In the books I looked and also in the script of my former Complex Analysis Prof. we proved the Cauchy Integral Formula for simply connected domains and closed curves in the following order:

  1. Goursat's Theorem for Triangles for functions continuous everywhere and holomorphic everywhere except at a single point
  2. Cauchy's Theorem for convex domains for functions continuous everywhere and holomorphic everywhere except at a single point
  3. Cauchy's Integral Formula for convex domains
  4. Cauchy's Integral Theorem for null homologous cycles
  5. Introduction of homotopic curves and proof of Cauchy's Integral Theorem for those curves.

What I really don't like is two things: First this artificial single point in (1) and (2). It is of course nice to have such a weak version of Goursat and Cauchy Integral Theorem, because then the Integral Formula is easy to derive. But as a student is seems very artificial at first to exclude a single point out of the domain of holomorphy, and then see later that this makes sense and is handy. Second, I don't like to go first over nice domains like convex or starlike domains and afterwards end up in arbitrary domains.

So, what I know is the following version of Cauchy's Integral Theorem:

Let $\Omega$ be a simply connected domain, let $f$ be holomorphic in $\Omega$ and let $\gamma$ be a closed curve (piecewise $C^1$) in $\Omega$. Then \begin{align*} \int_\gamma f(z) dz=0. \end{align*}

I would like to deduce from this the following version of the Cauchy Integral Formula:

Let $\Omega$ be a simply connected domain, let $f$ be holomorphic in $\Omega$ and let $\gamma$ be a closed curve (piecewise $C^1$) in $\Omega$. Then \begin{align*} n(\gamma, z)f(z)=\frac{1}{2\pi i}\int_\gamma \frac{f(w)}{w-z} dw,\qquad z\in\Omega\backslash\gamma. \end{align*} Here, $n(\gamma,z)$ is the winding number of $\gamma$ at $z$.

The only way i can think of to achieve my goal is to fix some $z\in\Omega\backslash\gamma$ and cut a line connecting $z$ and the boundary of $\Omega$ out of $\Omega$ to obtain a simply connected domain. Then I need to adjust $\gamma$ so that it does not run trough this slit anymore and goes around $z$. After that I must show that this detour goes to zero if I move along the slit sufficiently close. This is certainly a way to go, but I really don't like how technical this gets if one wants to write down a rigorous proof.

My question is: Is there a more elegant way?

Once again, I don't know that holomorphic functions have a continuous derivative, that they can be developed in a power series expension and so on. I only have the abovementioned version of Cauchy's Theorem.

  • $\begingroup$ how does that help? provided that $0\in\Omega$ then this integral is 0 for small r by Cauchy's Theorem... $\endgroup$
    – sranthrop
    Sep 25, 2016 at 11:42
  • $\begingroup$ Assume $f(z)$ is complex differentiable, and let $g(r) = \int_0^1 f(r\gamma(t)) \gamma'(t) dt$. Then $r g'(r) = r\int_0^1 \gamma(t) f'(r\gamma(t)) \gamma'(t) dt = \gamma(t)f(r\gamma(t)) dt|_0^1 -\int_0^1 \gamma'(t) f(r\gamma(t)) dt = -g(r)$. So with $h(r) = r g(r)$ you have $rg'(r) + g(r) = h'(r) = 0$ and $h(0) = 0 \implies h(r) = 0$. From this you get the Cauchy integral theorem for any convex domain, and easily get all the other cases. $\endgroup$
    – reuns
    Sep 25, 2016 at 12:38
  • $\begingroup$ I meant $f(z)$ is (piecewise) continuously complex differentiable, and this is the hard part, where Goursat theorem is needed : showing the Cauchy integral theorem for "non continuously" complex differentiable functions. $\endgroup$
    – reuns
    Sep 25, 2016 at 13:01
  • $\begingroup$ I still don't know how this helps. I only HAVE the Cauchy Integral Theorem, I want to derive from this the Cauchy Integral FORMULA in the abovementioned version ;) $\endgroup$
    – sranthrop
    Sep 25, 2016 at 16:31
  • $\begingroup$ (if $f(z)$ is continuously complex differentiable on $U$) From this you easily get that $\int_{\gamma} f(z)dz = \int_{\gamma'}f(z) dz$ if $\gamma$ is homotopically equivalent to $\gamma'$ in $U$, and hence if $U$ is simply connected containing $a$ : $\int_\gamma \frac{f(z)}{z-a}dz = \int_\gamma \frac{f(z)-f(a)}{z-a}dz+\int_\gamma \frac{f(a)}{z-a}dz = n(\gamma,a) \int_{|z|=r} \frac{f(a)}{z-a}dz = 2i\pi n(\gamma,a) f(a)$ (since $\frac{f(z)-f(a)}{z-a}$ is holomorphic on $U$) $\endgroup$
    – reuns
    Sep 25, 2016 at 16:53

