2.9: Branch Cuts and Function Composition

We often compose functions, i.e. $f(g(z))$. In general in this case we have the chain rule to compute the derivative. However we need to specify the domain for $z$ where the function is analytic. And when branches and branch cuts are involved we need to take care.

Example $\PageIndex{1}$

Let $f(z) = e^{z^2}$. Since $e^z$ and $z^2$ are both entire functions, so is $f(z) = e^{z^2}$. The chain rule gives us

$$f'(z) = e^{z^2} (2z). \nonumber$$

Example $\PageIndex{2}$

Let $f(z) = e^z$ and $g(z) = 1/z$. $f(z)$ is entire and $g(z)$ is analytic everywhere but 0. So $f(g(z))$ is analytic except at 0 and

$C$ - { $2\pi n i$, where $n$ is any integer }

The quotient rule gives $h'(z) = -e^z/(e^z - 1)^2$. A little more formally: $h(z) = f(g(z))$. where $f(w) = 1/w$ and $w = g(z) = e^z - 1$. We know that $g(z)$ is entire and $f(w)$ is analytic everywhere except $w = 0$. Therefore, $f(g(z))$ is analytic everywhere except where $g(z) = 0$.

Example $\PageIndex{3}$

It can happen that the derivative has a larger domain where it is analytic than the original function. The main example is $f(z) = \log (z)$. This is analytic on $C$ minus a branch cut. However

$$\dfrac{d}{dz} \log (z) = \dfrac{1}{z} \nonumber$$

is analytic on $C$ - {0}. The converse can’t happen.

Example $\PageIndex{4}$

Define a region where $\sqrt{1 - z}$ is analytic.

Solution

Choosing the principal branch of argument, we have $\sqrt{w}$ is analytic on

$C - \{ x \le 0, y = 0\}$, (see figure below.).

So $\sqrt{1 - z}$ is analytic except where $w = 1 - z$ is on the branch cut, i.e. where $w = 1 - z$ is real and $\le 0$. It's easy to see that

$w = 1 - z$ is real and $\le 0$ $\Leftrightarrow$ $z$ is real and $\ge 1$.

So $\sqrt{1 - z}$ is analytic on the region (see figure below)

$$C - \{x \ge 1, y = 0\} \nonumber$$

Note

A different branch choice for $\sqrt{w}$ would lead to a different region where $\sqrt{1 - z}$ is analytic.

The figure below shows the domains with branch cuts for this example.

Example $\PageIndex{5}$

Define a region where $f(z) = \sqrt{1 + e^z}$ is analytic.

Solution

Again, let's take $\sqrt{w}$ to be analytic on the region

$$C - \{x \le 0, y = 0\} \nonumber$$

So, $f(z)$ is analytic except where $1 + e^z$ is real and $\le 0$. That is, except where $e^z$ is real and $\le -1$. Now, $e^z = e^x e^{iy}$ is real only when $y$ is a multiple of $\pi$. It is negative only when $y$ is an odd mutltiple of $\pi$. It has magnitude greater than 1 only when $x > 0$. Therefore $f(z)$ is analytic on the region

$$C - \{x \ge 0, y = \text{odd multiple of } \pi \} \nonumber$$

The figure below shows the domains with branch cuts for this example.