Calculate $\int_{\gamma}\frac{1}{z}dz$ by definition.

Question

I'd like to calculate

$$\int_{\gamma}\frac{1}{z}dz$$

where $\gamma$ is the contour of a circle that doesn't contain the origin. I'd like to do it using the definition of integral along a curve, without Cauchy theorem. Thanks for the help.

Answer 1

Suppose $f$ is holomorphic in $U$ and $\gamma $ is a closed contour in $U$ from $z_1$ to $z_2.$ If $F' = f$ in $U,$ then $\int_\gamma f(z)\,dz = F(z_2)-F(z_1).$ This follows from the FTC from ordinary calculus. It follows that if $\gamma$ is closed, then the integral equals $0.$

A special case of this: $\log'(z) = 1/z$ in $\mathbb C \setminus (-\infty,0].$ Here $\log z$ is the principal value logarithm.

For $a> 0,r>0,$ define $\gamma (t) = a + r^{it}, t \in [-\pi,\pi].$ Suppose first $r<a.$ Then $\gamma$ lies in the open right half plane, which is contained in $\mathbb C \setminus (-\infty,0].$ By the above, $\int_\gamma (1/z)\,dz = 0.$

That was easy. But now let's look at the case $r>a.$ This time $0$ is inside $\gamma$ and $\gamma $ is not contained in $\mathbb C \setminus (-\infty,0].$ We get around this by defining, for small $\epsilon>0,$ $\gamma_\epsilon (t) = a + r^{it}, t \in [-\pi+\epsilon,\pi-\epsilon].$ Now each $\gamma_\epsilon$ is contained in $\mathbb C \setminus (-\infty,0],$ so we can use the first paragraph. Together with the fact that

$$\int_\gamma \frac{dz}{z} = \lim_{\epsilon\to 0}\int_{\gamma_\epsilon} \frac{dz}{z}$$

the answer $2\pi i$ falls out. There's some work to do here, but it comes out nicely.

I assumed $a>0$ but the argument for any $a$ is very similar. (In fact, by a change of variables, you can reduce the result for general $a$ to the $a>0$ case.)