Contents / Precalculus / Complex Numbers & Polar Coordinates

Chapter 7

Complex Numbers & Polar Coordinates

Complex arithmetic, the complex plane, polar form, De Moivre's theorem, polar and parametric curves.

Introduction

Complex arithmetic, the complex plane, polar form, De Moivre's theorem, polar and parametric curves.

7.1Complex number

Complex numbers extend the reals to solve equations like x²+1=0 and provide algebra for rotations and oscillations.

Complex Numbers

Definition 7.1 (Complex number). A complex number is an expression

z = a + bi,

where $a$ and $b$ are real numbers and $i$ is the imaginary unit satisfying $i^{2} = - 1$ .

Notation (Parts, modulus and conjugate).

Real part $Re (z) = a$
Imaginary part $Im (z) = b$
Modulus $∣ z ∣ = a^{2} + b^{2}$
Conjugate $\overset{z}{ˉ} = a - bi$ (note $\overline{z w} = \overset{z}{ˉ} \overset{w}{ˉ}$ and $∣ z ∣^{2} = z \overset{z}{ˉ}$ )

Arithmetic follows the distributive law, replacing $i^{2}$ with $- 1$ whenever it appears. Addition and subtraction act componentwise (add real parts, add imaginary parts); multiplication uses FOIL and then the rule $i^{2} = - 1$ .

Proposition 7.2 (The complex numbers form a field). Every nonzero complex number $z$ has a multiplicative inverse,

\frac{1}{z} = \frac{z ˉ}{∣ z ∣ ^{2}} .

The condition $z \neq = 0$ matters — dividing by zero is undefined here just as in the reals — and $∣ z ∣^{2} = a^{2} + b^{2}$ is a positive real, so the inverse is a genuine complex number.

Intuition. Complex numbers are "2D numbers": one real direction, one imaginary direction at right angles. Adding them is vector addition; multiplying scales and rotates. The conjugate is the mirror image across the real axis — flipping a number over and multiplying by it always lands you on the real axis (at $a^{2} + b^{2}$ ), which is the trick that clears $i$ out of a denominator.

Example 7.3 (Multiply (2 + 3i)(1 - i) and simplify). Distribute (FOIL): $2 \cdot 1 + 2 \cdot (- i) + 3 i \cdot 1 + 3 i \cdot (- i) = 2 - 2 i + 3 i - 3 i^{2}$ .

Solution.

Replace $i^{2} = - 1$ : $- 3 i^{2} = - 3 (- 1) = + 3$ .
Combine real parts and imaginary parts: $(2 + 3) + (- 2 + 3) i = 5 + i$ .
Sanity check with moduli: $∣2 + 3 i ∣ = 13$ and $∣1 - i ∣ = 2$ , so the product should have modulus $13 \cdot 2 = 26$ . Indeed $∣5 + i ∣ = 25 + 1 = 26$ ✓.

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Example 7.4 (Divide $\frac{3 + 2 i}{1 - i}$ and write in $a + bi$ form). Multiply numerator and denominator by the conjugate of the denominator, $1 + i$ :

\frac{3 + 2 i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{( 3 + 2 i ) ( 1 + i )}{( 1 - i ) ( 1 + i )} .

Denominator becomes real:

(1 - i) (1 + i) = 1^{2} - i^{2} = 1 - (- 1) = 2

. Numerator:

(3 + 2 i) (1 + i) = 3 + 3 i + 2 i + 2 i^{2} = 3 + 5 i - 2 = 1 + 5 i

. Divide each part by 2:

\frac{1 + 5 i}{2} = \frac{1}{2} + \frac{5}{2} i

. Sanity check: multiply back,

(\frac{1}{2} + \frac{5}{2} i) (1 - i) = \frac{1}{2} - \frac{1}{2} i + \frac{5}{2} i - \frac{5}{2} i^{2} = \frac{1}{2} + \frac{5}{2} + 2 i = 3 + 2 i

✓.

Common Mistakes

i² = -1 arithmetic: Forgetting to replace i² with -1 when multiplying complex numbers leads to wrong real/imaginary parts.
Inverse of complex: 1/(a+bi) requires multiplying numerator and denominator by conjugate; forgetting yields non-real denominator.
Modulus of product: |z w| = |z| |w| (magnitudes multiply); not |z + w|.
Conjugate of sum vs product: \bar (z+w) = \bar z + \bar w always; same for product.
Writing j instead of i: In some engineering contexts j is used, but in math i is standard.
Forgetting complex include the reals: When b=0, z=a is real.

