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I got stuck in it. Please help me.

Let $\lim\limits_{z\rightarrow a}(z-a)f(z)=\lambda$ and let $C$ be the arc $\theta_1\leq \theta\leq \theta_2$ of the circle $|z-a|=r$. Prove that $\lim\limits_{r\rightarrow 0}\int_C f(z)dz=i\lambda(\theta_2-\theta_1)$.

Can you please tell me how to proceed in this one?

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First of all you need to note that $a$ is a simple pole of $f$, then $f$ is analytic around $a$, thus it can be written as $$ f(z)=\frac{b_1}{z-a}+h(z) $$ where $b_1=\text{Res}(f,a)$, and $h$ is analytic, hence $$ \int_{C}f(z)dz = \int_{C}\frac{b_1}{z-a} dz + \int_{C}h(z)dz $$ Expressing $C$ in parametric form, we have $C(\theta)=a+re^{i \theta}$ with $\theta \in [\theta_1, \theta_2]$, therefore for all $r>0$ $$ \int_{C(\theta)}\frac{b_1}{z-a} dz = b_1 \int_{\theta_1}^{\theta_2} \frac{ire^{i\theta}}{re^{i\theta}}=i b_1(\theta_2-\theta_1) $$ Now, since $h$ is analytic, then around $a$ it must be bounded , say by $M>0$ $$ \left| \int_{C(\theta)}h(z)dz \right|\leq M\cdot long(C(\theta))= M\cdot r(\theta_2-\theta_1)\underset{ r \to 0}{\longrightarrow 0} $$ Then we have proved that $$ \lim_{r \to 0}\int_{C}f(z)dz =ib_1(\theta_2-\theta_1) + 0 $$ However, since $$ b_1=\text{Res}(f,a)=\lim_{z \to a}(z-a)f(z)=\lambda $$ the claim follows

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    $\begingroup$ @Anjan3 Since $\lambda \neq \infty$ and $\lambda = \lim_{z\to a} (z-a)f(z)$, this precisely means that $a$ is a simple pole of $f$. Do you see it now? if anything else is not clear please let me know. $\endgroup$ Jun 10, 2015 at 14:58
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    $\begingroup$ @Anjan3 Yes it follows. To prove it, take the Laurent expansion of $f$ about $z=a$, $f(z)=\sum_{n=-\infty}^\infty a_n (z-a)^n$, thus since $\lambda \neq 0, \infty $, $$ \lambda = \lim_{z \to a} (z-a) f(z) = \lim_{z \to a} \sum_{n=-\infty}^\infty a_n (z-a)^{n+1} = \lim_{z \to a} \sum_{n=-\infty}^{-1} a_n (z-a)^{n+1} + 0 $$ and hence, $a_n=0$ for all $n<-1$ and $a_{-1}=\lambda$. Therefore $$ f(z)=\frac{\lambda}{z-a} + \sum_{n=0}^{\infty} a_n(z-a)^n $$ which gives indeed that $z=a$ is a simple pole of $f$. $\endgroup$ Jun 10, 2015 at 20:59
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    $\begingroup$ @AlonsoDelfín is this a particular case of the proof you did? If $f(z)$ is continuous in the region $\vert z\vert\ge R_0, 0\le arg(z)\le\alpha (0<\alpha\le 2\pi)$ and the limit $\lim_{z\to \infty}zf(z)=A$ exist, then $\lim_{R\to\infty} \int_{T_R} f(z)dz=iA\alpha,$ where $T_R$ is the arc of the circle $\vert z\vert=R,$ which lies in the given region traversed in the positive direction with respect to the coordinate origin. $\endgroup$
    – user441848
    May 30, 2017 at 19:22
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    $\begingroup$ @AnneliseToft Yeah that's for the first equality. For the second one you probably need to be more careful and work the details. $\endgroup$ May 31, 2017 at 0:11
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    $\begingroup$ yep that's right $\endgroup$
    – user441848
    May 31, 2017 at 0:19

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