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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.

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    $\begingroup$ 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
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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 $$0=\int_{C^+(a,r)}f(z)\,dz=\int_0^{2\pi}f(a+re^{i\theta})ire^{i\theta}\,d\theta.$$

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