Poisson's integral formula for $f'(re^{i\theta})$ Would you please help me investigate Problem 72 in Chapter 5, "Cauchy's Integral Formulas and Related Theorems," of Complex Variables, Murray R. Spiegel, Schaum's Outline Series:

If $f(z)$ is analytic inside and on the circle $C$ defined by $|z| = R$, and if $z=re^{i\theta}$ is any point inside $C,$ show that
  $$f'(re^{i\theta}) = \frac i{2\pi} \int_0^{2\pi} \frac{R(R^2 - r^2)f(Re^{i\phi})\sin(\theta - \phi)}{(R^2 - 2Rr\cos(\theta - \phi) + r^2)^2}d\phi.$$

Specifically,


*

*Is the formula above false?

*If so, is there a typo in the formula?

*Again, if so, is my derivation below for $f'(re^{i\theta})$ correct?

*If so, is there a way to express my integrand to get a $\sin(\theta - \phi)$ factor in the numerator?  (I factored $R^2 - r^2$ out of the numerator, but that did not seem to lead to such an expression.)


Mark Viola's answer raises more questions:


*

*Is there a way to express Mark's integrand to get a $\sin(\theta - \phi)$ factor in the numerator?

*Assuming that Mark's and my formulas are correct (if not, find a flaw in one of the derivations or a counterexample), is there a way to show that they are equivalent without appealing to the fact that they are both derived correctly?


I think that Spiegel's formula is false because if I put $f(z) = z$, then $f'(z) = 1 = f'(0)$.  But if I put $r = 0, \theta = 0, R = 1, f(Re^{i\phi}) = e^{i\phi}$ in the formula above, I get $f'(0) = 1/2 \ne 1$.
My derivation for $f'(re^{i\theta})$ mimics the proof of Poisson's integral formula but starts with Cauchy's integral formula for the first derivative:
Because $z = re^{i\theta}$ is any point inside $C$, we have by Cauchy's integral formula
$$f'(z) = f'(re^{i\theta}) = \frac 1{2\pi i} \oint_C \frac{f(w)}{(w-z)^2}dw.$$
The inverse of the point $z$ with respect to $C$ is given by $R^2/\bar z$ and  lies outside $C$, so $f(w)/(w-R^2/\bar z)^2$ is analytic inside $C$.  Hence, by Cauchy's theorem,
$$0 = \frac 1{2\pi i} \oint_C \frac{f(w)}{(w-R^2/\bar z)^2}dw.$$
Subtracting those two equations, we find that
\begin{align}
f'(z) & = \frac 1{2\pi i} \oint_C \left( \frac 1{(w-z)^2} - \frac 1{(w-R^2/\bar z)^2} \right) f(w) dw\\
& = \frac 1{2\pi i} \oint_C \frac {R^4/\bar z^2 - 2R^2w/\bar z + 2wz - z^2}{(w - z)^2 (w - R^2/\bar z)^2} f(w) dw.
\end{align}
Now let $z = re^{i\theta}$ so that $1/\bar z = e^{i\theta}/r$, and $w = Re^{i\phi}$ so that $dw = Rie^{i\phi}d\phi$.  Then
\begin{align}
f'(re^{i\theta}) & = \frac 1{2\pi i} \int_0^{2\pi} \frac {R^4e^{2i\theta}/r^2 - 2R^2Re^{i\phi}e^{i\theta}/r + 2Re^{i\phi}re^{i\theta} - r^2e^{2i\theta}}{(Re^{i\phi} - re^{i\theta})^2 (Re^{i\phi} - R^2e^{i\theta}/r)^2} f(Re^{i\phi}) Rie^{i\phi}d\phi \cdot \frac{r^2}{r^2}\\
& = \frac 1{2\pi} \int_0^{2\pi} \frac {R^4e^{2i\theta} - 2R^3re^{i\phi}e^{i\theta} + 2Rr^3e^{i\phi}e^{i\theta} - r^4e^{2i\theta}}{(Re^{i\phi} - re^{i\theta})^2(re^{i\phi} - Re^{i\theta})^2 R} f(Re^{i\phi}) e^{i\phi}d\phi \cdot \frac{e^{-2i\theta}e^{-2i\phi}}{(e^{-i\theta}e^{-i\phi})^2}\\
& = \frac 1{2\pi R} \int_0^{2\pi} \frac {R^4e^{-i\phi} - 2R^3re^{-i\theta} + 2Rr^3e^{-i\theta} - r^4e^{-i\phi}}{(Re^{i\phi} - re^{i\theta})^2 (re^{-i\theta} - Re^{-i\phi})^2} f(Re^{i\phi}) d\phi\\
& = \frac 1{2\pi R} \int_0^{2\pi} \frac {(R^4 - r ^4)e^{-i\phi} - 2Rr(R^2 - r^2)e^{-i\theta}}{(Re^{i\phi} - re^{i\theta})^2 (Re^{-i\phi} - re^{-i\theta})^2} f(Re^{i\phi}) d\phi\\
& = \frac 1{2\pi R} \int_0^{2\pi} \frac {(R^4 - r ^4)e^{-i\phi} - 2Rr(R^2 - r^2)e^{-i\theta}}{(R^2 - 2Rr\cos(\theta - \phi) + r^2)^2} f(Re^{i\phi}) d\phi
\end{align}
I verified that both my and Mark's formulas yield correct values for a few simple functions (e.g., polynomials) $f$ and values for $R, r, \theta$.  When I test complicated functions (e.g., logarithm), Mark's and my formulas yield values that are different and both close to, but not equal to, the correct value; I suspect that those discrepancies are due to limitations in the numerical integration program I used rather than the formulas' being incorrect.  
 A: The formula as quoted from Spiegel's text is incorrect as discussed in the OP.  The correct formula can be obtained by using Cauchy's Integral Formula
$$f'(z)=\frac1{2\pi i}\oint_{|z|=R}\frac{f(z')}{(z'-z)^2}\,dz'\tag 1$$
for $|z|<R$.  

