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Let $f=u+iv$ be an analytic function in disk $\mathbb{D}$ and $0<r<1$. Can you help me to prove that

$$\pi{r}f'(0)=\int_{0}^{2\pi}\frac{u(re^{i\theta})}{e^{i\theta}}d\theta\;\;\;?$$

I tried with the Cauchy integral formula, but unsuccessfully.

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  • $\begingroup$ is $u$ function of (x,y) or what? $\endgroup$ – Salihcyilmaz Sep 8 '15 at 18:33
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An idea, (but perhaps not the best answer, there is some computations). Put $f(z)=\sum a_k z^k$, and $a_k=u_k+iv_k$ with $u_k, v_k \in \mathbb{R}$. Then $$u(r\exp(i\theta))=\sum (u_kr^k \cos(k\theta)-v_kr^k\sin(k\theta))$$

And these series of functions are normally convergent. Note that as the functions are periodic with period $2\pi$:$$\int_0^{2\pi}u(r\exp(i\theta))\exp(-i\theta)d\theta=\int_{-\pi}^{+\pi}u(r\exp(i\theta))\exp(-i\theta)d\theta$$ and that $$\int_{-\pi}^{+\pi}\cos(k\theta)\sin(\theta)d\theta= \int_{-\pi}^{+\pi}\sin(k\theta)\cos(\theta)d\theta=0$$ for all $k$ (odd functions) and $$\int_{-\pi}^{+\pi}\cos(k\theta)\cos(\theta)d\theta= \int_{-\pi}^{+\pi}\sin(k\theta)\sin(\theta)d\theta=0$$ if $k\not =1$, and their value is $\pi$ if $k=1$.

This gives $$\int_{-\pi}^{+\pi}u(r\exp(i\theta))\exp(-i\theta)d\theta=\pi r u_1+i\pi rv_1=\pi r a_1=\pi r f^{\prime}(0)$$

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By the Cauchy formula for derivatives: $$\int_0^{2\pi}\frac{f(re^{i\theta})}{e^{i\theta}}d\theta=\int_0^{2\pi}\frac{f(re^{i\theta})}{r^2e^{2i\theta}}\frac{rd(re^{i\theta})}{i}=2r\pi\frac{1!}{2i\pi}\int_{|z|=r}\frac{u(z)+iv(z)}{z^2}dz=2r\pi f'(0)$$

Now, $$\int_{|z|=r}\frac{u-iv}{z^2}dz=\int_{|z|=r}\frac{\bar{f}(z)}{z^2}dz=0$$ as is easily seen, e.g. considering the series $$\bar{f}(z)=\sum_{n}a_n \bar{z}^n$$ and the fact that $$\int_{|z|=r}z^a\bar{z}^bdz=0$$ unless $a=b$.

NB The Cauchy formula for derivatives is $$f^{(n)}(w)=\frac{n!}{2i\pi}\int_{|z-w|=r}\frac{f(z)}{(z-w)^{n+1}}dz$$

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  • $\begingroup$ @wisefool you're replaced $u=\operatorname{Re}f$ and $f$. $\endgroup$ – user195664 Nov 26 '14 at 16:39
  • $\begingroup$ Oh, sorry, I mixed up things ... I'll fix it. $\endgroup$ – wisefool Nov 26 '14 at 17:01
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You can take the power series of $f$ and integrate termwise.

If $f(z)=\sum\limits_{n=0}^\infty a_nz^n$ then $$ u(z)=\frac12\left(\sum_{n=0}^\infty a_nz^n+\sum_{n=0}^\infty \overline{a_n}\cdot \overline{z}^n \right) $$ so $$ \int_0^{2\pi} u(re^{it})e^{-it}dt = \int_0^{2\pi} \left(\sum_{n=0}^\infty \frac{ a_n r^ne^{nit} + \overline{a_n}r^ne^{-nit}}2 \right) e^{-it}dt = \\ = \sum_{n=0}^\infty \frac{a_nr^n}2 \int_0^{2\pi}e^{(n-1)it}dt + \sum_{n=0}^\infty \frac{\overline{a_n}r^n}2 \int_0^{2\pi}e^{(-n-1)it}dt= \\ = \sum_{n=0}^\infty \frac{a_nr^n}2 \begin{Bmatrix} 2\pi & \text{if $n-1=0$} \\ 0 & \text{if $n-1\ne0$} \\ \end{Bmatrix} + \sum_{n=0}^\infty a_n r^n \begin{Bmatrix} 2\pi & \text{if $-n-1=0$} \\ 0 & \text{if $-n-1\ne0$} \\ \end{Bmatrix} = \\ = \frac{a_1r}2 \cdot 2\pi = \pi r a_1 = \pi r f'(0). $$

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Here is a slightly more real-analytic proof: by the Cauchy-Riemann equations $$ f'=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}$$ and since $f'$ is harmonic (i.e. its value at any point equals the mean of its values on any ball centered at that point) we have $$f'(0)=\frac{1}{\pi r^2}\int_{\mathbb{D}_r}f'(x+iy)\,dx\,dy=\frac{1}{\pi r^2}\int_{\mathbb{D_r}}\left(\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}\right)=\frac{1}{\pi r^2}\int_{\partial\mathbb{D}_r}u(\nu_x-i\nu_y).$$ Here $\mathbb{D}_r=\{|z|<r\}$ and $\nu=(\nu_x,\nu_y)$ is the outward unit normal. In the last equality we used the divergence theorem. But if we parametrize $\partial\mathbb{D}_r$ by $\theta\mapsto re^{i\theta}$ (as usual) we notice that $\nu=e^{i\theta}$, so that $\nu_x-i\nu_y=\overline{\nu}=e^{-i\theta}$ and we finally get $$f'(0)=\frac{1}{\pi r}\int_0^{2\pi}u(re^{i\theta})e^{-i\theta}\,d\theta.$$

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Maybe, next perform is correct. Let $g(z)=\overline{f(\overline{z})}$. Than $g(z)$ is analytic in $\mathbb{D}$. By Cauchy integral formula we have

$$f'(0)=\frac{1}{2\pi{i}}\int_{\partial D(0,r)}\frac{f(z)}{z^2}dz=\frac{1}{2\pi{i}}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{r^2e^{i2\theta}}re^{i\theta}id\theta=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{re^{i\theta}}d\theta$$.

On the other hand we have

$$\frac{1}{\pi{r}}\int_{0}^{2\pi}\frac{u(re^{i\theta})}{e^{i\theta}}d\theta=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{re^{i\theta}}d\theta+\frac{1}{2\pi}\int_{0}^{2\pi}\frac{\overline{f(re^{i\theta})}}{re^{i\theta}}d\theta=$$

$$=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{re^{i\theta}}d\theta+\frac{1}{2\pi}\int_{0}^{2\pi}\frac{g(re^{-i\theta})}{re^{i\theta}}d\theta=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{re^{i\theta}}d\theta+\frac{1}{2\pi{r^2}i}\int_{\partial D(0,r)}g(z)dz=$$

$$=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(re^{i\theta})}{re^{i\theta}}d\theta=f'(0).$$

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  • 1
    $\begingroup$ Use the space bar more often. Double dollar signs opening and closing expressions will center them and will make them bigger. $\endgroup$ – Timbuc Nov 26 '14 at 18:34

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