Exponential form z = re^(iθ): Worked examples

You already met the machine $z = r e^{i θ}$ on the parent note. This page is the firing range: we walk through every kind of case that a complex number can hand you when you try to write it as $r e^{i θ}$ — every quadrant, the axes, the degenerate zero, huge powers, division, and a real-world twist. If a scenario exists, it lives in the matrix below and gets a fully worked example.

The scenario matrix

Below, $x$ is the real part, $y$ the imaginary part. "Ref" means the reference angle $α = arctan \frac{y}{x}$ — the acute angle the arrow makes with the real axis, always positive.

Cell	Signs of $(x, y)$	Where the arrow points	Correct $θ$	Example
A Quadrant I	$x > 0, y > 0$	up-right	$α$	Ex 1
B Quadrant II	$x < 0, y > 0$	up-left	$π - α$	Ex 2
C Quadrant III	$x < 0, y < 0$	down-left	$α - π$	Ex 3
D Quadrant IV	$x > 0, y < 0$	down-right	$- α$	Ex 4
E On an axis	one of $x, y = 0$	along an axis	$0, \frac{π}{2}, π, - \frac{π}{2}$	Ex 5
F Degenerate	$x = 0, y = 0$	nowhere	undefined	Ex 5
G Big power	any	—	De Moivre	Ex 6
H Division	any	—	subtract angles	Ex 7
I Word problem	any	AC circuit phasor	modulus & phase	Ex 8
J Exam twist	negative "radius"	flip by $π$	Ex 9

The map below shows exactly which region each quadrant cell owns and how the reference angle $α$ (yellow) turns into the true angle $θ$ (blue).

Prerequisites you can lean on: Modulus and argument of a complex number, Polar form (trigonometric) r(cosθ + i sinθ), De Moivre's theorem, Complex multiplication as rotation and scaling.

Cell A — Quadrant I

Worked example Ex 1 — Convert

$z = 3 + 3 i$ Forecast: Both parts positive → arrow points up-right → first quadrant. Guess: is $θ$ exactly $4 5^{\circ}$ ? Write your guess before reading on.

Step 1. $r = 3^{2} + 3^{2} = 18 = 32$ . Why this step? Modulus is the length of the arrow, $x^{2} + y^{2}$ — the Pythagorean distance from the origin to the point $(3, 3)$ .

Step 2. Reference angle $α = arctan \frac{3}{3} = arctan (1) = \frac{π}{4}$ . Why this step? $tan$ is "opposite over adjacent" on the right triangle whose legs are $x = 3$ and $y = 3$ . The ratio $y / x = 1$ tells us the arrow rises as fast as it runs — a perfect $4 5^{\circ}$ slope.

Step 3. $x > 0, y > 0$ is Cell A, so $θ = α = \frac{π}{4}$ — no correction. Why this step? $arctan$ already returns first-quadrant angles, so its raw output is right here.

Answer: $z = 32 e^{iπ /4}$ .

Verify: $r cos θ = 32 \cdot \frac{1}{2} = 3 = x$ ✓ and $r sin θ = 32 \cdot \frac{1}{2} = 3 = y$ ✓.

Cell B — Quadrant II (the classic trap)

Worked example Ex 2 — Convert

$z = - 2 + 23 i$ Forecast: $x < 0$ , $y > 0$ → up-left → second quadrant. The naive $arctan (y / x) = arctan (- 3)$ will hand back a negative angle. That is wrong; predict the true angle before checking.

Step 1. $r = (- 2)^{2} + (23)^{2} = 4 + 12 = 16 = 4$ . Why this step? Squaring kills the sign, so the modulus doesn't care that $x$ is negative.

Step 2. Reference angle $α = arctan \frac{2 3}{- 2} = arctan (3) = \frac{π}{3}$ . Why this step? We use the absolute ratio so $α$ is the honest acute angle the arrow makes with the real axis — $6 0^{\circ}$ here.

Step 3. Cell B correction: $θ = π - α = π - \frac{π}{3} = \frac{2 π}{3}$ . Why this step? Look at figure s01. In quadrant II the arrow is a mirror of the reference triangle across the imaginary axis, so its true angle is $18 0^{\circ}$ minus the reference angle. Blindly trusting $arctan (- 3) = - \frac{π}{3}$ would place the arrow in quadrant IV — the wrong half of the plane.

Answer: $z = 4 e^{i 2 π /3}$ .

Verify: $4 cos \frac{2 π}{3} = 4 \cdot (- \frac{1}{2}) = - 2 = x$ ✓ and $4 sin \frac{2 π}{3} = 4 \cdot \frac{3}{2} = 23 = y$ ✓.

