Use the Cauchy Integral Formula to evaluate the following integral:$\int_e{\frac{1}{z^2+4}}dz$

Question

I need to evaluate the following integral $$\int_e{\frac{1}{z^2+4}}dz$$ on the ellipse $4x^2+(y-2)^2=4$

The integrand does not exist for $z=\pm2i$. I see that $-2i$ lies outside of the ellipse, by the Cauchy Formula I know that this integral will be zero.

Knowing this I can write

$$\int_e{\frac{1}{z^2+4}}dz=\int_e{\frac{f_1(z)}{(z-2i)}}dz=2\pi i*f_1(2i)=\frac{\pi}{2}$$

Where $f_1(z)=\frac{1}{z+2i}$

Answer 1

$$I=\int_C \frac{dz}{z^2+4}, C: 4x^2+(y-2)^2=4$$ the integrand has poles at $z=\pm 2i$ of which only $z=2i$ lies in the given elliptic contour. So by residue theorem, we have $$Res[(z^2+4)^{-1}]_{z=2i}=\lim_{z\rightarrow 2i} \frac{(z-2i)}{(z-2i)(z+2i)}=\frac{1}{4i} \implies I= 2i\pi \frac{1}{4i}=\frac{\pi}{2}$$

Answer 2

I can write $$\frac{1}{z^2+4}=\frac{1}{4i}*\left( \frac{1}{z-2i} -\frac{1}{z+2i} \right)$$

Such that

$$\int_e{\frac{1}{z^2+4}}dz=\frac{1}{4i}*\left( \int_e{\frac{1}{z-2i}} -\int_e{\frac{1}{z+2i}} \right)=\frac{\pi}{2}$$

Where $$\int_e{\frac{1}{z-2i}}=2\pi i*f(2i)=2 \pi i$$ $$\int_e{\frac{1}{z+2i}}=0$$

Since the pole $-2i$ lies outside the elliptic contour