Exercises for §7.1 →

7.2Complex plane

Plotting complex numbers as points in the plane turns algebra into geometry and reveals multiplication as rotation+scaling.

The Complex Plane

Definition 7.5 (The complex plane). Plot $z = a + bi$ as the point $(a, b)$ in the Cartesian plane. The horizontal axis is the real axis; the vertical axis is the imaginary axis. Then $∣ z ∣$ is the distance from the origin to the point, and the argument $ar g (z)$ is the angle from the positive real axis to the ray from the origin through $z$ (principal value in $(- π, π]$ ).

Addition is the parallelogram law (vector sum). Multiplication is the more interesting operation.

Proposition 7.6 (Multiplication scales and rotates). Multiplication by a fixed complex number $w$ is a similarity transformation: it scales every point's distance from the origin by $∣ w ∣$ and rotates it by $ar g (w)$ .

This is why " $i$ is a $90°$ turn" — multiplying by $i = 1 \cdot (cos \frac{π}{2} + i sin \frac{π}{2})$ keeps the length the same ( $∣ i ∣ = 1$ ) and rotates by $\frac{π}{2}$ .

Intuition. The complex plane turns algebra into geometry: every complex number is a point or an arrow from the origin. Pure imaginaries live on the vertical axis, reals on the horizontal. Multiplying by a number is like "grab the whole plane, spin it by that number's angle, and stretch it by that number's length."

Example 7.7 (Plot -1 + 2i , compute its modulus and principal argument). Point $(- 1, 2)$ lives in quadrant II.

Solution.

$∣ z ∣ = (- 1)^{2} + 2^{2} = 1 + 4 = 5$ .
The reference angle is $arctan (\frac{∣ Im ∣}{∣ Re ∣}) = arctan \frac{2}{1}$ . In quadrant II the principal argument is $π$ minus this: $ar g (z) = π - arctan 2 \approx 3.1416 - 1.107 = 2.034$ rad.
Sanity check the quadrant: $2.034$ rad is between $\frac{π}{2} \approx 1.571$ and $π \approx 3.1416$ , i.e. in quadrant II, matching the plot ✓.

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Example 7.8 (Multiply $(2 + i)$ by $i$ and confirm it is a $90°$ rotation). Algebraically: $i (2 + i) = 2 i + i^{2} = - 1 + 2 i$ .

Solution.

Geometrically, multiplying by $i$ (which has $∣ i ∣ = 1$ , $ar g (i) = \frac{π}{2}$ ) should keep the length and rotate by $90°$ counterclockwise.
Check length: $∣2 + i ∣ = 5$ and $∣ - 1 + 2 i ∣ = 5$ — unchanged ✓.
Check the turn: the point $(2, 1)$ rotated $90°$ counterclockwise about the origin goes to $(- 1, 2)$ , which is exactly $- 1 + 2 i$ ✓.

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Common Mistakes

Principal argument range: arg usually taken in (-π, π]; using (0,2π] or other intervals can cause sign errors in polar form.
Plotting imag as x: Real is horizontal x, imag vertical y. 3-2i is (3,-2) not (-2,3).
arg of negative real: arg(-1) = π (or -π boundary), not 0.
Modulus vs argument mixup: |z| is length (always ≥0), arg is angle.
Forgetting conjugate is reflection over real axis: a-bi is (a,-b).
Vector addition vs complex: Same, but remember i direction is perpendicular.

Exercises for §7.2 →

7.3Polar form of complex numbers

Polar form turns multiplication and powers into arithmetic on the modulus r and argument θ. De Moivre follows immediately.

Polar Form of Complex Numbers

Definition 7.9 (Polar form). Any nonzero complex number $z$ can be written

z = r (cos θ + i sin θ), r = ∣ z ∣ \geq 0, θ = ar g (z) .

This is also written $r cis θ$ , shorthand for $cos θ + i sin θ$ .

Theorem 7.10 (Euler's formula). For every real $θ$ ,

e^{i θ} = cos θ + i sin θ,

so the polar form can be written more compactly as $z = r e^{i θ}$ .

Method 7.11 (Converting between rectangular and polar form).

Rectangular to polar: $r = ∣ z ∣$ and $θ = atan2 (Im, Re)$ .
Polar to rectangular: $a = r cos θ$ and $b = r sin θ$ .

