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My book claims that if a function $f(z)$ such that $z f(z) \rightarrow 0 $ uniformly whenever $z \rightarrow 0$ then the line integral along the semi-circular path is zero as $ R \rightarrow \infty$. $\lim\limits_{R \rightarrow \infty} \int_c f(z) = 0$ The same thing is claimed in here on Wikipedia. I don't understand this particular step

$ M_R:=\max_{\theta\in [0,\pi]} \bigl|g \bigl(R e^{i \theta}\bigr)\bigr| \to 0\quad \mbox{as } R \to \infty\,,\qquad(*)$ How does the Max value go to zero? And what does $e^{iaz}$ has to do with $zf(z)$ from my book.

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    I assume you mean $z\to\infty$ in the first line?2012-08-27

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First, in Wiki there's a confussion between $\,g\,\,\,and\,\,\,f\,$ in that part, second what you ask is an assumption: if that happens, then...the next line.

Finally, it is the same $|z|=R\Longleftrightarrow z=Re^{i\theta}\,\,,\,\text{for some }\,\ \theta\in[0,2\pi)$

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    It is not: **if** the limit of that maximum, when $\,R\to\infty\,$ is zero, then... For example, take on the semicircle $\,\gamma_R:=\{ z=Re^{i\theta}\;\;|\;\;\,0\leq \theta\leq \pi\}\,$ , the function $f(z):=\frac{e^{iz}}{z^2}\Longrightarrow max_\gamma|f(z)|\cdot \mathcal l(\gamma)=\frac{e^{-R\sin\theta}}{R^2}\cdot \pi R\xrightarrow [R\to\infty]{}0$since $\,iRe^{i\theta}=iR(\cos\theta+i\sin\theta)=-R\sin\theta+iR\cos\theta\,$ , and $\,|e^{ix}|=1\,\,,\,for\,\,\,x\in\Bbb R\,$2012-08-27