Argument (complex analysis)

Principal value

Because a complete rotation around the origin leaves a complex number unchanged, there are many choices which could be made for ${\displaystyle \phi }$ by circling the origin any number of times. This is shown in figure 2, a representation of the multi-valued (set-valued) function ${\displaystyle f(x,y)=\arg (x+iy)}$, where a vertical line (not shown in the figure) cuts the surface at heights representing all the possible choices of angle for that point.

When a well-defined function is required, then the usual choice, known as the principal value, is the value in the open-closed interval (−π, π] radians, that is from −π to π radians excluding −π radians itself (equiv., from −180 to +180 degrees, excluding −180° itself). This represents an angle of up to half a complete circle from the positive real axis in either direction.

Some authors define the range of the principal value as being in the closed–open interval [0, 2π).

Notation

The principal value sometimes has the initial letter capitalized, as in Arg z, especially when a general version of the argument is also being considered. Note that notation varies, so arg and Arg may be interchanged in different texts.

The set of all possible values of the argument can be written in terms of Arg as: $${\displaystyle \arg (z)=\{\mathrm{Arg}(z)+2\pi n\mid n\in \mathbb{Z}\}.}$$

Computing from the real and imaginary part

If a complex number is known in terms of its real and imaginary parts, then the function that calculates the principal value Arg is called the two-argument arctangent function, atan2: $${\displaystyle \mathrm{Arg}(x+iy)=\mathrm{atan2}(y,x)}$$ The atan2 function is available in the math libraries of many programming languages, sometimes under a different name, and usually returns a value in the range (−π, π]. See atan2 for further detail and alternative implementations, such as $${\displaystyle \mathrm{atan2}(y,x)=2\mathrm{atan}\frac{y}{\sqrt{x^2+y^2}+x},}$$ which works except when ⁠${\displaystyle y=0}$⁠ and ⁠${\displaystyle x\le 0}$⁠, in which case the value is instead π if ⁠${\displaystyle x<0}$⁠ or undefined when ⁠${\displaystyle x=0}$⁠.

Realizations of the function in computer languages

Wolfram language (Mathematica)

In Wolfram language, there's Arg[z]:

Arg[x + y I] ${\displaystyle =\{\begin{array}{ll}0& \text{if }x=0\text{ and }y=0,\\ \text{undefined}& \text{if }|x|=\mathrm{\infty }\text{ and }|y|=\mathrm{\infty },\\ 0& \text{if }x=\mathrm{\infty }\text{ and }|y|\ne \mathrm{\infty },\\ \pi & \text{if }x=-\mathrm{\infty }\text{ and }|y|\ne \mathrm{\infty },\\ \pm \frac{\pi }{2}& \text{if }y=\pm \mathrm{\infty }\text{ and }|x|\ne \mathrm{\infty },\\ \mathrm{Arg}(x+yi)& \text{otherwise}.\end{array}}$

or using the language's ArcTan:

Arg[x + y I] ${\displaystyle =\{\begin{array}{ll}0& \text{if }x=0\text{ and }y=0,\\ \text{ArcTan[x, y]}& \text{otherwise}.\end{array}}$

ArcTan[x, y] is ${\displaystyle \mathrm{atan2}(y,x)}$ extended to work with infinities. ArcTan[0, 0] is Indeterminate (i.e. it is still defined), while ArcTan[Infinity, -Infinity] doesn't return anything (i.e. it is undefined).

Maple

Maple's argument(z) behaves the same as Arg[z] in Wolfram language, except that argument(z) also returns ${\displaystyle \pi }$ if z is the special floating-point value −0.. Also, Maple doesn't have ${\displaystyle \mathrm{atan2}}$.

MATLAB

MATLAB's angle(z) behaves the same as Arg[z] in Wolfram language, except that it is

$${\displaystyle \{\begin{array}{ll}\frac{1\pi }{4}& \text{if }x=\mathrm{\infty }\text{ and }y=\mathrm{\infty },\\ -\frac{1\pi }{4}& \text{if }x=\mathrm{\infty }\text{ and }y=-\mathrm{\infty },\\ \frac{3\pi }{4}& \text{if }x=-\mathrm{\infty }\text{ and }y=\mathrm{\infty },\\ -\frac{3\pi }{4}& \text{if }x=-\mathrm{\infty }\text{ and }y=-\mathrm{\infty }.\end{array}}$$

Unlike in Maple and Wolfram language, MATLAB's atan2(y, x) is equivalent to angle(x + y*1i). That is, atan2(0, 0) is ${\displaystyle 0}$.

Identities

One of the main motivations for defining the principal value Arg is to be able to write complex numbers in modulus-argument form. Hence for any complex number z, $${\displaystyle z=\left|z\right|e^{i\mathrm{Arg}z}.}$$

This is only really valid if z is non-zero, but can be considered valid for z = 0 if ${\displaystyle \mathrm{Arg}0}$ is considered as an indeterminate form—rather than as being undefined.

