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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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    thanks ... i'll do it from onwards2012-08-25

2 Answers 2

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

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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.