I have tried but did not solve the following problem (Exercise $16.3$) from Apostol's Mathematical Analysis Book:

Let $f=u+i v$ be analytic on a disk $B(a;r)$. If $0<r<R$, prove that $$f^\prime (a)=\frac{1}{\pi r}\int\limits_0^{2\pi}{u(a+re^{i\theta})e^{-i\theta}} d\theta.$$

Here is my attempt: By the Cauchy integral formula $$f^\prime(a)=\frac{1}{2\pi i}\int\limits_{C_r}{\frac{f(z)}{(z-a)^2}}dz,$$ where $C_r$ is the circle $|z-a|=r$. Also since $\overline{f(z)}$ is analytic (with the help of Cauchy-Riemann equations) on $B(a;r)$ with $\overline{f(z)}^\prime=\overline{f^\prime(z)}$, again by the Cauchy integral formula we have $$\overline{f^\prime(a)}=\frac{1}{2\pi i}\int\limits_{C_r}{\frac{\overline{f(z)}}{(z-a)^2}}dz.$$ Switching to polar coordinates and adding these two, we have $$\frac{1}{\pi r}\int\limits_0^{2\pi}{u(a+re^{i\theta})e^{-i\theta}} d\theta=f^\prime(a)+\overline{f^\prime(a)}.$$ In the Right Hand Side I obtain an extra factor $\overline{f^\prime(a)}$, actually the RHS equals to $2u_x(a)$. Did I misunderstood something ? Help is appreciated.

  • 1
    $\begingroup$ Be careful. When you write $\overline{f(z)}$, you need $\overline{f(\bar z)}$ if you want an analytic function. So whereas $f'(a)+\overline{f'(a)}$ is twice the real part of $f'(a)$, this is not what your approach is calculating :) $\endgroup$ – Ted Shifrin Jan 10 '15 at 20:13
  • $\begingroup$ Thanks @Ted Shifrin, I understood what I did wrong! Since $\int f(z) dz=0$, by transforming this to polar coordinates it is easy to observe that $$\int\limits_0^{2 \pi}(u(.)\cos\theta-v(.)\sin\theta)d\theta=0=\int\limits_0^{2 \pi}(u(.)\sin\theta+v(.)\cos\theta)d\theta.$$ Using this and equating the real and imaginary parts of $f^\prime(a)$, it follows that $$u_x(a)=\frac{1}{\pi r}\int\limits_0^{2 \pi}{u(.)\cos\theta}$$ and hence $$\frac{1}{\pi r}\int {u(.)e^{-i\theta}d\theta}$$ $$=u_x(a)+iv_x(a)-\frac{i}{\pi r}\int(u(.)\sin\theta+v(.)\cos\theta)d\theta=f^\prime(a)-0=f^\prime(a).$$ $\endgroup$ – user149418 Jan 10 '15 at 22:21

I have a similar answer to yours. By the Cauchy's integral formula, $$\begin{align*} f'(a)= {} & \frac{1}{2\pi i}\int_{C^+(a,r)}\frac{f(\omega)}{(\omega-a)^2}\,d\omega=\frac{1}{2\pi r}\int_0^{2\pi}f(a+re^{i\theta})e^{-i\theta}\,d\theta \\ = {} & \frac{1}{2\pi r}\int_0^{2\pi}u(a+re^{i\theta})e^{-i\theta}\,d\theta+\frac{i}{2\pi r}\int_0^{2\pi}v(a+re^{i\theta})e^{-i\theta}\,d\theta. \end{align*}$$ Then we need to show that $$ \frac{i}{2\pi r}\int_0^{2\pi}v(a+re^{i\theta})e^{-i\theta}\,d\theta=\frac{1}{2\pi r}\int_0^{2\pi}u(a+re^{i\theta})e^{-i\theta}\,d\theta,$$ that is, $$\int_0^{2\pi}\overline{f(a+re^{i\theta})}e^{-i\theta}\,d\theta=0,$$ which is equivalent to $$\int_0^{2\pi}f(a+re^{i\theta})e^{i\theta}\,d\theta=0.$$ And this is clear because $$0=\int_{C^+(a,r)}f(z)\,dz=\int_0^{2\pi}f(a+re^{i\theta})ire^{i\theta}\,d\theta.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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