Primitive of $\frac{1}{z(z-1)}$ in $\mathbb{C}-[0,1]$

Question

I'm in a first course in complex analysis and we are using Remmert's Book: Theory of Complex Functions.

In chapter 6, section 3, there is an exercise which explicitly states:

Let $ G := \mathbb{C}-[0,1]$, $f: G \rightarrow \mathbb{C}$ the function $f(z) := \frac{1}{z(z-1)}$. Show that for every closed path $\gamma$ in $G$

\begin{equation*} \int_{\gamma}f(\xi)\ d\xi=0. \end{equation*}

I tried to used a criterion that is proved in the book that says that if $f$ is continuous in a domain $D$, then $\int_{\gamma}f(\xi)\ d\xi=0$ for every closed path $\gamma$ in $D$ is equivalent to $f$ being integrable in $D$ ($f$ is said to be integrable in $D$ if it has a primitive).

First I tried was splitting $f(z)$ in partial fractions, \begin{equation*} \frac{1}{z(z-1)}=\frac{1}{z-1}-\frac{1}{z}, \end{equation*} but neither $\frac{1}{z-1}$ or $\frac{1}{z}$ has a primitive in $\mathbb{C}-[0,1]$ because there is no logarithm function defined in that domain. (I think the greatest domain in which we can define a logarithm function is $\mathbb{C}$ removing at least one half line starting and 0, as the book does when defines the principal branch of logarithm, right?)

So I don't know how to proceed with this problem.

I appreciate any help or comment.

Answer 1

In calculus, the indefinite integral $\int{\frac{1}{x(x-1)}\,dx}$ is $\ln\left(\frac{x-1}{x}\right)+C$. So if $\frac{1}{z(z-1)}$ has a primitive, then $\ln\left(\frac{z-1}{z}\right)$ should be one such primitive. Let's try to show that the image of $g(z) = \frac{z-1}{z}$ is a domain where we can define a logarithm.

Notice that $g$ defines a bijection between $\mathbb{C}-\{0\}$ and $\mathbb{C}-\{1\}$, with an inverse defined by $g^{-1}(z) = \frac{1}{1-z}$. Furthermore, for every $z\in(0,1]$, we have that $\frac{z-1}{z}$ is real and nonpositive, i.e. $g(z)\in(-\infty,0]$, and conversely for every $z\in(-\infty,0]$ we have that $g^{-1}(z) = \frac{1}{1-z}\in(0,1]$. It follows that the image of $(0,1]$ under $g$ is precisely $(-\infty,0]$, and so $$g(G) = g(\mathbb{C}-[0,1]) = \mathbb{C}-\{1\} - g((0,1]) = \mathbb{C}-\{1\}-(-\infty,0]\subset \mathbb{C}-(-\infty,0].$$ Hence, the image of $g$ restricted to $G$ is contained in the complex plane minus the negative real line, on which a logarithm can be defined. If $\log:\mathbb{C}-(-\infty,0]\rightarrow\mathbb{C}$ is such a logarithm, then $F:\mathbb{C}-[0,1]\rightarrow\mathbb{C}$ defined by $F(z) = \log(g(z))$ is holomorphic, with $$F'(z) = \frac{d}{dz}\left(\log\left(\frac{z-1}{z}\right)\right) = \frac{1}{z(z-1)} = f(z) $$ i.e. $F$ is a primitive for $f$.

Answer 2

If you know that the value of the integral is independent of the path chosen you may choose as contour a circle of radius $R$ and consider the limit as $R\rightarrow \infty$.

Answer 3

Hint: What happens if you choose the branch cuts for $\int \frac{\mathrm{d}z}{z}$ and $\int \frac{\mathrm{d}z}{z-1}$ to run in the negative direction along the real axis? In $G$, the result of the subtraction will be continuous (because the jumps cancel) and differentiable (surprise!).