title: Cauchy's integral theorem
text: # Cauchy's integral theorem
  
  In mathematics, the Cauchy integral theorem (also known as the Cauchy–Goursat theorem) in complex analysis, named after Augustin-Louis Cauchy (and Édouard Goursat), is an important statement about line integrals for holomorphic functions in the complex plane. Essentially, it says that if $f(z)$ is holomorphic in a simply connected domain Ω, then for any simply closed contour $C$ in Ω, that contour integral is zero.
  
  $\int _{C}f(z)\,dz=0.$
  
  ## Statement
  
  ### Fundamental theorem for complex line integrals
  
  If f(z) is a holomorphic function on an open region U, and $\gamma$ is a curve in U from $z_{0}$ to $z_{1}$ then,
  
  $\int _{\gamma }f'(z)\,dz=f(z_{1})-f(z_{0}).$
  
  Also, when f(z) has a single-valued antiderivative in an open region U, then the path integral $∫_γf^′(z)dz{\textstyle$ \int _{\gamma }f'(z)\, dz} is path independent for all paths in U.
  
  **Formulation on simply connected regions**
  
  Let $U\subseteq \mathbb {C}$ be a simply connected open set, and let $f:U\to \mathbb {C}$ be a holomorphic function. Let $\gamma :[a,b]\to U$ be a smooth closed curve. Then:
  
  $\int _{\gamma }f(z)\,dz=0.$
  
  (The condition that $U$ be simply connected means that $U$ has no "holes", or in other words, that the fundamental group of $U$ is trivial.)
  
  **General formulation**
  
  Let $U\subseteq \mathbb {C}$ be an open set, and let $f:U\to \mathbb {C}$ be a holomorphic function. Let $\gamma :[a,b]\to U$ be a smooth closed curve. If $\gamma$ is homotopic to a constant curve, then:
  
  $\int _{\gamma }f(z)\,dz=0.$
  
  (Recall that a curve is homotopic to a constant curve if there exists a smooth homotopy (within $U$) from the curve to the constant curve. Intuitively, this means that one can shrink the curve into a point without exiting the space.) The first version is a special case of this because on a simply connected set, every closed curve is homotopic to a constant curve.
  
  **Main example**
  
  In both cases, it is important to remember that the curve $\gamma$ does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve:
  
  $\gamma (t)=e^{it}\quad t\in \left[0,2\pi \right],$
  
  which traces out the unit circle. Here the following integral:
  
  $\int _{\gamma }{\frac {1}{z}}\,dz=2\pi i\neq 0,$
  
  is nonzero. The Cauchy integral theorem does not apply here since $f(z)=1/z$ is not defined at $z=0$. Intuitively, $\gamma$ surrounds a "hole" in the domain of $f$, so $\gamma$ cannot be shrunk to a point without exiting the space. Thus, the theorem does not apply.
  
  ## Discussion
  
  As Édouard Goursat showed, Cauchy's integral theorem can be proven assuming only that the complex derivative $f'(z)$ exists everywhere in $U$. This is significant because one can then prove Cauchy's integral formula for these functions, and from that deduce these functions are infinitely differentiable.
  
  The condition that $U$ be simply connected means that $U$ has no "holes" or, in homotopy terms, that the fundamental group of $U$ is trivial; for instance, every open disk $U_{z_{0}}=\{z:\left|z-z_{0}\right|<r\}$, for $z_{0}\in \mathbb {C}$, qualifies. The condition is crucial; consider
  
  $\gamma (t)=e^{it}\quad t\in \left[0,2\pi \right]$
  
  which traces out the unit circle, and then the path integral
  
  $\oint _{\gamma }{\frac {1}{z}}\,dz=\int _{0}^{2\pi }{\frac {1}{e^{it}}}(ie^{it}\,dt)=\int _{0}^{2\pi }i\,dt=2\pi i$
  
  is nonzero; the Cauchy integral theorem does not apply here since $f(z)=1/z$ is not defined (and is certainly not holomorphic) at $z=0$.
  
  One important consequence of the theorem is that path integrals of holomorphic functions on simply connected domains can be computed in a manner familiar from the fundamental theorem of calculus: let $U$ be a simply connected open subset of $\mathbb {C}$, let $f:U\to \mathbb {C}$ be a holomorphic function, and let $\gamma$ be a piecewise continuously differentiable path in $U$ with start point $a$ and end point $b$. If $F$ is a complex antiderivative of $f$, then
  
  $\int _{\gamma }f(z)\,dz=F(b)-F(a).$
  
  The Cauchy integral theorem is valid with a weaker hypothesis than given above, e.g. given $U$, a simply connected open subset of $\mathbb {C}$, we can weaken the assumptions to $f$ being holomorphic on $U$ and continuous on $U^¯{\textstyle$ {\overline {U}}} and $\gamma$ a rectifiable simple loop in $U^¯{\textstyle$ {\overline {U}}}.
  
  The Cauchy integral theorem leads to Cauchy's integral formula and the residue theorem.
  
  ## Proof
  
  If one assumes that the partial derivatives of a holomorphic function are continuous, the Cauchy integral theorem can be proven as a direct consequence of Green's theorem and the fact that the real and imaginary parts of $f=u+iv$ must satisfy the Cauchy–Riemann equations in the region bounded by $\gamma$, and moreover in the open neighborhood U of this region. Cauchy provided this proof, but it was later proven by Goursat without requiring techniques from vector calculus, or the continuity of partial derivatives.
  
  We can break the integrand $f$, as well as the differential $dz$ into their real and imaginary components:
  
  $f=u+iv$
  
  $dz=dx+i\,dy$
  
  In this case we have
  
  $\oint _{\gamma }f(z)\,dz=\oint _{\gamma }(u+iv)(dx+i\,dy)=\oint _{\gamma }(u\,dx-v\,dy)+i\oint _{\gamma }(v\,dx+u\,dy)$
  
  By Green's theorem, we may then replace the integrals around the closed contour $\gamma$ with an area integral throughout the domain $D$ that is enclosed by $\gamma$ as follows:
  
  $\oint _{\gamma }(u\,dx-v\,dy)=\iint _{D}\left(-{\frac {\partial v}{\partial x}}-{\frac {\partial u}{\partial y}}\right)\,dx\,dy$
  
  $\oint _{\gamma }(v\,dx+u\,dy)=\iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}\right)\,dx\,dy$
  
  But as the real and imaginary parts of a function holomorphic in the domain $D$, $u$ and $v$ must satisfy the Cauchy–Riemann equations there:
  
  ${\frac {\partial u}{\partial x}}={\frac {\partial v}{\partial y}}$
  
  ${\frac {\partial u}{\partial y}}=-{\frac {\partial v}{\partial x}}$
  
  We therefore find that both integrands (and hence their integrals) are zero
  
  $\iint _{D}\left(-{\frac {\partial v}{\partial x}}-{\frac {\partial u}{\partial y}}\right)\,dx\,dy=\iint _{D}\left({\frac {\partial u}{\partial y}}-{\frac {\partial u}{\partial y}}\right)\,dx\,dy=0$
  
  $\iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}\right)\,dx\,dy=\iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial u}{\partial x}}\right)\,dx\,dy=0$
  
  This gives the desired result
  
  $\oint _{\gamma }f(z)\,dz=0$
  
  