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Contour radical integral

Consider the branch of \displaystyle f(z) =\sqrt{z^2-1} which is defined outside the segment [-1, 1] and which coincides with the positive square root \sqrt{x^2-1} for x>1. Let R>1 then evaluate the contour integral:

    \[\ointctrclockwise \limits_{\left | z\right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}}\]

Solution

It is a classic case of residue at infinity. Subbing z \mapsto \frac{1}{z} the counterclockwise contour integral rotates the northern pole of the Riemannian sphere to the southern one and the contour integral is transformed to a clockwise one. Hence:

    \begin{align*} \oint \limits_{\left | z \right |=R}\frac{{\rm d}z}{\sqrt{z^2-1}} &=-2\pi i \underset{z=\infty}{\mathfrak{Res}}\frac{1}{\sqrt{z^2-1}} \\ &=-2\pi i \underset{w=0}{\mathfrak{Res}} \frac{-1}{w^2\sqrt{w^{-2}-1}} \\ &=2\pi i \underset{w=0}{\mathfrak{Res}}\frac{1}{w\sqrt{1-w^2}} \\ &= 2\pi i \lim_{w\rightarrow 0}\frac{1}{\sqrt{1-w^2}}\\ &=2\pi i \end{align*}

The equality w\sqrt{w^{-2}-1}=\sqrt{1-w^2}does hold for all |w|<1 if we take the standard branch \sqrt{1-w^2}=\exp \left ( \frac{1}{2}\mathrm{Log} \left ( 1-w^2 \right ) \right ) , otherwise it is not that obvious why this holds, since we are dealing with a multi-valued function.

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Upper bound of max product

Let z_1, z_2 , \dots, z_n \in \mathbb{C} be the roots of the polynomial

    \[f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \mathbb{C}[x]\]

Prove that:

    \[\prod_{k = 1}^{n} \max (1,|z_k|) \leq \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2}\]

Solution

Suppose the roots of polynomial f are z_1, z_2, \dots, z_n where

    \[|z_1| \geq |z_2| \geq \cdots \geq |z_m| > 1 \geq |z_{m+1}| \geq \cdots \geq |z_n|\]

Let g(z)=z^nf \left(\frac{1}{z}\right) = 1 + a_{n-1}z + \cdots + a_0z^n. Then, the \{1/z_k\}_{1\leq k \leq m} are the zeros of g in the disk |z| \le r = 1-\epsilon < 1 where \epsilon is chosen such that g(re^{i\theta}) \neq 0 for \theta \in [0, 2\pi].

Jensen’s inequality implies that

    \begin{align*} \log|r^m z_1 \cdots z_m| = \frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d} \theta & \Leftrightarrow \\ |r^m z_1\cdots z_m| = \exp \left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d}\theta\right) &\Leftrightarrow \\ \exp \left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d}\theta\right) \le \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,\mathrm{d}\theta \end{align*}

Applying Cauchy – Schwartz yields,

    \begin{align*} \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,\mathrm{d}\theta &\leq \frac{1}{2\pi}\left(\int_0^{2\pi}\,\mathrm{d}\theta\int_0^{2\pi} |g(re^{i\theta})|^2\,\mathrm{d}\theta\right)^{1/2} \\ &= \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2} \end{align*}

Therefore,

    \[\left|r^m z_1\cdots z_m\right| \leq \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2}\]

Letting r\rightarrow 1 and \epsilon \rightarrow 0 we get the result.

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Contour integral

Let f be analytic in the disk |z|<2. Prove that:

    \[\frac{1}{2\pi i} \oint \limits_{\left | z \right |=1} \frac{\overline{f(z)}}{z-\alpha} \, \mathrm{d}z = \left\{\begin{matrix} \overline{f(0)} & , & \left | \alpha \right |<1 \\\\ \overline{f(0)} - \overline{f\left ( \frac{1}{\bar{\alpha}} \right )} & , & \left | \alpha \right |>1 \end{matrix}\right.\]

Solution

It follows from Taylor that f(z)=\sum \limits_{n=0}^{\infty} c_n z^n and the convergence is uniform. Hence,

    \begin{align*} \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &=\frac{1}{2\pi i }\oint \limits_{|z|=1} \sum_{n=0}^{\infty} \frac{\overline{c_n} \bar{z}^n}{z-\alpha} \,\mathrm{d}z \\ &= \sum_{n=0}^{\infty} \frac{\overline{c_n}}{2\pi i }\oint \limits_{|z|=1}\frac{ \bar{z}^n}{z-\alpha} \,\mathrm{d}z \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{\left | z \right |=1\Rightarrow \bar{z}^n = \frac{1}{z^n}}{=\! =\! =\! =\! =\! =\! =\!=\! =\!=\!}\sum_{n=0}^{\infty} \frac{\overline{c_n}}{2\pi i }\oint \limits_{|z|=1}\frac{1}{z^n(z-\alpha)} \,\mathrm{d}z \end{align*}

