Polar form — r(cos θ + i sin θ) = r·cis θ: Worked examples

We only need three tools, all built in the parent note and its prerequisites Modulus and Argument and Argand Diagram:

$r = a^{2} + b^{2}$ — the arrow's length (never negative).
reference angle $α = tan^{- 1} \frac{b}{a}$ — the acute angle the arrow makes with the real axis, ignoring direction.
a quadrant fix that turns $α$ into the true argument $θ$ .

Figure — Polar form — r(cos θ + i sin θ) = r·cis θ

The scenario matrix

Every complex number $z = a + bi$ falls into exactly one row below. Our examples cover all of them.

Cell	Case class	Signs $(a, b)$	Argument rule	Example
C1	Quadrant I	$(+, +)$	$θ = α$	Ex 1
C2	Quadrant II	$(-, +)$	$θ = π - α$	Ex 2
C3	Quadrant III	$(-, -)$	$θ = - (π - α)$	Ex 3
C4	Quadrant IV	$(+, -)$	$θ = - α$	Ex 4
C5	On an axis (degenerate)	one of $a, b = 0$	read off directly	Ex 5
C5b	Origin (undefined arg)	$a = b = 0$	argument undefined	Ex 5b
C6	Polar → Cartesian	given $r, θ$	$a = r cos θ, b = r sin θ$	Ex 6
C7	Limiting behaviour	$a \to 0^{+}$ vs $a \to 0^{-}$	watch the jump	Ex 7
C8	Word problem (rotation)	multiply moduli, add args	Ex 8
C9	Exam twist (power via polar)	De Moivre setup	Ex 9

The worked examples

Quadrant II

Worked example Ex 2 (cell C2) —

$z = - 3 + i$ Forecast: negative real, positive imaginary ⇒ arrow points up-and-left. The true angle is bigger than $\frac{π}{2}$ . Guess it.

Step 1 — length. $r = (- 3)^{2} + 1^{2} = 3 + 1 = 2$ . Why this step? Squaring kills the minus sign, so the length is honest about distance regardless of direction.

Step 2 — reference angle. $α = tan^{- 1} \frac{1}{- 3} = tan^{- 1} \frac{1}{3} = \frac{π}{6}$ . Why this step? We use the magnitude $\frac{1}{- 3} = \frac{1}{3}$ (bars kept!) so $α$ stays a clean acute angle; sign handling comes next.

Step 3 — quadrant fix. $(-, +)$ is Quadrant II, so $θ = π - α = π - \frac{π}{6} = \frac{5 π}{6}$ . Why this step? The naive $tan^{- 1}$ would report $- \frac{π}{6}$ (an arrow in Q IV). Subtracting $α$ from $π$ reflects that arrow across the imaginary axis to where it actually is. See the second wedge in the figure.

Result: $z = 2 cis \frac{5 π}{6}$ .

Verify: $2 cos \frac{5 π}{6} = 2 \cdot (- \frac{3}{2}) = - 3$ ✓ and $2 sin \frac{5 π}{6} = 2 \cdot \frac{1}{2} = 1$ ✓.

Quadrant III

Worked example Ex 3 (cell C3) —

$z = - 1 - i$ Forecast: both parts negative ⇒ arrow points down-and-left. Since Q III sits below the axis, the principal argument (recall: kept in $(- π, π]$ ) will be negative, past $- \frac{π}{2}$ . Guess it.

Step 1 — length. $r = (- 1)^{2} + (- 1)^{2} = 2$ . Why this step? Same as always — Pythagoras on the two legs fixes the arrow's length before we worry about direction.

Step 2 — reference angle. $α = tan^{- 1} \frac{- 1}{- 1} = tan^{- 1} 1 = \frac{π}{4}$ . Why this step? The bars turn $\frac{- 1}{- 1}$ into $1$ , giving a clean acute wedge; the true direction is handled by the quadrant next.

