The Principal Argument

In this text the notation $\textbf{arg} (z)$ is used to designate an arbitrary argument of $z,$ which means that $\textbf{arg} (z)$ is a set rather than a number. In particular, the relation $$\textbf{arg} (z_1) = \textbf{arg} (z_2)$$ is not an equation, but expresses equality of two sets.

As a consequence, two non-zero complex numbers $r_1 (\cos \varphi_1 + i \sin\varphi_1)$ and $r_2 (\cos \varphi_2 + i \sin\varphi_2)$ are equal if and only if

$$r_1=r_2,\quad \text{and}\quad \varphi_1 = \varphi_2+ 2 k \pi,$$

where $k \in \mathbb Z.$

In order to make the argument of $z$ a well-defined number, it is sometimes restricted to the interval $(-\pi, \pi].$ This special choice is called the principal value or the main branch of the argument and is written as $\textbf{Arg}(z).$

Note that there is no general convention about the definition of the principal value, sometimes its values are supposed to be in the interval $[0, 2\pi).$ This ambiguity is a perpetual source of misunderstandings and errors.

The principal value $\textbf{Arg}(z)$ of a complex number $z=x+iy$ is normally given by $$\Theta=\arctan\left(\frac{y}{x}\right),$$ where $y/x$ is the slope, and $\arctan$ converts slope to angle. But this is correct only when $x > 0,$ so the quotient is defined and the angle lies between $-\pi/2$ and $\pi/2.$ We need to extend this definition to cases where $x$ is not positive, considering the principal value of the argument separately on the four quadrants.

The function $\textbf{Arg}(z)$ $:\mathbb C \setminus \{0\} \rightarrow \left(-\pi,\pi\right]$ is defined as follows:

\begin{eqnarray*} \textbf{Arg}(z)= \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}r@{}} \arctan \frac{y}{x} & \;\text{if $x>0,$ $y\in \mathbb R$}\\ \arctan \frac{y}{x}+\pi & \;\text{if $x <0,$ $y\geq 0$}\\ \arctan \frac{y}{x}-\pi & \;\text{if $x < 0,$ $y < 0$}\\ \frac{\pi}{2} & \;\text{if $x=0,$ $y> 0$}\\ -\frac{\pi}{2} & \;\text{if $x=0,$ $y < 0$}\\ \text{undefined}& \;\text{if $x=0,$ $y=0$}\\ \end{array}\right. \end{eqnarray*}

Thus, if $z=r(\cos \Theta +i\sin \Theta),$ with $r>0$ and $-\pi < \Theta \leq \pi,$ then

\begin{eqnarray*} \textbf{arg}(z)= \{ \textbf{Arg}(z)+2n\pi \mid n \in \mathbb{Z} \}. \end{eqnarray*}

We can visualize the multiple-valued nature of $\textbf{arg}(z)$ by using Riemann surfaces. The following interactive shows some of the infinite values of $\textbf{arg}(z).$ Each branch is identified with a different color.

Arguments of products and quotients

If $z_1= r_1e^{\theta_1}$ and $z_2= r_2e^{\theta_2},$ then \begin{eqnarray}\label{product} z_1z_2 = \left(r_1r_2\right)e^{i\left(\theta_1 + \theta_2\right)} \end{eqnarray} implies \begin{eqnarray}\label{arg-product} \arg\left(z_1z_2\right) = \arg\left(z_1\right) + \arg\left(z_2\right). \end{eqnarray} We can easily prove (\ref{arg-product}) by letting $\theta_1$ and $\theta_2$ denote any values of $\arg\left(z_1\right)$ and $\arg\left(z_1\right),$ respectively. Then expression (\ref{product}) tells us that $\theta_1+\theta_2$ is a value of $\arg\left(z_1z_2\right).$ If the values of $\arg\left(z_1z_2\right)$ and $\arg\left(z_1\right)$ are specified, those values correspond to particular choices of $n$ and $k$ in the expressions \[ \arg\left(z_1z_2\right) = \left(\theta_1 + \theta_2\right)+ 2n\pi\quad (n\in \Z) \] and \[ \arg\left(z_1\right) =\theta_1 + 2k\pi\quad (k\in \Z) \] Now, since \[ \left(\theta_1 + \theta_2\right)+ 2n\pi = \theta_1 + 2k\pi + \left[\theta_2 + 2(n-k)\pi\right], \] equation (\ref{arg-product}) is satisfied when we choose the value \[ \arg\left(z_2\right) =\theta_2 + 2(n-k)\pi. \] The verification when values of $\arg\left(z_1z_2\right)$ and $\arg\left(z_2\right)$ are specified follows by symmetry.

Statement (\ref{arg-product}) is sometimes valid when $\arg$ is replaced everywhere by $\Arg.$ However, as the following example illustrates, that is not always the case.

Example: Consider $z_1 = -1$ and $z_2=i.$ Then \[ \Arg\left(z_1z_2\right) = \Arg(-i)=-\frac{\pi}{2} \] but \[ \Arg\left(z_1\right)+\Arg\left(z_2\right) = \pi +\frac{\pi}{2}= \frac{3\pi}{2}. \] However, if we take the values of $\arg (z_1)$ and $\arg (z_2)$ just used and select the value \[ \Arg\left(z_1z_2\right) + 2\pi = -\frac{\pi}{2}+ 2\pi = \frac{3\pi}{2} \] of $\arg (z_1z_2),$ we find that equation (\ref{arg-product}) is satisfied.

Exercise: Recall that $\dfrac{z_1}{z_2}=z_1z_2^{-1}.$ Use statemet (\ref{arg-product}) to show that \[ \arg\left(\frac{z_1}{z_2}\right) = \arg\left(z_1\right) -\arg\left(z_2\right). \]