Q.Find the argument of $z=1-i$, giving your answer in the range $(-\pi,\pi]$.

For a complex number $z=a+ib$ represented by the point $P(a,b)$ in the Argand plane, the modulus $|z|=r=\sqrt{a^2+b^2}$ is the straight-line distance $OP$ from the origin to $P$ (found via Pythagoras on the right triangle formed by the point and the two axes), and the argument $\theta=\arg(z)$ is the angle that $OP$ makes with the positive real axis, satisfying $\cos\theta=a/r$, $\sin\theta=b/r$, and $\tan\theta=b/a$ when $a\neq0$. Because the inverse tangent function alone only ever returns an angle in $(-\pi/2,\pi/2)$, it cannot on its own distinguish a point in quadrant I from one in quadrant III (or quadrant II from quadrant IV), so finding the argument in the standard range $0\le\theta<2\pi$ requires checking which quadrant (or axis) the point $(a,b)$ actually lies in and adding the appropriate correction: no correction in quadrant I, $+\pi$ in quadrants II and III, and $+2\pi$ in quadrant I …