Step 3 — quadrant fix. $(-, -)$ is Quadrant III. Principal value: $θ = - (π - α) = - (π - \frac{π}{4}) = - \frac{3 π}{4}$ . Why this step? Q III sits "opposite" Q I. We could write $π + α = \frac{5 π}{4}$ , but that exceeds $π$ and so is outside the principal window $(- π, π]$ ; subtracting a full turn gives the principal value $- \frac{3 π}{4}$ . Both name the same arrow — see the down-left wedge in the figure.

Result: $z = 2 cis (- \frac{3 π}{4})$ .

Verify: $2 cos (- \frac{3 π}{4}) = 2 \cdot (- \frac{1}{2}) = - 1$ ✓ and $2 sin (- \frac{3 π}{4}) = - 1$ ✓.

Quadrant IV

Worked example Ex 4 (cell C4) —

$z = 1 - i 3$ Forecast: positive real, negative imaginary ⇒ arrow points down-and-right. Here (and only here) the raw calculator $tan^{- 1}$ gives the right sign already.

Step 1 — length. $r = 1^{2} + (- 3)^{2} = 1 + 3 = 2$ . Why this step? Pythagoras on legs $1$ and $3$ ; the square erases the minus, so $r = 2$ regardless of the arrow dipping downward.

Step 2 — reference angle. $α = tan^{- 1} \frac{- 3}{1} = tan^{- 1} 3 = \frac{π}{3}$ . Why this step? The bars send $\frac{- 3}{1}$ to $3$ , so the wedge is the clean acute angle $\frac{π}{3}$ ; the downward direction is fixed in the next step.

Step 3 — quadrant fix. $(+, -)$ is Quadrant IV, so $θ = - α = - \frac{π}{3}$ . Why this step? The arrow dips below the real axis; measuring anticlockwise from the positive real axis means going clockwise a little, i.e. a negative angle — and $- \frac{π}{3}$ is safely inside $(- π, π]$ .

Result: $z = 2 cis (- \frac{π}{3})$ .

Verify: $2 cos (- \frac{π}{3}) = 2 \cdot \frac{1}{2} = 1$ ✓ and $2 sin (- \frac{π}{3}) = 2 \cdot (- \frac{3}{2}) = - 3$ ✓.

The degenerate cases (on an axis)

Worked example Ex 5 (cell C5) — the four points on the axes:

$5, 5 i, - 5, - 5 i$ Forecast: these sit exactly ON a line, so $α$ is $0$ or the arrow is vertical and $\frac{b}{a}$ divides by zero. You cannot blindly use the quadrant table — you read the angle straight off the axis.

Step 1 — $z = 5$ (positive real axis). $r = 5$ , arrow points right, $θ = 0$ . ⇒ $5 cis 0$ . Why this step? $b = 0$ so $\frac{b}{a} = 0$ gives $α = 0$ ; the arrow lies along the very axis it measures from, so no adjustment is possible or needed.

Step 2 — $z = 5 i$ (positive imaginary axis). $a = 0$ , so $\frac{b}{a}$ is undefined — you must NOT feed it to $tan^{- 1}$ . The arrow points straight up: $θ = \frac{π}{2}$ . ⇒ $5 cis \frac{π}{2}$ . Why this step? Division by zero is a signpost: "vertical arrow, angle is $\pm \frac{π}{2}$ , decide by the sign of $b$ ." Here $b > 0$ so we take $+ \frac{π}{2}$ .

Step 3 — $z = - 5$ (negative real axis). Arrow points left: $θ = π$ . ⇒ $5 cis π$ . Why this step? The negative real axis is the seam of the window $(- π, π]$ ; by convention we take the included endpoint $+ π$ , not $- π$ .

Step 4 — $z = - 5 i$ (negative imaginary axis). Arrow points down: principal value $θ = - \frac{π}{2}$ . ⇒ $5 cis (- \frac{π}{2})$ . Why this step? Vertical arrow again, but now $b < 0$ , so we take the $- \frac{π}{2}$ branch, which lies inside $(- π, π]$ .

Verify: $5 cos \frac{π}{2} = 0, 5 sin \frac{π}{2} = 5$ (gives $5 i$ ) ✓; $5 cos π = - 5$ ✓; $5 sin (- \frac{π}{2}) = - 5$ (gives $- 5 i$ ) ✓.