2 Answers 2


Every proof I have every seen of Cauchy's formula from Cauchy's theorem applies a version of Cauchy's theorem to the function $g$ defined by $g(w) = \dfrac{f(w) - f(z)}{w-z}$ for $w \neq z$ and $g(z) = f'(z)$. Notice $g$ is continuous wherever $f$ is continuous, and $g$ is holomorphic wherever $f$ is holomorphic, expect possibly at $z$.

As you noticed, you can't apply your version of Cauchy's theorem to $g$ because of lack of holomorphic-ness at $z$, so people establish versions of Cauchy's theorem that allow for the lack of holomorphic-ness at a single point.

I agree that this seems a bit tedious. I don't know if there is a fundamentally different method, but I think there is a way to make it look less tedious.

I suggest you have a look at the book Complex Variables and Applications by Churchill and Brown. You will find:

  • A direct proof of Cauchy's theorem that does not first go through special regions like triangles or convex sets. Section title: Cauchy-Goursat Theorem.
  • The statement of Cauchy's theorem in simply connected domains. Section title: Simply Connected Domains (or Simply and Mulitply Connected Domains if you have an older edition).
  • Cauchy's theorem for multiply connected domains. The proof is just to draw some lines and use cancellation of contour integrals in opposite directions. Section title: Multiply Connected Domains (or Simply and Multiply Connected Domains if you have an older edition)
  • Cauchy's formula in simply connected domains. Section title: Cauchy Integral Formula.

You can find this book online easily. Try googling Library Genesis.

  • $\begingroup$ Thank you! I will have a look at it. $\endgroup$
    – sranthrop
    Sep 25, 2016 at 17:28

I think this is only going part-way to what you want, because I still want to use a ``nice domain" first, but I'd like to make a case for the fact that using a convex domain in this way is not an artificial device.

Step 1: (this is the bit you're not going to like) We use Cauchy's theorem for disks to show a special case of Cauchy's integral formula: that is, we show that for $z \in B(a,r)$ the open disk of radius $r$ centred at $a$ we have $f(z) = \frac{1}{2\pi i} \int_{\gamma_{a,r}} \frac{f(w)}{w-z}dw$, where $\gamma_{a,r} = a+r\exp(2\pi i.t)$ is the closed path given by traversing the boundary of $D(a,r)$ anticlockwise.

This follows easily from Cauchy's theorem: by translating and scaling one can assume that $a =0$ and $z=s>0$ is a positive real number, so that $s\in (0,1)$. If we write $\gamma_{[z_0,z_1]}(t) = z_0+t(z_1-z_0)$ for the line segment path from $z_0$ to $z_1$, and $u_{a,r}(t) = a+ r\exp(\pi i.t)$, $l_{a,r}(t) = a-r\exp(\pi i.t)$ for the semicircular paths from $a+r$ to $a-r$ and $a-r$ to $a+r$ respectively, then if we fix $r<(1-s)/2$ and, for a path $\gamma\colon [0,1]\to \mathbb C$, write $\gamma^-$ for the path $\gamma^-(t) = \gamma(1-t)$, define

$$ \begin{split} \Gamma_+ &= u_{0,1}*\gamma_{[-1,r-s]}*u_{s,r}^-*\gamma_{[s+r,1]} \\ \Gamma_- &= \gamma_{[s+r,1]}^-*l_{s,r}^-*\gamma_{[-1,s-r]}^-*l_{0,1} \end{split} $$ with $*$ denoting concatenation of paths.

Thus $\Gamma_{\pm}$ are "indented" versions of the closed paths obtained by traversing the interval $[-1,1]$ and one of semicircles joining $-1$ and $1$, with semicircular ``indentations" of radius $r$ centred at $s$ ensuring that they avoid $s$. As a result, each of the paths $\Gamma_{\pm}$ are closed contours lying in a simply-connected domain in which the function $g(z) = \frac{f(z)}{z-s}$ is holomorphic, and so the integral over each is zero.

