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- Feb 5, 2012

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**Samantha128's question from Math Help Forum,**

Hi Samantha128,Hi in my textbook there is the following question and my teacher said one similar is likely to be in the final exam. Can anyone help?

let f(z) = (z^2 + 2z -5)/((z^2+4)(z^2+2z+2)) If C is the circle |z|=R show that lim (from R to infinity) of the cyclic integral f(z) dz=0

I don't really know where to start

I hope you want to show, \(\displaystyle\lim_{R\rightarrow \infty}\oint_{c}f(z)\,dz=0\). For this let us first find, \(\displaystyle\oint_{c}f(z)\,dz\)

\[f(z) = \frac{z^2 + 2z -5}{(z^2+4)(z^2+2z+2)}\]

The points where the denominator become zero are, \(z=\pm 2i\mbox{ and }z=-1\pm i\). These are the points of discontinuities of the function \(f\). For \(R\neq 2, \sqrt{2}\) you can use the Estimation lemma. Then you will get,

\[\left|\oint_{c}f(z)\,dz\right|\leq\frac{2\pi R(R^2 + 2R -5)}{(R^2+4)(R^2+2R+2)}\]

By the Squeeze theorem,

\[\lim_{R\rightarrow \infty}\left|\oint_{c}f(z)\,dz\right|=0\]

\[\Rightarrow\lim_{R\rightarrow \infty}\oint_{c}f(z)\,dz=0\]