Let $z$ be expressed in polar coordinates, $(r,\theta)$, in $(1)$ as $z=re^{i\theta}$.  
Then, with the parameterization $z'=Re^{i\phi}$ with $dz'=iRe^{i\phi}\,d\phi$, we have 
$$\begin{align}
f'(re^{i\theta})&=\frac1{2\pi i}\oint_{|z|=R}\frac{f(z')}{(z'-re^{i\theta})^2}\,dz'\\\\
&=\frac1{2\pi i}\int_0^{2\pi}\frac{iRe^{i\phi}f(Re^{i\phi})}{(Re^{i\phi}-re^{i\theta})^2}\,d\phi\tag2
\end{align}$$
Next, multiplying the numerator and denominator of the integrand of $(2)$ by the complex conjugate of the denominator reveals
$$\begin{align}
f'(re^{i\theta})&=\frac R{2\pi }\int_0^{2\pi}\frac{e^{i\phi}(Re^{-i\phi}-re^{-i\theta})^2f(Re^{i\phi})}{|Re^{i\phi}-re^{i\theta}|^4}\,d\phi\\\\
&=\frac R{2\pi }\int_0^{2\pi}\frac{e^{i\phi}(Re^{-i\phi}-re^{-i\theta})^2f(Re^{i\phi})}{(R^2+r^2-2rR\cos(\theta-\phi))^2}\,d\phi\\\\
&=\frac R{2\pi }\int_0^{2\pi}\frac{(R^2e^{-i\phi}-2rRe^{-i\theta}+r^2e^{-i(2\theta-\phi)})f(Re^{i\phi})}{(R^2+r^2-2rR\cos(\theta-\phi))^2}\,d\phi\tag3
\end{align}$$
Note that $(3)$ does not match the formula given in the OP.
