Is $\int_{C[-2i,r]} \frac{dz}{z^2+1} = 0 , 1 < r < 3$? I got $-\pi$.

Question

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 4.33

It is given that $r \ne 1,3$ and that the answer is $0 \ \forall r$. I got $0 \ \forall \ r \in(0,1) \cup (3,\infty)$ and $-\pi \ \forall \ r \in(1,3)$. I have 2 solutions:

1. $$\int_{C[-2i,r]} \frac{dz}{z^2+1} = \frac{-i}{2}\int_{C[-2i,r]} \frac{1}{z-i} - \frac{1}{z+i}dz = \frac{-i}{2}[0-2\pi i] = - \pi$$

2. $$\int_{C[-2i,r]} \frac{dz}{z^2+1} = \int_{C[-2i,r]} \frac{\frac{1}{z-i}}{z+i} dz = 2\pi i \frac{1}{z-i}|_{z=-i} = 2\pi i \frac{1}{-2i} = - \pi$$

I even computed $-\pi$ on Wolfram Alpha for $r=2$, $r=2.5$ and $r=e$.

And yet the answer is simply $0$

Why is it 0 please?

Note: Textbook has been previously wrong: Is $\int \frac{e^{z^2}}{z^3} dz=\pi i$? I got $2 \pi i$.

Answer 1

You are right and your textbook is wrong: the answer is $-\pi$ if $r\in(1,3)$.