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This problem is from Brown/Churchill Complex Variables and Applications, $8$th edition $2009$.

Section $52$, exercise $2$, subsection (a)

How do I show that the integral of the function $g(z) = (z^2+4)^{-1}$ along the circular contour $|z-i| = 2$ is $\frac{\pi}2$?

I believe usage of Cauchy's Integral is necessary.

Cauchy's integral formula states that if $f(z)$ is analytic on and within a simple closed countour $C$ oriented in the positive direction and the point $z_0$ is interior to the contour then

$$2 \pi if(z_0) = \int_c \frac{1}{z-z_0}f(z)\,dz$$

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  • $\begingroup$ What is your question? $\endgroup$ Commented Mar 12, 2013 at 0:49

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$$\oint\limits_{|z-i|=2}\frac{1}{z^2+4}dz=\oint\limits_{|z-i|=2}\frac{\frac{1}{z+2i}}{z-2i}dz=\left.2\pi i\left(\frac{1}{z+2i}\right)\right|_{z=2i}=2\pi i\frac{1}{4i}=\frac{\pi}{2}$$

Now, why did I do the above the way I did? Draw a picture of the circle $\,|z-i|=2\,$ and try to locate the poles of the integrand function inside the circle.

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  • $\begingroup$ the pole for the function $\frac{1}{z+2i}$ is at $-2i$ if I'm not mistaken. the point $-2i$ lies outside the circle centered at $i$ with radius $2$. $\endgroup$ Commented Mar 12, 2013 at 1:13
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    $\begingroup$ Yes @franklin, and because of that the function $\,f(z):=\frac{1}{z+2i}\,$ is analytic on $\,|z-i|\le 2\,$ ... $\endgroup$ Commented Mar 12, 2013 at 2:13

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