Argument of z in Different Quadrants/Axes

Argument of z in Different Quadrants/Axes

Because $\theta=\tan^{-1}(b/a)$ alone only ever gives a value in $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, correctly finding $\arg(z)$ in the standard range $0\le\theta<2\pi$ requires checking which quadrant (or which axis) the point $z=a+ib$ actually lies in, and adding the matching correction. The accompanying table (built from Figs. corresponding to each case) is the complete reference used throughout the rest of the chapter:

• $a>0,\ b=0$ (positive real axis, e.g. $z=3$): $\theta=0$.
• $a>0,\ b>0$ (Quadrant I, e.g. $z=1+i$): $\theta=\tan^{-1}\left(\dfrac{b}{a}\right)$, with $0<\theta<\dfrac{\pi}{2}$. For $z=1+i$: $\theta=\tan^{-1}(1)=\dfrac{\pi}{4}$.
• $a=0,\ b>0$ (positive imaginary axis, e.g. $z=5i$): $\theta=\dfrac{\pi}{2}$.
• $a<0,\ b>0$ (Quadrant II, e.g. $z=-\sqrt3+i$): $\theta=\tan^{-1}\left(\dfrac{b}{a}\right)+\pi$, with $\dfrac{\pi}{2}<\theta<\pi$. For $z=-\sqrt3+i$: $\theta=\tan^{-1}\left(-\dfrac{1}{\sqrt3}\right)+\pi=-\dfrac{\pi}{6}+\pi=\dfrac{5\pi}{6}$.
• $a<0,\ b=0$ (negative real axis, e.g. $z=-6$): $\theta=\pi$.
• $a<0,\ b<0$ (Quadrant III, e.g. $z=-1-\sqrt3i$): $\theta=\tan^{-1}\left(\dfrac{b}{a}\right)+\pi$, with $\pi<\theta<\dfrac{3\pi}{2}$. For $z=-1-\sqrt3i$: $\theta=\tan^{-1}(\sqrt3)+\pi=\dfrac{\pi}{3}+\pi=\dfrac{4\pi}{3}$.
• $a=0,\ b<0$ (negative imaginary axis, e.g. $z=-2i$): $\theta=\dfrac{3\pi}{2}$.
• $a>0,\ b<0$ (Quadrant IV, e.g. $z=1-i$): $\theta=\tan^{-1}\left(\dfrac{b}{a}\right)+2\pi$, with $\dfrac{3\pi}{2}<\theta<2\pi$. For $z=1-i$: $\theta=\tan^{-1}(-1)+2\pi=-\dfrac{\pi}{4}+2\pi=\dfrac{7\pi}{4}$. …