What is the Average Value of |dφ/dz|^2 over the Open Unit Disk?

[Euge]

Here is this week's POTW:

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Suppose $a$ is a fixed complex number in the open unit disk $\Bbb D$. Consider the holomorphic mapping $\phi : \Bbb D \to \Bbb D$ given by $\phi(z) := (z - a)/(1 - \bar{a}z)$. Find, with proof, the average value of $\left\lvert\frac{d\phi}{dz}\right\rvert^2$ over $\Bbb D$, i.e., the integral $$\frac{1}{\pi}\iint_{\Bbb D} |\phi'(x + yi)|^2\, dx\, dy$$-----

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[Euge]

I'm giving another week for members to solve this POTW. As a hint, consider writing the integral in polar coordinates, and note that the integrand can be written as an expression of the Poisson kernel.

 

[Euge]

This week's problem was solved correctly by Opalg. You can read his solution below.

Spoiler

By the quotient rule, $\phi'(z) = \dfrac{1-\bar{a}z + \bar{a}(z-a)}{(1-\bar{a}z)^2} = \dfrac{1 - \bar{a}a}{(1-\bar{a}z)^2}$, and therefore $$|\phi'(z)|^2 = \phi'(z)\overline{\phi'(z)} = \frac{(1 - \bar{a}a)^2}{(1-\bar{a}z)^2(1-a\bar{z})^2} = \frac{(1 - \bar{a}a)^2}{(1-2\text{re}(\bar{a}z) + \bar{a}a\bar{z}z)^2}.$$ So we want to evaluate \(\displaystyle \frac1\pi \iint_{\Bbb D}\frac{(1 - \bar{a}a)^2}{(1-2\text{re}(\bar{a}z) + \bar{a}a\bar{z}z)^2}\,dA\), where $A$ denotes area measure. The substitution $z\mapsto e^{i\arg a}z$ leaves the integral unchanged (it just represents a rotation of the disc, which preserves the measure), and has the effect of replacing $a$ by $|a|$. For the rest of the proof I will write $a$ instead of $|a|$, so that $a$ is real, positive, and less than $1$.

Next, write the integral on terms of polar coordinates as $$ \frac{(1 - a^2)^2}\pi \iint_{\Bbb D}\frac{1}{(1-2a\,\text{re}(z) + a^2\bar{z}z)^2}\,dA = \frac{(1 - a^2)^2}\pi \int_0^1\int_0^{2\pi}\frac{1}{(1-2ar\cos\theta + a^2r^2)^2}\,r\,d\theta \,dr.$$ Taking the theta integral first, substitute $w = e^{i\theta}$ to get $$ \begin{aligned}\int_0^{2\pi}\frac{1}{(1-2ar\cos\theta + a^2r^2)^2}\,d\theta &= \oint_{\partial\Bbb D}\frac1{\bigl((1+a^2r^2) - ar(w+w^{-1})\bigr)^2}\frac{dw}{iw} \\ &= \oint_{\partial\Bbb D}\frac{-iw}{(arw^2 - (1+a^2r^2)w + ar)^2}dw \\ &= \oint_{\partial\Bbb D}\frac{-iw}{a^2r^2(w - ar)^2\bigl(w - \frac1{ar}\bigr)^2}dw. \end{aligned} $$ By the Cauchy Integral Formula (and the fact that $ar$ lies inside $\Bbb D$), that last integral is equal to $$ \begin{aligned}\frac{2\pi i}{a^2r^2}\frac d{dw}\biggl(\frac{-iw}{(w - \frac1{ar})^2}\biggr)\bigg|_{w=ar} &= \frac{2\pi}{a^2r^2}\frac{(w - \frac1{ar})^2 - 2ar(w - \frac1{ar})}{(w - \frac1{ar})^4}\bigg|_{w=ar} \\ &= \frac{2\pi}{a^2r^2}\frac{ar - \frac1{ar} - 2ar}{(ar - \frac1{ar})^3} \\ &= \frac{2\pi(1 + a^2r^2)}{(1 - a^2r^2)^3}.\end{aligned}$$ Now plug that into the $r$-integral and then make the substitution $s = a^2r^2$, to get $$ \begin{aligned} \frac1\pi \iint_{\Bbb D}\frac{(1 - \bar{a}a)^2}{(1-2\text{re}(\bar{a}z) + \bar{a}a\bar{z}z)^2}\,dA &= \frac{(1 - a^2)^2}\pi \int_0^1\frac{2\pi r(1+a^2r^2)}{(1 - a^2r^2)^3}dr \\ &= \frac{(1 - a^2)^2}{a^2}\int_0^{a^2}\frac{1+s}{(1-s)^3}ds \\ &= \frac{(1 - a^2)^2}{a^2}\int_0^{a^2}\biggl(\frac2{(1-s)^3} - \frac1{(1-s)^2} \biggr)ds \\ &= \frac{(1 - a^2)^2}{a^2}\biggl[\frac1{(1-s)^2} - \frac1{1-s}\biggr]_0^{a^2} \\ &= \frac{(1 - a^2)^2}{a^2}\frac{a^2}{(1 - a^2)^2} \\ &= \large 1 \quad! \end{aligned}$$

Edit. Euge has pointed out that my solution does not work when $a=0$. To deal with that case, note that the function then becomes $\phi(z) = z$. So $\phi'(z)$ is the constant $1$, and the average value of its square over $\Bbb D$ is also $1$. Thus the result \(\displaystyle \frac{1}{\pi}\iint_{\Bbb D} |\phi'(x + yi)|^2\, dx\, dy = 1\) still holds in that case.