The contours <span class=$\Gamma_{\pm}$" />

Now since the line segment paths cancel in the sum $\Gamma_+ + \Gamma_-$ and the semicircular paths combine to give $\gamma_{0,1} + \gamma_{s,r}^-$, we deduce that the integral $\int_{\gamma_{0,1}} \frac{f(z)}{z-s}dz = \int_{\gamma_{s,r}}\frac{f(z)}{z-s}dz$. Since it is easy to see just using the continuity of $f$ at $s$ that $\lim_{r\to 0} \int_{\gamma_{s,r}} \frac{f(z)}{z-s}dz = 2\pi i.f(s)$, it follows that $\frac{1}{2\pi i}\int_{\gamma_{0,1}} \frac{f(z)}{z-s} dz = f(s)$.

Step 2: Once you have this simple form of Cauchy's Integral Formula, you can prove all the good "local" properties of holomorphic functions, in particular it is easy to see that holomorphic functions are analytic, that is, if $f$ is holomorphic on an open set $U$ and $a \in U$, $f$ is equal to its Taylor series in any ball $B(a,r)$ strictly contained in $U$. Using ``nice domains" to prove local properties of holomorphic functions seems reasonable to me, because $\mathbb C$ (indeed any vector space) is locally convex, or "locally nice" so to speak.

Step 3: So now, finally, for what hopefully counts as the "good bit": if you know that holomorphic functions are analytic, you can give a proof of the "holomorphic except at one point implies holomorphic" statement which is (to my mind) much more satisfying than the method you outlined, because it feels more conceptual:

To clarify what we want to do, recall that if $f$ is holomorphic on a domain $D$ and $B(z_0,r)\backslash \{z_0\} \subseteq D$ and $f$ is bounded on $B(z_0,r)\backslash\{z_0\}$, then $f$ extends to a holomorphic function on all of $B(z_0,r)$ (so in particular, if $f(z)$ is continuous at $z_0$ then it was in fact already holomorphic there). The reason this is easy to believe when you know holomorphic implies analytic, is that it is easy to "squash" the singularity at $z_0$: Since $f$ is bounded near $z_0$, the function $(z-z_0).f(z)$ vanishes as $z\to z_0$. Thus if we set $h(z) = (z-z_0)^2f(z)$ for $z \neq z_0$ and $h(z_0)=0$ we have

$$ \lim_{z\to z_0} \frac{h(z)-h(z_0)}{z-z_0} = \lim_{z\to z_0} (z-z_0).f(z) \to 0, \text{ as } z\to z_0, $$ so that $h'(z_0)=0$. Clearly $h(z)$ is holomorphic everywhere $f$ is, and hence it follows that $h(z)$ is holomorphic in $B(z_0,r)$. But then we know that, near $z_0$ (say on $B(z_0,r/2)$) the function $h(z)$ is given by a power series, that is $h(z) = \sum_{n\geq 0} a_n(z-z_0)^n$, and moreover because $h(z)=h'(z)=0$, standard properties of power series show that $a_0=a_1=0$, and so $h(z) = (z-z_0)^2.\sum_{n\geq 0} a_{n+2}(z-z_0)^n$. But then $f(z) = \sum_{n\geq 0} a_{n+2}(z-z_0)^n$ is analytic and hence holomorphic on $B(z_0,r/2)$, so that in particular if we set $f(z_0) = a_2$, then $f$ is holomorphic at $z_0$.

Step 4: Once we have the removable singularity result, we can easily obtain the integral formula from Cauchy's theorem using the secants of $f$: if $f(z)$ is holomorphic in a simply-connected domain $\Omega$, and $w\in \Omega$, the function $g(z) = \frac{f(z)-f(w)}{z-w}$ is clearly holomorphic in $\Omega\backslash\{w\}$, and as $z \to w$ clearly $g(z) \to f'(w)$, hence it is bounded. By the above, it follows that $g(z)$ is holomorphic on all of $\Omega$, whence $\int_{\gamma} g(z)dz = 0$ by Cauchy's theorem. But then it follows that $$ \int_{\gamma} \frac{f(z)}{z-w} - \frac{f(w)}{z-w} dz =0, $$ so that $$ \frac{1}{2\pi i}\int_{\gamma} \frac{f(z)}{z-w}dz = \frac{f(w)}{2\pi i}\int_{\gamma} \frac{1}{z-w} dz = n(\gamma,w).f(w) $$ where $\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z-w} = n(\gamma,w)$ is the winding number by definition for some people, but is readily identified with it however it is defined.

In conclusion, if you can put up with disks for a bit then this approach removes the need to consider any artificial points in the domain of $f$.


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