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# 2.9: Branch Cuts and Function Composition

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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

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

## 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

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

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)

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

## 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

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

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

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

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

2.9: Branch Cuts and Function Composition is shared under a not declared license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.