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In the last section we computed the path integral of $f(z) = 1/z$ over the path $\gamma(t) = e^{it}$ parameterizing the unit circle and got a non-zero answer \[\int\limits_\gamma f = \int\limits_0^{2\pi} \frac{1}{e^{it}} i e^{it} \intd t = 2 \pi i\] from which we concluded $f$ does not have an antiderivative on $\C \setminus \{0\}$. (If it did have an antiderivative on $\C \setminus \{0\}$ the path integral above would have to be zero!)
We can write \[\frac{1}{z} = \frac{1}{1 - (1-z)} = \sum_{n=0}^\infty (-1)^n (z-1)^n\] for all $|z-1| < 1$ using our knowledge of the geometric series. This gives a power series representation of $f$ on $\ball(1,1)$. Let's define \[F(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n+1} (z-1)^{n+1}\] which is the term-by-term antiderivative of our power series representing $f$ on $\ball(1,1)$. We can check (for example, using the ratio test) that its radius of convergence is $1$. As a power series in its own right, it is holomorphic on $\ball(1,1)$ and certainly $F'(z) = f(z)$ for all $z$. Thus $f(z) = 1/z$ does have an antiderivative on $\ball(1,1)$! We now have two questions.
We will have to wait a little to answer the first question. For the second question, recall first of all from real analysis that \[\bigg( \ln(x) \bigg)' = \frac{1}{x}\] so that our antiderivative $F$ has (at least on the real axis) something to do with the natural logarithm. In real analysis the natural logarithm can be defined as the functional inverse of the exponential function. Let's try to define an inverse of our exponential function $\exp$.
Fix $w \in \C \setminus \{0\}$. We would like to solve the equation $w = \exp(z)$ for $z$. Writing $z = x+iy$ gives \[w = e^x(\cos(y) + i \sin(y))\] and we see that $|w| = e^x$ and that $y$ is an argument of $w$. Thus the solutions are \[ z = \ln(|w|) + i ( \Arg(w) + 2n \pi ) \] for any $n \in \Z$.
In particular, every non-zero complex number $z$ has a (indeed many) natural logarithms: a complex number $w$ such that $\exp(z) = w$. The quantity $\Arg(z)$ was defined as the argument of $z$ lying in the interval $(-\pi,\pi]$. The function $z \mapsto \Arg(z)$ is not continuous, so we cannot define a continuous function $z \mapsto \ln(|z|) + i \Arg(z)$ on all of $\C \setminus \{0\}$. However, if we exclude the negative real axis \[ (-\infty,0] = \{ x + iy : x \le 0 \textup{ and } y = 0 \} \] where the output of $\Arg$ jumps, we do get a continuous function.
The principal logarithm of $z \in \C \setminus (-\infty,0]$ is \[\Log(z) = \ln(|z|) + i \Arg(z)\] and this defines a function $\Log : \C \setminus (-\infty,0] \to \C$.
What is $\Log(i)$? From $\Arg(i) = \pi/2$ we calculate \[\Log(i) = \ln(|i|) + i \Arg(i) = i \pi /2\] which is reasonable because $\exp(i \pi / 2) = i$.
Having defined a logarithm, we can define complex powers of complex numbers as follows.
For $b,z \in \C$ with $b \ne 0$ define \[b^z = \exp(z \Log(b))\] to be the principal value of $b^z$.
Since $z \mapsto |z|$ is continuous on $\C$ and $z \mapsto \Arg(z)$ is continuous on $\C \setminus (-\infty,0]$ the principal logarithm is continuous on its domain. In fact, more is true.
The principal logarithm is holomorphic on its domain and \[\Log'(z) = \dfrac{1}{z}\] for all $z \in \C \setminus (-\infty,0]$.
Fix $z \in \C \setminus (-\infty,0]$. We need to evalue the limit \[\lim_{h \to 0} \frac{\Log(z+h) - \Log(z)}{h}\] and will do so using the relationship between $\Log$ and $\exp$. Put $w = \Log(z)$ and $k(h) = \Log(z+h) - w$. Since $\Log$ is continuous the limit $h \to 0$ implies that $k(h) \to 0$. We may therefore calculate \[\begin{aligned} & \lim_{h \to 0} \frac{w + k(h) - w}{\exp(w+k(h)) - \exp(w)} \\ = & \lim_{h \to 0} \frac{k(h)}{\exp(w+k(h)) - \exp(w)} \\ = & \dfrac{1}{\lim_{h \to 0} \dfrac{\exp(w+k(h)) - \exp(w)}{k(h)}} \\ = & \dfrac{1}{\exp(w)} = \dfrac{1}{z} \end{aligned}\] as desired. $\square$
We have established the following about the function $f(z) = 1/z$.
What difference between the domains $\C \setminus \{0\}$ and $\C \setminus (-\infty,0]$ is responsible for the two statements above? We will uncover the full answer to this question next week.