complex integral with cauchys integral formula

Question

Integrate $$ \int_{|z-i|=1} \frac{1}{4z^2+1}dz $$ I have used the Cauchy's integral formula and got the answer $\pi/2$. However, my solution manual tells me its $i\pi/2$, who is right?

Answer 1

You are correct. Well done!

We have

$$\frac{1}{4z^2+1}=\frac14\frac{1}{(z+i/2)(z-i/2)}$$

The only singularity inside the disc $|z-i|\le 1$ is at $z=i/2$.

The residue is $\frac {1}{4i}$ and therefore the integral is $\frac{\pi}{2}$.

Alternatively, using Cauchy's Integral Theorem with $f(z)=\frac{1}{4(z+i/2)}$ gives

$$f(i/2)=\frac{1}{2\pi i}\oint_{|z-i|=1}\frac{f(z)}{z-i/2}\,dz$$

and therefore $$\oint_{|z-i|=1}\frac{f(z)}{z-i/2}\,dz=\frac{\pi}{2}$$as expected!