Evaluating $\int_0^{2\pi} e^{2i\theta} f(e^{i\theta})\,d\theta$ using Cauchy's Theorem

Question

If $f(z)$ is analytic on the disk $|z| \leq 2$, evaluate $$\int_0^{2\pi} e^{2i\theta} f(e^{i\theta})\,d\theta.$$

This problem comes from Mathematical Methods in the Physical Sciences 2nd edition by Mary L. Boas in the section on contour integrals covering Cauchy's Theorem and Cauchy's Integral Formula. The previous question in the section is more straightforward, asking for the evaluation of $\int_0^{2\pi} e^{i\theta} f(e^{i\theta})\,dx$ where $f(z)$ is analytic on and inside the circle $|z| \le 1$. The integral is $0$ by applying Cauchy's Theorem and $z = e^{i\theta}$, $dz = ie^{i\theta}d\theta$. However, the extra $2$ is throwing me off in the question above, and I am not sure where to start here. How does one evaluate this integral?

Answer 1

Substitute

$$z=2e^{it}\;,\;\;0\le t\le 2\pi\implies dz=2izdt\implies dt=\frac{dz}{2iz}\implies$$

$$\int_0^{2\pi}e^{2i}f(e^{it})dt=\oint_{|z|=2}\frac{z^2}4f\left(\frac z2\right)\frac{dz}{2iz}=\frac1{8i}\oint_{|z|=2}zf\left(\frac z2\right)\,dz=0$$

since $\;zf\left(\frac z2\right)\;$ is analytic on $\;|z|\le2\;$

Answer 2

You have $$ e^{2i\theta}\,f(e^{i\theta})=e^{i\theta}\,e^{i\theta}\,f(e^{i\theta}), $$ so you will apply Cauchy's Theorem to the function $g(t)=tf(t)$.