The origin — where polar form breaks

Worked example Ex 5b (cell C5b) —

$z = 0$ Forecast: the arrow has no length at all. If there's no arrow, there's no direction — so the angle should be undefined, not zero. Predict what $r$ and $θ$ do.

Step 1 — modulus. $r = 0^{2} + 0^{2} = 0$ . ⇒ the length is genuinely zero. Why this step? Pythagoras still works and gives a perfectly good answer: the point sits exactly on the origin, distance $0$ .

Step 2 — argument. Both $a = 0$ and $b = 0$ , so $\frac{b}{a} = \frac{0}{0}$ — not just undefined, but indeterminate. Geometrically there is no arrow to point anywhere. ⇒ $ar g 0$ is undefined. Why this step? Every non-zero direction $θ$ satisfies $0 = 0 \cdot cis θ$ equally well, so no single principal value can be chosen. The definition of polar form in the parent note deliberately said "any non-zero complex number".

Result: $z = 0$ has $r = 0$ and no argument. You may write $0 = 0 cis θ$ for any $θ$ , but you never assign it a specific one.

Verify: for any $θ$ , $0 \cdot cos θ = 0$ and $0 \cdot sin θ = 0$ , recovering $0 + 0 i$ — confirming the angle is free/undefined.

Polar → Cartesian (the reverse trip)

Limiting behaviour — the jump across the negative axis

Worked example Ex 7 (cell C7) — what happens to

$ar g z$ as the arrow crosses the negative real axis? Take a point near the negative real axis just above it, then just below it: $z_{+} = - 1 + 0.01 i$ and $z_{-} = - 1 - 0.01 i$ .

Forecast: the two points are almost on top of each other, so you'd expect nearly equal arguments. But the principal argument lives in $(- π, π]$ , and the negative real axis is where that window wraps around. Predict a sudden jump of about $2 π$ .

Step 1 — above the axis. $z_{+}$ is in Q II, so $θ_{+} = π - tan^{- 1} \frac{0.01}{- 1} = π - tan^{- 1} (0.01) \approx π - 0.01 \approx 3.1316$ . Why this step? Q II's rule is $θ = π - α$ , and the bars make the reference angle $α = tan^{- 1} \frac{0.01}{- 1} = tan^{- 1} (0.01)$ , a tiny positive number.

Step 2 — below the axis. $z_{-}$ is in Q III, so (principal value) $θ_{-} = - (π - tan^{- 1} \frac{- 0.01}{- 1}) = - (π - 0.01) \approx - 3.1316$ . Why this step? Q III's rule is $θ = - (π - α)$ with the same tiny $α = tan^{- 1} (0.01)$ ; being below the axis forces the negative branch of the window.

Result: $θ_{+} - θ_{-} \approx 2 π - 0.02 \approx 6.263$ — a near-full-turn jump for two points a hair apart. What it means: the principal argument is discontinuous across the negative real axis. That's why $ar g (- 1) = + π$ (we take the $+ π$ end of the window), not $- π$ .

Verify: compute $θ_{+} - θ_{-}$ and confirm it is close to $2 π$ (within $0.03$ ).

Word problem — rotation of a drone

Worked example Ex 8 (cell C8) — spinning a position vector

A drone sits at position $z = 4 + 3 i$ (units: metres east and north). It flies in a circle about the origin, turning anticlockwise by $9 0^{\circ} = \frac{π}{2}$ . Where does it end up?

Forecast: rotating keeps the distance from the origin the same and just adds to the angle. So the length stays $5$ m; only the direction changes.

Step 1 — polar form of the start. $r = 4^{2} + 3^{2} = 25 = 5$ . Q I, so $θ = tan^{- 1} \frac{3}{4} \approx 0.6435$ rad. So $z = 5 cis 0.6435$ . Why this step? Rotation is "add to the angle," which only makes sense once the number is written in polar form.

