# Different possible branch cuts for the principal argument, principal logarithm and principal square root functions

For some work I was doing with a student, I was trying to find different ways of proving the familiar result that the complex square root function $f(z) = \sqrt{z}$ is discontinuous everywhere on the negative real axis. As I was working on alternative proofs it became very clear to me how sensitive’ all the proofs were to the particular definition of the principal argument I was using, namely that the principal argument $\theta = \text{Arg}z$ is the unique argument of $z$ satisfying $-\pi < \theta \leq \pi$. In a sense, this definition manufactures’ the discontinuity of the complex square root function on the negative real axis, because the principal argument function itself is discontinuous here: the principal argument of a sequence of points approaching the negative real axis from above will tend to $\pi$, whereas the principal argument of a sequence approaching the same point on the negative real axis from below will tend to $-\pi$. I realised that all the proofs I was coming up with were exploiting this discontinuity of the principal argument function. However, this particular choice of principal argument function is completely arbitrary. An alternative could be to say that the principal argument of $z$ is the unique argument satisfying $0 \leq \theta < 2\pi$ which we can call $\text{Arg}_{2\pi} z$. The effect of this choice of principal argument function is to make the complex square root function discontinuous everywhere on the positive real axis! It turns out that we can choose an infinite number of different lines to be lines of discontinuity for the complex square root function, simply by choosing different definitions of the principal argument function. The same applies to the complex logarithm function. In this note I want to record some of my thoughts about this.

The reason for having to specify principal argument functions in the first place is that we need to make complex functions of complex variables single-valued rather than multiple-valued, to make them well-behaved with regard to operations like differentiation. Specifying a principal argument function in order to make a particular complex function single-valued is called choosing a branch of the function. If we specify the principal argument function to be $f(z) = \text{Arg} z$ where $-\pi < \text{Arg} z \leq \pi$ then we define the principal branch of the logarithm function to be

$\text{Log} z = \text{log}_e |z| + i \text{Arg} z$

for $z \in \mathbb{C} - \{0\}$, and the principal branch of the square root function to be

$z^{\frac{1}{2}} = \text{exp}\big(\frac{1}{2} \text{Log} z \big)$

for $z \in \mathbb{C}$ with $z \neq 0$.

If we define the functions $\text{Log} z$ and $z^{\frac{1}{2}}$ in this way they will be single-valued, but the cost of doing this is that they will not be continuous on the whole of the complex plane (essentially because of the discontinuity of the principal argument function, which both functions inherit’). They will be discontinuous everywhere on the negative real axis. The negative real axis is known as a branch cut for these functions. Using this terminology, what I want to explore in this short note is the fact that different choices of branch for these functions will result in different branch cuts for them.

To begin with, let’s formally prove the discontinuity of the principal argument function $f(z) = \text{Arg} z$, $z \neq 0$, and then see how this discontinuity is inherited’ by the principal logarithm and square root functions. For the purposes of the proof we can consider the sequence of points

$z_n = |\alpha| \text{e}^{(-\pi + 1/n)i}$

where

$\alpha \in \{ x \in \mathbb{R}: x < 0 \}$

Clearly, as $n \rightarrow \infty$, we have $z_n \rightarrow -|\alpha| = \alpha$. However,

$f(z_n) = \text{Arg} \big( |\alpha| \text{e}^{(-\pi + 1/n)i}\big)$

$= -\pi + \frac{1}{n}$

$\rightarrow -\pi$

whereas

$f(\alpha) = \text{Arg}\big(|\alpha| \text{e}^{\pi i} \big) = \pi$

Therefore $f(z_n) \not \rightarrow f(\alpha)$, so the principal argument function is discontinuous at all points on the negative real axis.

