this made me think about what happens at the boundary.

complex-analysis cauchy-integral-formula

share|cite|improve this question

asked Mar 28 at 1:52

JakeJake

565314

$endgroup$

share|cite|improve this question

asked Mar 28 at 1:52

JakeJake

565314

share|cite|improve this question

asked Mar 28 at 1:52

JakeJake

565314

asked Mar 28 at 1:52

JakeJake

565314

asked Mar 28 at 1:52

JakeJake

565314