# not following a step in ash and novinger example of analytic but does not have primitive

I'm trying to self-study complex analysis and am currently reading the book "complex analysis" book by ash and novick. 0n the top of page 14, they write that , if $f(z) = \frac{1}{z}$ and the path integral is over the path, $\exp(it)$, for $0 \le t \le 2\pi$, then $\int_{\gamma} = \int_{0}^{2\pi} \frac{1}{z} dz = [_{0}^{2\pi}\frac{i \exp(it)}{\exp(it)} dt = 2\pi$.

That was clear. But then they explain that, for this example, the theorem that $\int_{\gamma} f(z) dz = 0$ when the path $\gamma$ is closed does not hold because this is an example where $f(z)$ is analytic on $C - \{0\}$ but it does not have a primitive on $C - \{0\}$. I thought having a primitive and being analytic was the same thing so could someone explain the difference ? Even more importantly, could someone explain what's going on with this example that makes the path integral not be zero. Thanks a lot.

Having a primitive function is more restrictive than being analytic. All analytic functions will have primitive functions locally, but the local primitive functions may not match together to form a global primitive function. Your function $f(z)={1\over z}$ is the simplest example of this. If your path of integration hadn't encircled the origin, it would have been possible to choose a branch of the logarithm as a primitive function. If there had been a primitive function $F(z)$, the value of the integral would have been $F(\gamma(2\pi))-F(\gamma(0))=0$. But any branch of the logarithm will have a discontinuity somewhere along $\gamma$, and the jump will be exactly the value $2\pi {\rm i}$ of your integral.
• Thanks Per Manne. That's a great explanation. Just one last related question: Is it because the derivative of f(z) blows up at z = (0,0) ? I guess I'm confused because the derivative of $\frac{1}{z} = - \frac{1}{z^2}$ but does that have anything to do with the issue of no primitive ? How can I plug in $(0,0)$ to z ? or is it really $\frac{1}{t}$ and then you plug in $0$. thanks. – mark leeds Jul 26 '14 at 8:27
• It is relevant that $f(z)$ has a singularity at the origin $z=0$, but it is not the whole explanation. The function $g(z)={1\over z^2}$ also has a singularity at the origin, but $g(z)$ still has a primitive function $G(z)=-{1\over z}$, so $\int_{\gamma}g(z)\,{\rm d}z=0$. (The difference between $f(z)$ and $g(z)$ lies in their residues at $z=0$.) – Per Manne Jul 26 '14 at 8:41