Prove that $\int_{\gamma} f(z)\ dz =0$.

Question

Let $f$ be continuous in $B(z_0 ; R)$ and analytic in $B(z_0 ; R) \setminus \{a\}$ for some $a \in B(z_0 ; R)$. Then $\int_{\gamma} f(z)\ dz =0$ for every closed contour $\gamma$ in $B(z_0 ; R)$.

I find lots of difficulty to prove the above result. Please help me in proving it.

Thank you in advance.

EDIT $:$

Before proving the result we have to first prove a lemma.

Lemma $:$

If $f$ is analytic in a region $G$ except at only one point then for any triangle $T$ in $G$ $$\int_{\partial T} f(z)\ dz = 0.$$

Proof $:$

If all the vertices of $T$ are in $G \setminus \{a\}$ then we directly apply Cauchy-Goursat to have $$\int_{\partial T} f(z)\ dz = 0.$$ Othewise one of the vertices of $T$ should be $a$. Then we divide the triangle into $4^n=(2^n)^2$ smaller triangles by adjoining the midpoints then $$\int_{\partial T} f(z)\ dz = \sum_{j=1}^{2^n} {\sum_{k=1}^{2^n} \int_{\partial T_{jk}} f(z)\ dz}$$ since the dividing segments cancel in pairs. Now if $T_{jk}$ doesn't contain $a$ then by Cauchy-Goursat theorem $$\int_{\partial T_{jk}} f(z)\ dz = 0.$$ Otherwise by M-L inequality we have $$\left | \int_{\partial T_{jk}} f(z)\ dz \right | \leq \frac {ML(\partial T)} {2^n}.$$ Where $M = \underset {z \in \partial T} {\sup} f(z)$.Now in $4^n$ subdivision $a$ can be on atmost $6$ triangles as one of their vertex.So we have $$\int_{\partial T} f(z)\ dz = \sum_{a\ is\ a\ vertex\ of\ T_{jk}} \int_{\partial T_{jk}} f(z)\ dz \leq \frac {6ML(\partial T)} {2^n} \rightarrow 0$$ as $n \rightarrow \infty$.

Hence the result follows.

Now let us prove the required result with the help of the above lemma.

Let $B=B(z_0;R)$.Here $a \in B$. Then since $B$ is convex so for any $z \in B$ the line segment $[a,z]$ is lying in $B$. Let us consider a function $F$ on $B$ defined by $$F(z) = \int_{[a,z]} f(\zeta)\ d{\zeta},$$ $z \in B$. Then take any $z \in B \setminus \{a\}$ and a $h \in \mathbb C$ with $|h|$ sufficiently small such that ${\bar B} (z;|h|) \subset B$ and hence $[z,z+h] \subset B$. Let us consider the triangle $T=[z,a,z+h]$. Since $B$ is open and $f$ is analytic in $B$ except at only one point $a$ so by the previous lemma we can deduce that $$F(z+h) - F(z) = \int_{[z,z+h]} f(\zeta)\ d{\zeta}.$$ From here it is not very tough to deduce that $F'(z) = f(z)$ for all $z \in B$ by the continuity of $f$.Now Fundamental theorem of contour integration completes the proof.

Answer 1

If you are to prove it from scratch then you could e.g. first show that the value of the integral along $\partial B(a;r)$ does not depend upon $0<r<\delta=R-|a-z_0|$, i.e. that for two different values of $r$ the difference is zero (note that the two circles bounds an annulus).

Then let $r\rightarrow 0$ and show that the value is zero.

Now, use this to show that the integral around the boundary of any triangle in $B(z_0,R)$ (possibly including $a$) is zero, and you may use your own argument to construct a primitive of $f$.

Answer 2

$f$ has an isolated singularity at $a$. Since $f$ is continuous at $a$, $f$ has a removable singularity at $a$ (Riemann !). Hence $f$ is analytic on $B(z_0 ; R)$.