Show that $\int_{|z|=3} \frac{1}{z^2-1} dz = 0$

Question

Here's a homework problem I'm having some trouble with:

Show that $$ \int_{|z|=3} \frac{1}{z^2-1} dz = 0$$

So far, I've shown using Cauchy's Integral Formula that $$ \int_{|z-1|=1} \frac{1}{z^2-1} dz = \pi i$$ and $$\int_{|z+1|=1} \frac{1}{z^2-1} dz = - \pi i$$ where $|z-1|=1, |z+1|=1$ are circles of radius $1$ centered at $1$ and $-1$, respectively.

Now, I'm trying to use these results to solve the problem. It seems as if I will somehow add the integrals up to obtain the result $0$, but I'm having trouble extending the two circles to the circle of radius $3$ about the origin. My thought was to split the circle $|z|=3$ into two semicircles, separated on the imaginary axis, but I'm not quite sure how to do it. Any help would be greatly appreciated.

Answer 1

Yes, split the path $p$ of $|z| = 3$ into two semicircles as you said to create left semicircle $p_1$ and right semicircle $p_2$. Since

$$\int_{-3i}^{3i} f(z)~dz + \int_{3i}^{-3i} f(z) ~ dz = 0$$

you will get that (in the case that $p_1$ and $p_2$ are both defined to be counter clockwise paths):

$$\begin{align} \int_p f(z)~dz &= \int_{\text{left arc}} f(z)~dz + \int_{\text{right arc}} f(z)~dz \\ &= \int_{\text{left arc}} f(z)~dz + \int_{-3i}^{3i} f(z)~dz + \int_{3i}^{-3i} f(z)~dz + \int_{\text{right arc}} f(z)~dz\\ &= \int_{p_1} f(z)~dz + \int_{p_2} f(z)~dz \end{align}$$

From Cauchy's Integral Formula we know that:

• The analytic integral around closed path $q$ only depends on the poles inside of $q$
• $p_1$ and $|z+1|=1$ contain the same poles, because the second is inside the first
• $p_2$ and $|z-1|=1$ contain the same poles, because the second is inside the first

So

$$\begin{align}\int_p f(z)~dz &= \int_{p_1} f(z)~dz + \int_{p_2} f(z)~dz \\ &= \int_{|z+1|=1} f(z)~dz + \int_{|z-1|=2} f(z)~dz \\ &= -\pi i + \pi i \\ &= 0\end{align}$$

Answer 2

Write $\frac{1}{z^2-1}= \frac{1}{2}\left(\frac{1}{z-1}-\frac{1}{z+1}\right)$ then use Cauchy's integral formula