Cell C — Quadrant III

Worked example Ex 3 — Convert

$z = - 1 - i$ Forecast: Both parts negative → down-left → third quadrant. Here $y / x = \frac{- 1}{- 1} = + 1$ , so $arctan$ gives $+ \frac{π}{4}$ — pointing up-right, the opposite corner! Predict how far off it is.

Step 1. $r = (- 1)^{2} + (- 1)^{2} = 2$ .

Step 2. Reference angle $α = arctan \frac{- 1}{- 1} = arctan (1) = \frac{π}{4}$ . Why this step? Same acute $4 5^{\circ}$ triangle as Ex 1 — the magnitude of the slope is identical; only the direction differs.

Step 3. Cell C correction: $θ = α - π = \frac{π}{4} - π = - \frac{3 π}{4}$ . Why this step? In quadrant III the arrow points into the down-left corner. Subtracting $π$ from the reference angle rotates it a half-turn into the correct quadrant. (You could equally write $θ = π + α = \frac{5 π}{4}$ ; the two differ by a full turn $2 π$ and name the same direction. We keep arguments in $(- π, π]$ , so $- \frac{3 π}{4}$ is the principal value.)

Answer: $z = 2 e^{- i 3 π /4}$ .

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

Cell D — Quadrant IV

Cells E & F — On the axes, and the degenerate zero

Worked example Ex 5 — Four points on axes, plus

$z = 0$ Forecast: On an axis one component is zero, so $arctan$ either gives $0$ or divides by zero. Guess each angle from the picture before computing.

Look at figure s02: four arrows lying flat along the axes, plus a dot at the origin.

(i) $z = 5$ (positive real axis). $r = 5$ , arrow points along $+$ real, so $θ = 0$ . $5 e^{i 0} = 5$ . Why? No imaginary part means no turn from the reference direction.

(ii) $z = 4 i$ (positive imaginary axis). $r = 4$ , arrow points straight up, so $θ = \frac{π}{2}$ . $4 e^{iπ /2}$ . Why this step matters? $arctan (y / x) = arctan (4/0)$ is undefined (division by zero) — you must read the angle off the picture, not the formula.

(iii) $z = - 7$ (negative real axis). $r = 7$ , arrow points left, so $θ = π$ . $7 e^{iπ}$ . Why? A half-turn from the positive real axis. (This is the seed of Euler's identity e^(iπ)+1=0: $e^{iπ} = - 1$ .)

(iv) $z = - 3 i$ (negative imaginary axis). $r = 3$ , arrow points down, so $θ = - \frac{π}{2}$ . $3 e^{- iπ /2}$ .

(v) $z = 0$ (Cell F, degenerate). $r = 0$ . There is no arrow, so no direction — $ar g (0)$ is undefined. We simply write $z = 0$ ; it has no exponential form because you can't ask "which way does a point of zero length face?"

Verify: $5 cos 0 = 5$ ✓; $4 sin \frac{π}{2} = 4$ ✓; $7 cos π = - 7$ ✓; $3 sin (- \frac{π}{2}) = - 3$ ✓.

Cell G — Big powers (De Moivre)

Worked example Ex 6 — Compute

$(- 1 - i)^{10}$ Forecast: Expanding a 10th power by binomial is brutal. Exponential form turns it into "power the length, multiply the angle." Guess: will the answer be real, imaginary, or neither?

Step 1. From Ex 3, $- 1 - i = 2 e^{- i 3 π /4}$ . Why this step? Once in exponential form, the power rule from De Moivre's theorem applies: $(r e^{i θ})^{n} = r^{n} e^{in θ}$ .

Step 2. $(- 1 - i)^{10} = (2)^{10} e^{- i \cdot 10 \cdot 3 π /4} = 2^{5} e^{- i 30 π /4} = 32 e^{- i 15 π /2}$ . Why this step? $(2)^{10} = (2^{1/2})^{10} = 2^{5} = 32$ and the angle simply scales by $10$ .

Step 3. Reduce the angle mod $2 π$ : $- \frac{15 π}{2} = - \frac{15 π}{2} + 8 π = - \frac{15 π}{2} + \frac{16 π}{2} = \frac{π}{2}$ . Why this step? Adding whole turns ( $2 π = \frac{4 π}{2}$ each; we added four turns $= 8 π$ ) doesn't move the arrow but lands $θ$ in the principal range.

Answer: $32 e^{iπ /2} = 32 i$ — purely imaginary.