We have that \displaystyle \mathrm{Res}\left( \frac{1}{z^n(z-\alpha)};\alpha\right) = \frac{1}{\alpha^n} and for n \geq 1 we also have that \displaystyle \displaystyle \mathrm{Res}\left( \frac{1}{z^n(z-\alpha)};0\right) = -\frac{1}{\alpha^n} due to

    \[\frac{1}{z^n(z-\alpha)} = -\frac{1}{\alpha z^n} \frac{1}{1-z/\alpha} = -\frac{1}{\alpha z^n}\left(1 + \frac{z}{\alpha} + \frac{z^2}{\alpha^2} + \cdots \right)\]

So if |\alpha|<1 then \alpha lies within the disk |z|=1; hence the integral equals \overline{c_0} = \overline{f(0)} whereas if |\alpha|>1 then \alpha lies outside the disk |z|=1; hence

    \begin{align*} \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &= -\sum_{n=1}^{\infty} \frac{\overline{c_n}}{\alpha^n} \\ &= \overline{c_n} - \sum_{n=0}^{\infty} \frac{\overline{c_n}}{\alpha^n} \\ &= \overline{f(0)} - \overline{\left ( \sum_{n=0}^{\infty} \frac{c_n}{\bar{\alpha}^n} \right )}\\ &= \overline{f(0)} - \overline{f\left ( \frac{1}{\bar{\alpha}} \right )} \end{align*}

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Integral of Jacobi Theta function

Let \vartheta_4(z;q) denote one of the Jacobi Theta functions. Prove that

    \[\int_{0}^{1}\vartheta_4\left ( 0;q \right ) \, \mathrm{d}q = \frac{\pi}{\sinh \pi}\]

Solution

We have successively,

    \begin{align*} \int_{0}^{1} \vartheta_4\left ( 0;q \right )\, \mathrm{d}q &= \int_{0}^{1} \sum_{n=-\infty}^{\infty} (-1)^n q^{n^2} \, \mathrm{d}q \\ &= \sum_{n=-\infty}^{\infty} (-1)^n \int_{0}^{1} q^{n^2} \, \mathrm{d}q\\ &=\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} \\ &= \frac{\pi}{\sinh \pi} \end{align*}

The sum is evaluated as follows. Consider the function

    \[f(z) = \frac{\pi \csc \pi z}{z^2+1}\]

and integrate it around a square \Gamma_N with vertices \left ( N+\frac{1}{2} \right )\left ( \pm 1\pm i \right ). The function f has poles at every integer z=n with residue \frac{(-1)^n}{n^2+1} as well as at z=\pm i with residues -\frac{\pi}{2 \sinh \pi}. We also note that as N \rightarrow +\infty the contour integral of f tends to 0. Thus,

    \begin{align*} \frac{1}{2\pi i}\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z &= \sum_{n=-N}^{N} \mathfrak{Res}_{z=n} \frac{\pi \csc \pi z}{z^2+1} + \mathfrak{Res}\left ( f ; i \right ) + \mathfrak{Res}\left ( f;-i \right ) \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{2\sinh \pi} -\frac{\pi}{2 \sinh \pi} \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \end{align*}

Hence,

    \begin{align*} 0 &=\lim_{N \rightarrow +\infty} \oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z \\ &= \lim_{N \rightarrow +\infty} \left ( \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \right )\\ &= \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} - \frac{\pi}{ \sinh \pi} \end{align*}

and the exercise is complete.

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Harmonic function

Consider the points O(0, 0) and  A(1, 0). Let \Gamma(x, y) be a point of the plane such that y>0. Set \varphi(x, y) to be the angle that is defined by O\Gamma and A \Gamma. ( the one that is less than \pi.) Prove that the function \varphi(x, y) is harmonic.

Solution [by Demetres Skouteris]

The complex function \log \left( 1 - \frac{1}{z} \right) has a holomorphic branch in the half plane \mathfrak{Im}(z)>0 and its imaginary part is the desired angle. Hence, the function of the angle is harmonic.

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