Determination of the argument of a complex number Consider a nonzero complex number \(z=x+\mathrm{i}\,y\) that can be written in the form \(z = r\,(\cos\varphi+\mathrm{i}\,\sin\varphi)\). For the relationship between \(x\), \(y\), \(r\) and \(\varphi\) holds in general: \[\begin{aligned}x=r\,\cos\varphi &,\quad y=r\,\sin\varphi\\ r=\sqrt{x^2+y^2} &,\quad \tan\varphi=\frac{y}{x}\end{aligned}\] Often one concludes that the value of the argument is given by the formula \[\arg(z)=\arctan \left(\frac{\mathrm{Im}(z)}{\mathrm{Re}(z)}\right)\tiny.\] This is only true if \(\mathrm{Re}(z)\gt 0\) and in other cases not (check this yourself after for the complex number \(-1-\mathrm{i}\)).

The argument is only determined modulo \(2\pi\). In practice the above equations are sufficient to determine \(\varphi\), but there's a snake in the grass. If \(x=y=0\), then \(\varphi\) is not defined. If \(x=0, y>0\), then \(\varphi =\tfrac{1}{2}\!\pi\), and if \(x=0, y<0\), then \(\varphi =-\tfrac{1}{2}\!\pi\). In all other cases, you can calculate \(\varphi\) using the arctangent (inverse function of tangent), but you have to into account that the artangent always yields a number between \(-\tfrac{1}{2}\!\pi\) and \(\tfrac{1}{2}\!\pi\) . If \(x<0\), that is, when the complex number has a negative real part, you have to to add or subtract \(\pi\) to/from the outcome of \(\arctan\left(\dfrac{y}{x}\right)\).

In summary, we have therefore: \[\begin{aligned}\text{if } x=0, y>0\;&\text{then }\varphi=\tfrac{1}{2}\!\pi+2k\pi\\ \text{if } x=0, y<0\;&\text{then }\varphi=-\tfrac{1}{2}\!\pi+2k\pi\\ \text{if } x<0, y>0\;&\text{then }\varphi=\arctan\left(\dfrac{y}{x}\right)+\pi+2k\pi\\ \text{if } x<0, y<0\;&\text{then }\varphi=\arctan\left(\dfrac{y}{x}\right)-\pi+2k\pi\\ \text{if } x>0\;&\text{then }\varphi=\arctan\left(\dfrac{y}{x}\right)+2k\pi\end{aligned}\] for some integer \(k\), because the argument is only determined up to an integer multiple of \(2\pi\). We can always select the value in the interval \((-\pi,\pi]\); We then speak of the principal value of the argument and denote this, as with complex numbers on the unit circle, with \(\mathbf{Arg}(z)\). In the above formulation, we have written down these formulas in such a way that the principal value is obtained by omitting the modulo-term with the multiple of \(2\pi\).

\[\mathrm{Arg}(z)=\begin{cases}\pi&\text{if } z\text{ lies on the negative real axis}\\ 2\arctan\left(\dfrac{\mathrm{Im}(z)}{|z|+\mathrm{Re} (z)}\right)&\text{otherwise }\end{cases}\]

Calculation Schedule for the principle value of the argument The quadrants of the complex plane are numbered in the figure below.

The following algorithms show how to compute the principal value \(\mathrm{Arg}(z)\) of the complex number \(z=x+y\,\ii\) on the basis of the position of the corresponding point \((x,y)\) in the complex plane. \[\begin{array}{|c|c|c|} \hline \textit{quadrant} & \textit{sign of }x\textit{ and }y & \mathrm{Arg}(z)\\ \hline\hline \mathrm{I} & x>0,\quad y>0 & \arctan(y/x) \\ \hline \mathrm{II} & x<0,\quad y>0 & \pi+\arctan(y/x) \\ \hline \mathrm{III} & x<0,\quad y<0 & -\pi+ \arctan(y/x)\\ \hline \mathrm{IV} & x>0,\quad y<0 & \arctan(y/x) \\ \hline \end{array}\] For points on the edges of the quadrants you can use the following scheme. \[\begin{array}{|c|c|c|c|} \hline \textit{edge} & \textit{type of complex number }z & \textit{conditions for }x\textit{ and }y & \mathrm{Arg}(z)\\ \hline\hline \mathrm{IV/I} & \text{real positive} & x>0,\quad y=0 & 0 \\ \hline \mathrm{I/II} & \text{pure imaginary with Im}(z)>0 & x=0,\quad y>0 & \tfrac{1}{2}\pi \\ \hline \mathrm{II/III} & \text{real positive} & x<0,\quad y=0 & \pi\\ \hline \mathrm{III/IV} & \text{pure imaginary with Im}(z)<0 & x=0,\quad y<0 & -\tfrac{1}{2}\pi\\ \hline \mathrm{origin} & \text{zero}& x=0,\quad y=0 & \text{undetermined} \\ \hline \end{array}\]