Show that the function $f(z)=\frac{z}{z^{2}-1}-\frac{1}{z}$ has a primitive

Question

Show that the function $$f(z)=\frac{z}{z^{2}-1}-\frac{1}{z}$$ has a primitive, i.e., there exists a function $F'(z)=f(z)$, on the region $\Omega=\{z\in\mathbb{C};\ \lvert z\rvert>3\}$.

I know that, the function $f$ has a primitive $F$ in $\Omega$ implies that, the integral of in any contour closed, is $0$. Any help please!

Answer 1

You probably also know the converse to result you state in your question, namely:

If $f$ is continuous on $\Omega$ and $\int_\gamma f(z)\,dz = 0$ for all (simple) closed curves $\gamma$ in $\Omega$, then $f$ has an anti-derivative on $\Omega$.

For your question, let $\Omega = \{ z : |z| > 3 \}$. A simple closed curve in $\Omega$ either has its interior contained in $\Omega$, or it encloses the circle $|z|=3$.

If the interior of $\gamma$ lies inside $\Omega$, then $\int_\gamma f(z)\,dz = 0$ by Cauchy's integral theorem.

If $\gamma$ encloses $|z|=3$, then by partial fractions and Cauchy's integral formula (or the residue theorem if you prefer): \begin{align} \int_\gamma f(z)\,z &= \int_\gamma \bigg(\frac{z}{z^2-1}-\frac1z \bigg)\,dz \\ &= \int_\gamma \bigg(\frac12 \frac{1}{z+1} + \frac12 \frac{1}{z-1}-\frac1z \bigg)\,dz \\ &= 2\pi i \big( \frac12 + \frac12 - 1 \big) = 0. \end{align}

In either case, the integral is $0$, so there is an anti-derivative of $f$ on $\Omega$.