How to prove $ \iint_{\mathbb D}\frac{1}{\left|1-\bar{z}\zeta\right|^4}\mathrm dx\mathrm dy = \frac{\pi}{(1-|\zeta|^2)^2} $?

Question

I'm stuck trying to prove the following identity, given $\zeta \in \mathbb{D}$, where $\mathbb{D} \subset \mathbb{C}$ is the unit disk. Here, $z = x +iy$.

$$ \iint_{\mathbb{D}}\frac{1}{\left|1-\bar{z}\zeta\right|^4}\mathrm dx\mathrm dy = \frac{\pi}{\left(1-|\zeta|^2\right)^2} $$

I was hoping just to expand in $x,y$ and then convert to polar, but things got real nasty and it wasn't working out. I'm now convinced there's some simple trick that I'm missing.

Answer 1

Let $a = \left|{\zeta}\right|$. We have

\begin{equation}I = \int_{\mathbb{D}}^{}\frac{1}{{\left|1-\overline{z} a\right|}^{4}} d x d y = \int_{\mathbb{D}}^{}\frac{1}{{\left({\left(1-a x\right)}^{2}+{\left(a y\right)}^{2}\right)}^{2}} d x d y\end{equation}

Using polar coordinates gives

\begin{equation}I = \int_{0}^{1}r d r \left(\int_{-\pi}^{{\pi}}\frac{1}{{\left(1-2 a r \cos {\theta}+{a}^{2} {r}^{2}\right)}^{2}} d {\theta}\right) \equiv \int_{0}^{1}J \left(a r\right) r d r\end{equation}

Now using the substitution $t = \tan \left({\theta}/2\right)$, one gets

\begin{equation}\renewcommand{\arraystretch}{2.0} \begin{array}{rcl}J \left(k\right)&=&\displaystyle \int_{-\infty}^{\infty }\frac{2 \left(1+{t}^{2}\right)}{{\left(\left(1+{k}^{2}\right) \left(1+{t}^{2}\right)-2 k \left(1-{t}^{2}\right)\right)}^{2}} d t\\ &=&\displaystyle \int_{-\infty}^{\infty }\frac{2 \left(1+{t}^{2}\right)}{{\left({\left(1+k\right)}^{2} {t}^{2}+{\left(1-k\right)}^{2}\right)}^{2}} d t \end{array}\end{equation}

Now using the substitution $t = u \left(1-k\right)/\left(1+k\right)$, this leads to

\begin{equation}\renewcommand{\arraystretch}{2.0} \begin{array}{rcl}J \left(k\right)&=&\displaystyle \frac{2}{{\left(1-k\right)}^{3} {\left(1+k\right)}^{3}} \int_{-\infty}^{\infty }\frac{{\left(1+k\right)}^{2}+{u}^{2} {\left(1-k\right)}^{2}}{{\left(1+{u}^{2}\right)}^{2}} d u\\ &=&\displaystyle \frac{2}{\left(1-k\right) {\left(1+k\right)}^{3}} \int_{-\infty}^{\infty }\frac{1}{1+{u}^{2}} d u+\frac{8 k}{{\left(1-k\right)}^{3} {\left(1+k\right)}^{3}} \int_{-\infty}^{\infty }\frac{1}{{\left(1+{u}^{2}\right)}^{2}} d u\\ &=&\displaystyle \frac{1+{k}^{2}}{{\left(1-{k}^{2}\right)}^{3}} {2 \pi} \end{array}\end{equation}

Hence

\begin{equation}\renewcommand{\arraystretch}{2.0} \begin{array}{rcl}I&=&\displaystyle \frac{{2\pi}}{{a}^{2}} \int_{0}^{a}\frac{1+{k}^{2}}{{\left(1-{k}^{2}\right)}^{3}} k d k\\ &=&\displaystyle \frac{{2\pi}}{{a}^{2}} {\left[\frac{{k}^{2}}{2 {\left(1-{k}^{2}\right)}^{2}}\right]}_{0}^{a}\\ &=&\displaystyle \frac{{\pi}}{{\left(1-{a}^{2}\right)}^{2}} \end{array}\end{equation}

Answer 2

Though you can use polar coordinates (see below) it is easier to use Stokes’ theorem. Let the $2$-form $\omega$ be given by

$$\omega = \frac{\mathrm{d}x \wedge \mathrm{d}y}{\lvert 1 - z \overline{\zeta} \rvert^4} = \frac1{2 \mathrm{i}} \frac{\mathrm{d} \overline{z} \wedge \mathrm{d}z}{(1-z \overline{\zeta})^2 (1 - \overline{z} \zeta)^2}$$ and the $1$-form $\phi$ by $$ \phi = \frac1{2 \mathrm{i}} \frac{\mathrm{d}z}{\zeta (1 - z \overline{\zeta})^2 (1 - \overline{z} \zeta)}.$$

Then $\mathrm{d}\phi = \omega$ and by Stokes

$$\iint_{\mathbb{D}} \omega = \iint_{\mathbb{D}} \mathrm{d}\phi = \int_{\partial \mathbb{D}} \phi = \frac1{2 \mathrm{i}} \int_{\lvert z \rvert = 1} \frac{\mathrm{d}z}{\zeta (1 - z \overline{\zeta})^2 (1 - \overline{z} \zeta)}.$$ Since $\lvert z \rvert=1$ the last integral can be rewritten as $$ \frac1{2 \mathrm{i}} \int_{\lvert z \rvert = 1} \frac{z \mathrm{d}z}{\zeta (1 - z \overline{\zeta})^2 (z - \zeta)} = \frac{\pi}{(1-\lvert \zeta \rvert^2)^2}.$$

Alternative: In polar coordinates: $$\int_0^1 r \int_0^{2 \pi} \frac{\mathrm{d}t}{(1-r e^{-\mathrm{i} t} \zeta)^2 (1- r e^{\mathrm{i} t} \overline{\zeta})^2} \mathrm{d} r.$$ Substitute $z \leftarrow r e^{\mathrm{i}t}$: $$\int_0^1 \frac{r}{\mathrm{i}}\int_{\lvert z \rvert = r} \frac{z \mathrm{d}z}{(z-r^2\zeta)^2 (1- z \overline{\zeta})^2} \mathrm{d} r.$$ The inner integral follows from the residue at the only pole at $r^2 \zeta$: $$\int_0^1 2 \pi r \frac{1+ r^2 \lvert \zeta \rvert^2}{(1-r^2 \lvert \zeta \rvert^2)^3}\mathrm{d} r= \frac{\pi}{(1-\lvert \zeta \rvert^2)^2}. $$