Here $atan2 (b, a)$ is the two-argument arctangent that returns the angle in the correct quadrant (unlike $arctan (b / a)$ , which cannot tell quadrant I from III or II from IV).

Proposition 7.12 (Multiplication and division in polar form). Polar form makes multiplication, division, and powers trivial, because it converts them into arithmetic on $r$ and $θ$ : to multiply, multiply the moduli and add the arguments; to divide, divide the moduli and subtract the arguments.

Intuition. Instead of (x,y) coordinates, describe the number by "how far from origin" and "which direction." Multiplication then becomes "multiply the distances and add the angles" — like two dancers whose spins add up.

Example 7.13 (Write -2 - 2i in polar form).

r = (- 2)^{2} + (- 2)^{2} = 4 + 4 = 8 = 22

Solution.

The point $(- 2, - 2)$ is in quadrant III. Its reference angle is $arctan \frac{2}{2} = \frac{π}{4}$ , so the principal argument (in $(- π, π]$ ) is $θ = - π + \frac{π}{4} = - \frac{3 π}{4}$ (equivalently $\frac{5 π}{4}$ if you prefer $[0, 2 π)$ ).
So $- 2 - 2 i = 22 (cos (- \frac{3 π}{4}) + i sin (- \frac{3 π}{4}))$ .
Sanity check by converting back: $cos (- \frac{3 π}{4}) = - \frac{2}{2}$ , $sin (- \frac{3 π}{4}) = - \frac{2}{2}$ , so $22 (- \frac{2}{2} - \frac{2}{2} i) = - 2 - 2 i$ ✓.

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Example 7.14 (Multiply $z = 2 cis \frac{π}{6}$ and $w = 3 cis \frac{π}{4}$ using polar arithmetic). Multiply the moduli: $∣ z w ∣ = 2 \cdot 3 = 6$ .

Solution.

Add the arguments: $ar g (z w) = \frac{π}{6} + \frac{π}{4} = \frac{2 π}{12} + \frac{3 π}{12} = \frac{5 π}{12}$ .
So $z w = 6 cis \frac{5 π}{12}$ .
Sanity check the modulus rule against $∣ z w ∣ = ∣ z ∣∣ w ∣$ : $∣ z ∣ = 2$ , $∣ w ∣ = 3$ , product $6$ ✓, and $\frac{5 π}{12} = 75°$ lies in quadrant I, as expected for two quadrant-I factors.

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Common Mistakes

Principal argument range: arg usually taken in (-π, π]; using (0,2π] or other intervals can cause sign errors in polar form.
Polar to Cartesian conversion sign: cos( quadrant II ) is negative; students often drop the sign when converting back.
r negative in polar plots: Forgetting that negative r means go opposite direction produces points in the wrong half-plane.
Forgetting r=0 is allowed: z=0 has r=0, θ arbitrary (but usually 0).
cis or e form not recognized: Equivalent to the cos + i sin.
θ in degrees when using formulas: The trig functions in the form expect radians for calculus, but identities hold either.

Exercises for §7.3 →

7.4De Moivre's theorem

De Moivre turns powers of complex in polar form into simple scaling of r and adding to the angle. It is the key to roots of unity and polynomial roots.

De Moivre's Theorem

Theorem 7.15 (De Moivre's theorem). For integer $n$ and real $r \geq 0$ ,

[r (cos θ + i sin θ)]^{n} = r^{n} (cos (n θ) + i sin (n θ)) .

Proof. It follows from the polar multiplication rule applied $n$ times — each factor multiplies the modulus by $r$ and adds $θ$ to the argument — and extends to negative integers via the reciprocal.∎

Powers are therefore easy: just raise $r$ to the $n$ and multiply the angle by $n$ . Running the theorem backwards gives the roots.

Corollary 7.16 (The $n$ -th roots of a complex number). For $r > 0$ and a positive integer $n$ , the $n$ -th roots of $r e^{i θ}$ are

n r exp (i \frac{θ + 2 πk}{n}), k = 0, 1, \dots, n - 1,

where $n r$ is the ordinary real positive $n$ -th root. The roots all have the same modulus $n r$ , so they lie on a circle, equally spaced every $\frac{2 π}{n}$ .

This is the key to finding roots of polynomials and understanding roots of unity. Note $k$ runs over exactly $n$ consecutive integers because $k$ and $k + n$ give the same root (their angles differ by $2 π$ ), so there are exactly $n$ distinct $n$ -th roots.

