Problem of the Week #271 - May 29, 2018

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Euge

MHB Global Moderator
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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

MHB Global Moderator
Staff member
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

MHB Global Moderator
Staff member
This week's problem was solved correctly by Opalg . You can read his solution below.

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.

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