Let $f$ be entire . Evaluate $\int ^{2\pi}_0 f(z_0+re^{i\theta)}e^{i\theta} d\theta$

Question

Let $f$ be entire . Evaluate $\int ^{2\pi}_0 f(z_0+re^{i\theta)}e^{i\theta} d\theta$

my attempt : $z=z_0+re^{i\theta}$

$dz=rie^{i\theta }d\theta$ then

$\int ^{2\pi}_0 f(z_0+re^{i\theta)}e^{i\theta} d\theta=\int^{2\pi}_0 \frac{f(z)}{ir}dz$

how to processed for further

Answer 1

Hint: You made a mistake in your last integral. Letting $\gamma (t) = z_0 + re^{it}, t\in [0,2\pi],$ it should be

$$\frac{1}{ir}\int_\gamma f(z)\,dz.$$

Answer 2

By the residue theorem or the Cauchy integral formula the answer is clearly zero, but you don't even need such results to be able to state it. Since $f$ is entire, $$ f(z_0+z) = f(z_0) + f'(z_0)(z+z_0)+ f''(z_0)\frac{(z+z_0)^2}{2!}+\ldots $$ holds uniformly over any compact subset of the complex plane. It follows that we are allowed to replace $z$ with $r e^{i\theta}$, multiply both sides by $e^{i\theta}$ and perform a termwise integration over $(0,2\pi)$ with respect to $d\theta$. Since for any $n\in\mathbb{N}^+$ we have $$\int_{0}^{2\pi} e^{ni\theta}\,d\theta = 0, $$ the conclusion is straightforward.