Some further identities follow. If z₁ and z₂ are two non-zero complex numbers, then $${\displaystyle \begin{array}{cc}\mathrm{Arg}(z_1{z}_2)& \equiv \mathrm{Arg}{z}_1+\mathrm{Arg}{z}_2(\mathrm{mod}\left(-\pi ,\pi \right]),\\ \mathrm{Arg}\frac{z_1}{z_2}& \equiv \mathrm{Arg}{z}_1-\mathrm{Arg}{z}_2(\mathrm{mod}\left(-\pi ,\pi \right]),\end{array}}$$ where ${\displaystyle (\mathrm{mod}\left(-\pi ,\pi \right])}$ means to add or subtract any integer multiple of 2π if necessary to bring the value into the interval of (−π, π] radians. In particular, $${\displaystyle \mathrm{Arg}(-z)\equiv \mathrm{Arg}z+\pi \{\begin{array}{ll}1,& \mathrm{Im}z<0,\\ \mathrm{sgn}z,& \mathrm{Im}z=0,\\ -1,& \mathrm{Im}z>0.\end{array}}$$

If z ≠ 0 and n is any integer, then $${\displaystyle \mathrm{Arg}\left(z^n\right)\equiv n\mathrm{Arg}z(\mathrm{mod}\left(-\pi ,\pi \right]).}$$

Example

$${\displaystyle \mathrm{Arg}\left(\frac{-1-i}{i}\right)=\left(\mathrm{Arg}(-1-i)-\mathrm{Arg}i=-\frac{3\pi }{4}-\frac{\pi }{2}=-\frac{5\pi }{4}\right)(\mathrm{mod}\left(-\pi ,\pi \right])=\frac{3\pi }{4}.}$$

Using the complex logarithm

From ${\displaystyle z=|z|e^{i\mathrm{Arg}z}}$, we get ${\displaystyle i\mathrm{Arg}z=\ln \frac{z}{|z|}}$, alternatively ${\displaystyle \mathrm{Arg}(z)=\mathrm{Im}\ln \frac{z}{|z|}=\mathrm{Im}\ln z}$. As we are taking the imaginary part, any normalization by a real scalar will not affect the result. This is useful when one has the complex logarithm available.

Differentiation

The argument function is not complex differentiable, as the limit $${\displaystyle \underset{z\to z_0}{lim}\frac{\arg z-\arg z_0}{z-z_0}}$$ does not exist for any ${\displaystyle z_0\in \mathbb{C}}$ and on any branch (indeed, for any chosen branch the numerator takes the same value). However, the Wirtinger derivatives may be applied to it. Begin from the logarithmic identity $${\displaystyle \ln z=\mathrm{Ln}|z|+i\arg z.}$$ Use that ${\displaystyle |z{|}^2=z\overline{z}}$: $${\displaystyle \ln z=\frac{1}{2}\mathrm{Ln}z+\frac{1}{2}\mathrm{Ln}\overline{z}+i\arg z.}$$ Performing different branch cuts lets one apply the derivatives to all points of the principal logarithm in two goes. First, apply ${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }z}}$: $${\displaystyle \frac{1}{z}=\frac{1}{2}\frac{1}{z}+0+i\frac{\mathrm{\partial }}{\mathrm{\partial }z}\arg z.}$$ Rearranging gives ${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }z}\arg z=\frac{-i}{2z}}$. Now apply ${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }\overline{z}}}$: $${\displaystyle 0=0+\frac{1}{2}\frac{1}{\overline{z}}+i\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{z}}\arg z.}$$ This shows that ${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }\overline{z}}\arg z=\frac{i}{2\overline{z}}}$. As both Wirtinger derivatives exist, one can then apply the Dolbeault derivatives, $${\displaystyle \mathrm{\partial }\arg (z)=\frac{-i}{2z}dz,}$$ $${\displaystyle \overline{\mathrm{\partial }}\arg (z)=\frac{i}{2\overline{z}}d\overline{z}.}$$ This then means the exterior derivative exists, given by their sum, $${\displaystyle d\arg (z)=\frac{-i}{2z}dz+\frac{i}{2\overline{z}}d\overline{z}.}$$ This is an important differential form, as it is precisely the generator of the first de Rham cohomology of the circle, ${\displaystyle H_{dR}^1(S^1)}$. In other words, $${\displaystyle d\arg (z)=\frac{-i}{2z}dz+\frac{i}{2\overline{z}}d\overline{z}=\frac{xdy-ydx}{x^2+y^2},}$$ making this form closed but not exact in the usual sense (a byproduct of being multivalued): $${\displaystyle \underset{S^1}{\oint }d\arg (z)=2\pi >0}$$ with the standard orientation.