I am trying assignments of my complex analysis course which will not be discussed and I got struck on this question:
Let D be the open unit disc centred at origin and $f : D \to \mathbb{C}$ be an analytic function. Let f=u+iv . If f(z)=$\sum_{n=0}^{\infty} a_n z^n $ is the power series of f and for any closed curve p in D , $\int_{p} \frac{ f(z) dz}{(z-a)^2} =0 $ for all a in D with $|a|\geq 1/2$ then f must be constant.
Cauchy theorem implies that $a_{1}=0$ for any defined power series with a as centre such that $|a|\geq 1/2$. But I don't have any reason to believe that f is constant and so I am unable to understand which result I should use.
Kindly help me to prove the deduction.
It is my first course in complex analysis.
Observe that Cauchy's formula gives for all $|z| \geq 1/2$
$$f'(z) = \frac{1}{2\pi i} \int_{p} \frac{f(\zeta)}{(\zeta-z)^2} d\zeta=0,$$
which shows that $f$ is constant in the annulus $1/2 < |z| < 1$.
Now, you can use the maximum modulus principle on $f'$ for the disc $|z| < 1/2$. Since $f'$ is itself analytic, it achieves its maximum at the boundry of this disc, which we have shown to be 0. Thus, $f'(z)=0$ for all $z \in D$, which shows that $f$ is constant.