Proving the residue integral vanishes as R approaches infinity

[Sudharaka]

Samantha128's question from Math Help Forum,

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

Hi Samantha128,

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,

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

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

 

[vacuum]

I hope this helps. Let me know if you have any further questions or if you need clarification on any steps. Good luck on your exam!