Now consider how the following proof of the discontinuity of $f(z) = z^{\frac{1}{2}}$ on the negative real axis depends crucially on the discontinuity of $\text{Arg} z$. We again consider the sequence of points

$z_n = |\alpha| \text{e}^{(-\pi + 1/n)i}$

where

$\alpha \in \{ x \in \mathbb{R}: x < 0 \}$

so that $z_n \rightarrow -|\alpha| = \alpha$. However,

$f(z_n) = z_n^{\frac{1}{2}} = \text{exp}\big(\frac{1}{2} \text{Log} z_n \big)$

$= \text{exp}\big( \frac{1}{2} \text{log}_e |z_n| + \frac{1}{2} i \text{Arg} z_n \big)$

$= \text{exp}\big( \frac{1}{2} \text{log}_e |\alpha| + \frac{1}{2} i (- \pi + \frac{1}{n}) \big)$

$\rightarrow |\alpha|^{\frac{1}{2}} \text{e}^{-i \pi /2} = - i |\alpha|^{\frac{1}{2}}$

whereas

$f(\alpha) = \big( |\alpha| \text{e}^{i \pi}\big)^{\frac{1}{2}}$

$= |\alpha|^{\frac{1}{2}} \text{e}^{i \pi/2} = i |\alpha|^{\frac{1}{2}}$

Therefore $f(z_n) \not \rightarrow f(\alpha)$, so the principal square root function is discontinuous at all points on the negative real axis.

Now suppose we choose a different branch for the principal logarithm and square root functions, say $\text{Arg}_{2\pi} z$ which as we said earlier satisfies $0 \leq \text{Arg}_{2\pi} z < 2\pi$. The effect of this is to change the branch cut of these functions to the positive real axis! The reason is that the principal argument function will now be discontinuous everywhere on the positive real axis, and this discontinuity will again be `inherited’ by the principal logarithm and square root functions.

To prove the discontinuity of the principal argument function $f(z) = \text{Arg}_{2\pi} z$ on the positive real axis we can consider the sequence of points

$z_n = \alpha \text{e}^{(2 \pi - 1/n)i}$

where

$\alpha \in \{ x \in \mathbb{R}: x > 0 \}$

We have $z_n \rightarrow \alpha$. However,

$f(z_n) = \text{Arg}_{2\pi} \big(\alpha \text{e}^{(2\pi - 1/n)i}\big)$

$= 2\pi - \frac{1}{n}$

$\rightarrow 2\pi$

whereas

$f(\alpha) = \text{Arg}_{2\pi}(\alpha) = 0$

Therefore $f(z_n) \not \rightarrow f(\alpha)$, so the principal argument function is discontinuous at all points on the positive real axis.

We can now again see how the following proof of the discontinuity of $f(z) = z^{\frac{1}{2}}$ on the positive real axis depends crucially on the discontinuity of $\text{Arg}_{2\pi} z$ there. We again consider the sequence of points

$z_n = \alpha \text{e}^{(2\pi - 1/n)i}$

where

$\alpha \in \{ x \in \mathbb{R}: x > 0 \}$

so that $z_n \rightarrow \alpha$. However,

$f(z_n) = z_n^{\frac{1}{2}} = \text{exp}\big(\frac{1}{2} \text{Log} z_n \big)$

$= \text{exp}\big( \frac{1}{2} \text{log}_e |z_n| + \frac{1}{2} i \text{Arg}_{2\pi} z_n \big)$

$= \text{exp}\big( \frac{1}{2} \text{log}_e |\alpha| + \frac{1}{2} i (2 \pi - \frac{1}{n}) \big)$

$\rightarrow \alpha^{\frac{1}{2}} \text{e}^{i 2 \pi /2} = - \alpha^{\frac{1}{2}}$

whereas

$f(\alpha) = \alpha^{\frac{1}{2}}$

Therefore $f(z_n) \not \rightarrow f(\alpha)$, so the principal square root function is discontinuous at all points on the positive real axis.

There are infinitely many other branches to choose from. In general, if $\tau$ is any real number, we can define the principal argument function to be $f(z) = \text{Arg}_{\tau} z$ where

$\tau \leq \text{Arg}_{\tau} < \tau + 2\pi$

and this will give rise to a branch cut for the principal logarithm and square root functions consisting of a line emanating from the origin and containing all those points $z$ such that $\text{arg}(z) = \tau$ modulo $2\pi$.