0
$\begingroup$

For $f$ be analytic on the disc $\overline{D}(0,R)$ centred at $0$ with radius $R>0$ and such that $f(0)\neq 0$, then the following formula is well-known

\begin{align} \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-i\theta}\log(|f(Re^{i\theta})|)d\theta=\frac{1}{2}R\frac{f'(0)}{f(0)}+\frac{1}{2}\sum_{\rho}\left(\frac{R}{\rho}-\frac{\overline{\rho}}{R} \right) \end{align}

where $\rho$ are the roots in $D(0,R)$

I have been stuck trying to calculate the following integral where the above formula cannot be applied (directly). Suppose $z_0 \in \mathbb{C}$ and $M=|z_0|+\epsilon$, where $\epsilon$ is as usual positive and small. What is then the value of the following integral

\begin{align} \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-i\theta}\log(|z_0+2M \Im(z_0 e^{-i \theta})-Me^{i\theta}|)d\theta \end{align}

I would be glad if anyone has suggestions, or can give some advice on reference

$\endgroup$
4
  • $\begingroup$ What is the “well known formula” in the earlier case? $\endgroup$ Commented Nov 29, 2023 at 17:19
  • $\begingroup$ \begin{align*} \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-i\theta}\log(|f(Re^{i\theta})|)d\theta=\frac{1}{2}R\frac{f'(0)}{f(0)}+\frac{1}{2}\sum_{\rho}\left(\frac{R}{\rho}-\frac{\overline{\rho}}{R} \right) \end{align*} where $\rho$ are the roots in $D(0,R)$ I edit the post $\endgroup$
    – 12321
    Commented Nov 29, 2023 at 17:25
  • $\begingroup$ with $w=e^{i\theta}$ use that $\Im(z_0 \bar w)=\frac{z_0/w-\bar z_0 w}{2i}$ and $\log |w|=0$ to transform the log integrand to $\log |z_0w-Miz_0+Mi\bar z_0 w^2-Mw^2|$ and the inside function $f(w)=z_0w-Miz_0+Mi\bar z_0 w^2-Mw^2$ is a quadratic so clearly analytic so you can apply the result above with $R=1$ and need find the roots of the quadratic inside the unit disc... $\endgroup$
    – Conrad
    Commented Nov 30, 2023 at 3:03
  • $\begingroup$ @Conrad thank you for the comment! $\endgroup$
    – 12321
    Commented Nov 30, 2023 at 10:33

0

You must log in to answer this question.

Browse other questions tagged .