Using Cauchy's integral formula to evaluate a function 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 $$2 \pi if(z_0) = \int_c \frac{1}{z-z_0}f(z)\,dz$$

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

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.