Verify: $32 cos \frac{π}{2} = 0$ (real part) and $32 sin \frac{π}{2} = 32$ (imag part), so $32 i$ ✓. Cross-check numerically: $(- 1 - i)^{2} = 2 i$ , so $(- 1 - i)^{10} = ((- 1 - i)^{2})^{5} = (2 i)^{5} = 32 i^{5} = 32 i$ ✓.

Cell H — Division (subtract angles)

Cell I — Real-world word problem (AC phasor)

Worked example Ex 8 — Two AC voltages combine

Two voltage sources at the same frequency are written as phasors (complex numbers whose modulus is amplitude and argument is phase): $V_{1} = 10 e^{i 0}, V_{2} = 10 e^{iπ /2} volts.$ The total voltage is $V = V_{1} + V_{2}$ . Find its amplitude and phase.

Forecast: $V_{1}$ is $10$ V "in phase", $V_{2}$ is $10$ V " $9 0^{\circ}$ ahead". Adding two equal, perpendicular arrows — guess the total amplitude (bigger or smaller than $20$ ?).

Step 1. Turn each phasor back to rectangular form to add: $V_{1} = 10 (cos 0 + i sin 0) = 10$ ; $V_{2} = 10 (cos \frac{π}{2} + i sin \frac{π}{2}) = 10 i$ . Why this step? Exponential form multiplies beautifully but does not add directly — to sum arrows you need their $x, y$ components, exactly like adding vectors.

Step 2. $V = 10 + 10 i$ .

Step 3. Amplitude $r = 1 0^{2} + 1 0^{2} = 102 \approx 14.14$ V. Phase: quadrant I, $θ = arctan \frac{10}{10} = \frac{π}{4} = 4 5^{\circ}$ . Why this step? The resultant of two equal perpendicular phasors is $2$ times one of them, tilted halfway between — not the naive $20$ V, because they are out of phase.

Answer: $V = 102 e^{iπ /4}$ V — amplitude $\approx 14.14$ V, phase $4 5^{\circ}$ .

Verify (units + geometry): Both inputs are volts; $V^{2} + V^{2} =$ V ✓. Sanity: $102 \approx 14.1 < 20$ , correct because perpendicular arrows partly "waste" each other's contribution ✓.

Cell J — Exam twist (a "negative radius" in disguise)

Worked example Ex 9 — Simplify

$z = - 3 e^{iπ /6}$ into standard $r e^{i θ}$ with $r > 0$ Forecast: A modulus can never be negative, yet here it looks like $r = - 3$ . What's really going on? Guess the true angle.

Step 1. Factor the minus sign as a rotation: $- 1 = e^{iπ}$ . Why this step? A leading $- 1$ is a half-turn of the plane ( $18 0^{\circ}$ ), i.e. $e^{iπ}$ (Cell E, point (iii)). Absorbing it into the angle keeps the modulus honest and positive.

Step 2. $z = e^{iπ} \cdot 3 e^{iπ /6} = 3 e^{i (π + π /6)} = 3 e^{i 7 π /6}$ . Why this step? Exponents of the same base add.

Step 3. Put in principal range $(- π, π]$ : $\frac{7 π}{6} - 2 π = \frac{7 π - 12 π}{6} = - \frac{5 π}{6}$ . Why this step? $\frac{7 π}{6} > π$ , so subtract one full turn to name the same arrow with the standard argument.

Answer: $z = 3 e^{- i 5 π /6}$ (modulus $3$ , argument $- \frac{5 π}{6}$ , quadrant III).

Verify: Compare rectangular parts. Original: $- 3 (cos \frac{π}{6} + i sin \frac{π}{6}) = - 3 (\frac{3}{2} + \frac{1}{2} i) = - \frac{3 3}{2} - \frac{3}{2} i$ . New: $3 (cos (- \frac{5 π}{6}) + i sin (- \frac{5 π}{6})) = 3 (- \frac{3}{2} - \frac{1}{2} i) = - \frac{3 3}{2} - \frac{3}{2} i$ ✓ — identical.

Active recall

Recall Which cell needs which correction?

Quadrant I correction? ::: none, $θ = α$ Quadrant II correction? ::: $θ = π - α$ Quadrant III principal angle? ::: $θ = α - π$ Quadrant IV correction? ::: $θ = - α$ Argument of $z = 0$ ? ::: undefined (no direction) How to remove a leading minus sign? ::: multiply by $e^{iπ}$ (add $π$ to the angle)

Connections

3.5.07 Exponential form z = re^(iθ) (Hinglish)
Modulus and argument of a complex number
Polar form (trigonometric) r(cosθ + i sinθ)
De Moivre's theorem
Roots of unity
Euler's identity e^(iπ)+1=0
Complex multiplication as rotation and scaling