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If $f : U \rightarrow \mathbb{C}$ is differentiable in $U$ open subset of $ \mathbb{C}$ then $f$ is analytic in $U$.

(In the proof you consider a triangle $T_1$ in $U$, subdivide it in a sequence of smaller triangles $T_n $, observe that $ \displaystyle \bigcap_{n=1} ^{\infty} T_n = \lbrace z_0 \rbrace $, show that $ \displaystyle \int _{\partial T_1} f(z) dz = 0 $ and then apply Morera's theorem).

At a certain point in the proof, since $f$ is differentiable it is stated correctly that $$ |f(z) - f(z_0) - f'(z_0)(z-z_0)| \leq \varepsilon_n (z-z_0)$$ with $\varepsilon_n \rightarrow 0$ as $n \rightarrow \infty$; then $$ \left| \int_{\partial T_n} f(z) dz \right| = \left| \int_{\partial T_n} f(z) - f(z_0) - f'(z_0)(z-z_0) dz \right| $$ How is this obtained?

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Hi, welcome to math.SE! The way to mark a question as solved is to click the grey checkmark next to the answer. Cheers. – Rahul Jun 3 '12 at 22:31
@Alex The best way to indicate that the issue has been resolve is the accept the answer that solved it. If you worked it out by yourself, please consider taking a moment to sketch the solution by writing up your own answer. You can accept that answer then. – Sasha Jun 3 '12 at 22:32
up vote 2 down vote accepted

$$\int_{\partial T_n} f(z) dz = \int_{\partial T_n} f(z_0) dz + \int_{\partial T_n} f'(z_0)(z-z_0) dz + \int_{\partial T_n} \left(f(z) - f(z_0) - f'(z_0)(z-z_0) \right) dz$$

Now the constant term $f(z_0)$ and the linear term $f'(z_0)(z-z_0)$ have primitives and hence you can integrate them out to get $0$.

Hence, $$\int_{\partial T_n} f(z) dz = \int_{\partial T_n} \left(f(z) - f(z_0) - f'(z_0)(z-z_0) \right) dz$$

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Thank you. Forgot that $\partial T_n$ is a closed curve... – user32847 Jun 3 '12 at 22:19
@Alex You do not need to write [solved] on the question title. You can instead accept the answer. – user17762 Jun 3 '12 at 22:20

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