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.
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.