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Consider the integral

$$I=\int_0^{2\pi}\log\left|re^{it}-a\right|\,dt$$

where $a$ is a complex number and $0<r<|a|$. We have

$$I=\operatorname{Re}\left(\int_0^{2\pi}\log\left|re^{it}-a\right|\,dt\right)$$

Let $\gamma=\partial D(0,r)$. Then

$$\begin{align}\int_\gamma\frac{\log(z-a)}{iz}\,dz&=\int_0^{2\pi}\frac{\log\left(re^{it}-a\right)} {ire^{it}}rie^{it}\,dt\\ &=\int_0^{2\pi}\log\left(re^{it}-a\right)\,dt\end{align}$$

Thus

$$I=\operatorname{Re}\left(\int_{\gamma}\frac{\log(z-a)}{iz}\,dz\right)$$

Now my problem is that $\log(z-a)$ is not holomorphic in $D(0,r)$, so i can't use Cauchy's integral formula to compute $I$. How can I solve this?

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  • 2
    $\begingroup$ If $z$ is in the disk of radius $r$ around $0$, then $z-a$ is in the disk of radius $r$ around $a$. Since $r\lt|a|$, this disk does not enclose $0$. So are you sure $\log(z-a)$ is not holomorphic? $\endgroup$ – Gerry Myerson Nov 15 '12 at 22:35
  • $\begingroup$ yes, $0$ is not in $D(a,r)$, but this, i think, ensures only that $log(z-a)$ is well defined. But for the holomorphicity, shouldn'i i require some more? For example, if $log(z-a)=Log(z-a)$ is the principal branch of logarithm, i think i should exclude a line from the complex plane, say the negative real numbers. But if i take $z$ and $a$ with the same imaginary part, then $z-a$ lies on that line, where Log is not holomorphic. $\endgroup$ – Federica Maggioni Nov 16 '12 at 8:40
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    $\begingroup$ Just choose a branch of the logarithm that is holomorphic on your disc (not necessarily the principal branch), which is possible since $a \notin D(0,r)$. $\endgroup$ – mrf Nov 16 '12 at 9:27
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Because $0<r<|a|,$ $$\ln(a-z)=\ln{a\left(1-\dfrac{z}{a}\right)}=\ln{a}+\ln{\left(1-\dfrac{z}{a}\right)}=\ln{a}+\sum\limits_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot\left(\frac{z}{a} \right)^k }.$$ Therefore, Laurent expansion for $\frac{\ln(a-z)}{iz}=-i\frac{\ln(a-z)}{z}$ is $$-i\frac{\ln{a}}{z}-i\sum\limits_{k=1}^{\infty}{\frac{(-1)^{k-1}}{k}\cdot\frac{z^{k-1}}{a^k } }.$$ Using the residue theorem for integral $\int\limits_{\gamma}\frac{\log(z-a)}{iz}\,dz$ gives $$\int\limits_{\gamma}\frac{\log(z-a)}{iz}\,dz=2\pi{i}\cdot(-i\ln{a})=2\pi\ln{a}.$$ Taking the real part, $$\operatorname{Re}\left(\int_{\gamma}\frac{\log(z-a)}{iz}\,dz\right)=\operatorname{Re}(2\pi\ln{a})=2\pi\ln{|a|}.$$

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    $\begingroup$ Note that this is homework and the OP had an almost complete solution by themselves. Also, how do you justify the logarithmic identities on line 1? (In general, it is not true that $\log(zw) = \log w + \log w$ for complex numbers, if you have fixed a branch of $\log$.) $\endgroup$ – mrf Nov 16 '12 at 9:44
  • $\begingroup$ ok, i thank you all for your help. I think there exists a definition of log such that log is holomorphic in every simply connected open set not containing the origin. But i really can't understand this fact in practice. What branch of log i take in order to be sure that it is holomorphic in the disc $D(0,r)$. I'm quite confused about this. Is it sufficient to say: ok, $z-a\neq 0$ hence log is holomorphic? $\endgroup$ – Federica Maggioni Nov 16 '12 at 12:03
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{a \in {\mathbb C}}$.

Lets $\ds{z \equiv r\expo{\ic t}\ \imp\ \dd z = r\expo{\ic t}\ic\,\dd t\ \imp\dd t = {\dd z \over \ic z}}$:

\begin{align} I&=\left.\int_{0}^{2\pi}\ln\pars{\verts{re^{\ic t} - a}}\,\dd t \,\right\vert_{\,0\ <\ r\ <\ \verts{a}}\ =\ \Re\int_{0}^{2\pi}\ln\pars{re^{\ic t} - a}\,\dd t \\[5mm] & = \Re\oint_{0\ <\ \verts{z}\ =\ r\ <\ \verts{a}} \ln\pars{z - a}\,{\dd z \over \ic z} = \Re\pars{2\pi\ic\lim_{z\ \to\ 0}\bracks{z\,{\ln\pars{z - a} \over \ic z}}} = 2\pi\,\Re\pars{\ln\pars{-a}} \\[5mm] &= \bbox[10px,border:1px groove navy]{2\pi\ln\pars{\verts{a}}} \end{align}

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  • $\begingroup$ Hi, I was looking at your answer, but could you explain how you got from the integral to the limit? Thanks! $\endgroup$ – user68472 Mar 16 '18 at 8:38
  • $\begingroup$ @user68472 That's the Residue Theorem. That's the way to evaluate the residue of $\displaystyle{\ln\left(z - a\right) \over \mathrm{i}z}$ at $\displaystyle z = 0$. Thanks. $\endgroup$ – Felix Marin Mar 16 '18 at 17:46

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