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Here is another problem from Complex Analysis. I think this is the most common question we ever see in exams.

If $f$ is continuous on a Jordan arc $\gamma$, prove that the function: $$F(z)=\int_{\gamma} {f{(\theta})\over\theta-z}d\theta$$ is analytic for all $z$ not in $\gamma$.

What I think is I can prove $F(z)$ is differentiable using definition of the derivative at some arbitrary point $z_{0}$, and the continuity of $F(z)$ is trivially hold.

But showing differentiable using definition seems kind of funky to me. I was wondering if anyone have a better way without using the definition of the derivative.

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Dear Deepak, I think you mean "for all $z$ not in $\gamma$". Regards, – Matt E Dec 21 '12 at 2:48
First show $\min_{t\in [0,1]} |\gamma(t)-z| >0$. Then show for sufficiently small $h$, $\frac{1}{\theta-z-h}-\frac{1}{\theta-z} = \frac{h}{(\theta-z)^2(1+\frac{h}{\theta-z})} = \frac{1}{(\theta-z)^2} h (1-\frac{h}{\theta-z}+\cdots)$, – copper.hat Dec 21 '12 at 3:46
Morera's theorem. – mrf Dec 21 '12 at 6:46
@MattE, you are right. My bad. – Deepak Dec 21 '12 at 19:02
@MattE, I just edited that. Thank you for pointing out. – Deepak Dec 21 '12 at 19:13
up vote 2 down vote accepted

Let $z_0 \notin \text{im} \, \gamma$. Then $r:=d(z_0,\text{im} \, \gamma)>0$. Let $z \in B(z_0,r)$, then

$$\frac{1}{\theta-z} = \frac{1}{\theta-z_0} \cdot \frac{1}{1-\frac{z-z_0}{\theta-z_0}} = \frac{1}{\theta-z_0} \cdot \sum_{n=0}^\infty \left( \frac{z-z_0}{\theta-z_0} \right)^n$$

for $\theta \in \text{im} \, \gamma$. The series is absolutely uniform convergent for $\theta \in \text{im} \, \gamma$ since $\left| \frac{z-z_0}{\theta-z_0} \right| \leq \frac{|z-z_0|}{r} < 1$. Hence

$$F(z) = \int_\gamma \lim_{N \to \infty} \sum_{n=0}^N \frac{(z-z_0)^n}{(\theta-z_0)^{n+1}} \cdot f(\theta) \, d\theta = \sum_{n=0}^\infty (z-z_0)^n \cdot \underbrace{\int_\gamma f(\theta) \cdot \frac{1}{(\theta-z_0)^{n+1}} \, d\theta}_{=:a_n}$$

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