Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Dear Deepak, I think you mean "for all $z$ not in $\gamma$". Regards, –  Matt E Dec 21 '12 at 2:48
1  
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
add comment

1 Answer

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.