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I don't understand this step on the Wikipedia article on Jordan's Lemma:

$$ \begin{align} I_R:=\biggl|\int_{C_R} f(z)\, dz\biggr| &\le R\int_0^\pi\bigl|g(Re^{i\theta})\,e^{aR(i\cos\theta-\sin\theta)}\,ie^{i\theta} \bigr|\,d\theta\\ &=R\int_0^\pi \bigl|g(Re^{i\theta})\bigr|\,e^{-aR\sin\theta}\,d\theta\,. \end{align} $$

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    Don't you think it'd be nice and cosy if you told us what $\,C_R\,,\,f\,,g\,,\,etc.\,$ are?! I'm guessing $\,C_R\,$ is an upper semicircle of radius R centered at the origin, but why not to specify this?2012-08-25
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    how to copy LaTeX from Wikipeida??2012-08-25
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    In Wikipedia, go to Edit Article (link is on the upper right corner of every WP article), then you'll find the source code of the WP article. Browse to the equation you want, and copy-paste.2012-08-25
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    thanks ... i'll do it from onwards2012-08-25

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The very last step follows from $$|i|=|e^{ix}|=1\,\,,\,\,x\in\Bbb R$$

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    I think I did: the inequaluty step is completely trivial in that Wiki proof, so I assumed the OP meant the very last equality step...2012-08-25
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    $$ |f(z) e^{i \theta} = |f(z)| ??$$2012-08-25
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    I just answered this in my answer, @MonkeyD.Luffy!2012-08-25
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    I meant to ask if I did that correctly?2012-08-25
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    Well, what you wrote above is missing the right | sign, but other than it is correct...*if* $\,\theta\in\Bbb R\,$, which in your OP indeed is.2012-08-25
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Remember that norms multiply: That is $|ab|=|a||b|$ for two complex numbers $a$ and $b$.

$$|g(Re^{i\theta})e^{aR(i\cos\theta-\sin\theta)}ie^{i\theta}|$$ $$=|g(Re^{i\theta})||e^{iaR\cos\theta}||e^{-aR\sin\theta}||i||e^{i\theta}|$$

$|i|=1$, and $|e^{ix}|=1$ for all $x$. That means that the second, fourth and fifth pieces are all $1$, leaving

$$|g(Re^{i\theta})||e^{-aR\sin\theta}|$$

I assume $a,R,\theta$ are all real, meaning that $|e^{-aR\sin\theta}|=e^{-aR\sin\theta}$ (which is true for all non-negative real numbers). This leaves you with

$$|g(Re^{i\theta})|e^{-aR\sin\theta}$$

as your integrand.