Polar (Mod–Arg) Form

Any non-zero complex number $z = a + bi$ can be described by its position vector $\vec{OZ}$: a length $r = |z|$ and an angle $\theta = \arg z$ measured anticlockwise from the positive real axis. Reading the components of $\vec{OZ}$ along the two axes gives:

$a = r\cos\theta$  and  $b = r\sin\theta$, so $z = a + bi = r\cos\theta + i\,r\sin\theta = r(\cos\theta + i\sin\theta)$.

• $r = |z| = \sqrt{a^2 + b^2}$ — the modulus is the distance from the origin to $z$.
• $\theta = \arg z$ — the angle of the position vector from the positive real axis. NESA convention: take $\theta \in (-\pi, \pi]$ (the principal argument).
• $\tan\theta = b/a$ — but only after checking the quadrant; otherwise the calculator will return an angle that is $\pi$ off.

Worked through the hook: $z = 1 + \sqrt{3}\,i$. Then $r = \sqrt{1 + 3} = 2$. The point $(1, \sqrt{3})$ is in quadrant 1 (both coordinates positive), so $\theta$ is acute with $\tan\theta = \sqrt{3}/1 = \sqrt{3}$, giving $\theta = \pi/3$. So $z = 2(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3})$.

$z = r(\cos\theta + i\sin\theta)$, $r = |z|$, $\theta = \arg z$ · $a = r\cos\theta$, $b = r\sin\theta$ · $r = \sqrt{a^2 + b^2}$; $\tan\theta = b/a$ with quadrant check · Principal argument: $\theta \in (-\pi, \pi]$

Pause — copy polar form $z = r(\cos\theta+i\sin\theta)$, the component formulas $a = r\cos\theta$, $b = r\sin\theta$, and the principal argument range $(-\pi,\pi]$ into your book.

We just saw that polar form $z = r(\cos\theta + i\sin\theta)$ uses $r = |z|$ and $\theta = \arg z$, with $a = r\cos\theta$ and $b = r\sin\theta$. That raises a question: how do we reliably convert in both directions — especially avoiding the $\arctan$ quadrant trap? This card answers it → sketch first, find the reference angle $\alpha = \arctan(|b/a|)$, then choose $\theta \in (-\pi,\pi]$ by quadrant.

Rectangular to polar ($a + bi \to r(\cos\theta + i\sin\theta)$):

1. Sketch $(a, b)$ on the Argand diagram to see the quadrant.
2. Compute $r = \sqrt{a^2 + b^2}$.
3. Find the reference angle $\alpha$ from $\tan\alpha = |b/a|$.
4. Place $\theta$ in the correct quadrant in $(-\pi, \pi]$: Q1 gives $\theta = \alpha$; Q2 gives $\theta = \pi - \alpha$; Q3 gives $\theta = -(\pi - \alpha)$; Q4 gives $\theta = -\alpha$.

Polar to rectangular ($r(\cos\theta + i\sin\theta) \to a + bi$): expand directly: $a = r\cos\theta$, $b = r\sin\theta$. No quadrant fuss needed.

Rect $\to$ polar: sketch, find $r$, find reference angle, choose quadrant, write $\theta \in (-\pi, \pi]$ · Polar $\to$ rect: $a = r\cos\theta$, $b = r\sin\theta$, then $z = a + bi$ · $\tan^{-1}$ alone gives Q1/Q4 — manually fix for Q2/Q3 by adding/subtracting $\pi$ · Reference angle $\alpha$ is always acute and computed from $|b/a|$

Pause — copy the two-direction conversion procedure: rect→polar (sketch, $r$, reference angle, quadrant choice) and polar→rect ($a = r\cos\theta$, $b = r\sin\theta$, write $a+bi$), and the warning that $\arctan$ alone gives Q1/Q4 only into your book.