Step 2 — rotate. A rotation by $\frac{π}{2}$ is multiplication by $cis \frac{π}{2} = i$ (multiply moduli, add arguments — the parent note's payoff). New angle $= 0.6435 + \frac{π}{2} \approx 2.2143$ , modulus still $5$ . Why this step? Turning the arrow $\frac{π}{2}$ anticlockwise is exactly adding $\frac{π}{2}$ to its argument while leaving its length untouched.

Step 3 — back to Cartesian. Easiest: multiply directly. $i \cdot (4 + 3 i) = 4 i + 3 i^{2} = - 3 + 4 i$ . Why this step? We want the drone's new east/north coordinates, and $i \cdot z$ delivers them without re-doing trig.

Result: the drone is at $- 3 + 4 i$ , i.e. $3$ m west and $4$ m north.

Verify: $(- 3)^{2} + 4^{2} = 5$ ✓ (distance preserved). And $ar g (- 3 + 4 i) = π - tan^{- 1} \frac{4}{- 3} \approx 2.2143$ ✓ (matches Step 2).

Exam twist — a power via polar form

Worked example Ex 9 (cell C9) — compute

$(1 + i)^{8}$ Forecast: multiplying $1 + i$ by itself eight times in Cartesian form is a nightmare. In polar form, powers just multiply the angle (this is De Moivre's Theorem). Since $ar g (1 + i) = \frac{π}{4}$ , eight copies give $8 \cdot \frac{π}{4} = 2 π$ — a full turn back to the positive real axis. So the answer should be a positive real number.

Step 1 — polar form. $1 + i = 2 cis \frac{π}{4}$ (from Ex 1's pattern). Why this step? Powers are cheap in polar form but expensive in Cartesian form, so we convert first.

Step 2 — raise to the 8th. $(1 + i)^{8} = (2)^{8} cis (8 \cdot \frac{π}{4}) = 16 cis 2 π$ . Why this step? De Moivre: $(r cis θ)^{n} = r^{n} cis (n θ)$ . The length is raised to the power; the angle is multiplied by $n$ .

Step 3 — simplify. $cis 2 π = cos 2 π + i sin 2 π = 1 + 0 i = 1$ . So $(1 + i)^{8} = 16$ . Why this step? $2 π$ is a full turn, which brings the arrow back to the positive real axis — exactly the real answer the forecast predicted.

Result: $(1 + i)^{8} = 16$ .

Verify: $(2)^{8} = 2^{4} = 16$ , and expanding directly $(1 + i)^{2} = 2 i$ , $(2 i)^{2} = - 4$ , $(- 4)^{2} = 16$ ✓.

Recall

Recall In which single quadrant does the raw calculator

$tan^{- 1} (b / a)$ need no correction (besides Q I)? Quadrant IV. There $θ = - α$ , which is exactly what $tan^{- 1}$ of a negative ratio returns. Every other quadrant needs a fix (Q II adds/subtracts $π$ , Q III too, and the axes bypass the formula).

Recall What is the argument and modulus of

$z = 0$ ? The modulus is $r = 0$ , but the argument is undefined — a point with no length has no direction. Polar form is only defined for non-zero $z$ .

Recall Why does

$ar g z$ jump by nearly $2 π$ across the negative real axis (Ex 7)? Because the principal argument is confined to $(- π, π]$ ; the negative real axis is the seam where the window wraps. Points just above sit near $+ π$ , points just below sit near $- π$ .

Recall What is the fast way to compute

$(1 + i)^{8}$ ? Convert to polar $2 cis \frac{π}{4}$ , apply De Moivre: modulus $(2)^{8} = 16$ , angle $8 \cdot \frac{π}{4} = 2 π$ , so answer $= 16$ .

Connections

Argand Diagram — the plane each arrow lives in.
Modulus and Argument — the two numbers we extract in every example.
Multiplication and Division of Complex Numbers — the rotation trick in Ex 8.
De Moivre's Theorem — the power trick in Ex 9.
Euler's Formula — the deeper reason angles add.
Roots of Complex Numbers — the reverse of Ex 9.