Intuition. Raising a complex number to a power is "stretch by r each time and turn by θ each time." After n times you have turned nθ and scaled r^n. The n-th roots are the n numbers that, when turned and stretched n times, land back on the starting number — like asking "which n equally-spaced spokes on a wheel, spun n times, all point back home."

Example 7.17 (Compute $(1 + i)^{4}$ with De Moivre). Put $1 + i$ in polar form: $r = 1^{2} + 1^{2} = 2$ , and $θ = arctan \frac{1}{1} = \frac{π}{4}$ (quadrant I). So $1 + i = 2 cis \frac{π}{4}$ .

Solution.

Apply De Moivre with $n = 4$ : $(1 + i)^{4} = (2)^{4} cis (4 \cdot \frac{π}{4}) = 4 cis π$ .
Evaluate: $4 (cos π + i sin π) = 4 (- 1 + 0 i) = - 4$ .
Sanity check by squaring twice: $(1 + i)^{2} = 1 + 2 i + i^{2} = 2 i$ , and $(2 i)^{2} = 4 i^{2} = - 4$ ✓.

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Example 7.18 (Find the three cube roots of 8 (i.e., solutions to z^3 = 8 = 8 + 0i)). Write $8 = 8 cis 0$ (here $r = 8$ , $θ = 0$ ).

Solution.

Roots: $38 cis (\frac{0 + 2 πk}{3}) = 2 cis \frac{2 πk}{3}$ for $k = 0, 1, 2$ .
$k = 0$ : $2 cis 0 = 2$ .
$k = 1$ : $2 cis \frac{2 π}{3} = 2 (- \frac{1}{2} + i \frac{3}{2}) = - 1 + i 3$ .
$k = 2$ : $2 cis \frac{4 π}{3} = 2 (- \frac{1}{2} - i \frac{3}{2}) = - 1 - i 3$ .
Sanity check one root: $(- 1 + i 3)^{3}$ . First $(- 1 + i 3)^{2} = 1 - 2 i 3 + 3 i^{2} = 1 - 2 i 3 - 3 = - 2 - 2 i 3$ . Then $(- 2 - 2 i 3) (- 1 + i 3) = 2 - 2 i 3 + 2 i 3 - 2 i^{2} \cdot 3 = 2 + 6 = 8$ ✓. The three roots are equally spaced $\frac{2 π}{3} = 120°$ apart on the circle of radius $2$ .

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Common Mistakes

De Moivre for fractional powers: n must be integer for the basic statement; roots require the k=0 to n-1 branches.
Forgetting the n distinct roots: There are always n roots counting multiplicity in complex; the formula gives all.
Arg not adding multiples of 2π: The +2π k /n generates the different branches; k mod n.
r negative: Take |r| and adjust arg by π.
Leaving roots in cis form when a+bi wanted: Convert back using cos sin of the angles.
n=0 or negative: Powers defined for negative integer via inverse; roots for positive integer n.

Exercises for §7.4 →

7.5Polar coordinates

Polar coordinates locate points by distance and direction from origin. They simplify many curves with rotational symmetry.

Polar Coordinates

Definition 7.19 (Polar coordinates). In the plane, a point can be located by $(r, θ)$ — a distance and a direction — instead of $(x, y)$ , the two descriptions being linked by

x = r cos θ, y = r sin θ,

r^{2} = x^{2} + y^{2}, θ = atan2 (y, x) .

Polar equations often simplify circles, cardioids, roses, and spirals that have ugly Cartesian forms.

Proposition 7.20 (Polar coordinates are not unique). A single point has infinitely many polar representations. Negative $r$ and angles outside $[0, 2 π)$ are allowed, and

(r, θ) \equiv (- r, θ + π) \equiv (r, θ + 2 πk) for any integer k .

The origin is special: it is $(0, θ)$ for every $θ$ .

Intuition. Polar is "go distance r in direction θ from the origin." Perfect for anything with a center and rotational symmetry (wheels, flowers, radar). A negative r means "face direction θ, then walk backwards."

Example 7.21 (Convert the Cartesian point (-3, 4) to polar (r>0, principal θ)).

r = (- 3)^{2} + 4^{2} = 9 + 16 = 25 = 5

Solution.

The point is in quadrant II ( $x < 0$ , $y > 0$ ). Reference angle is $arctan \frac{4}{3}$ , so the principal angle is $θ = π - arctan \frac{4}{3}$ .
Exact answer: $(5, π - arctan \frac{4}{3})$ .
Sanity check: $arctan \frac{4}{3} \approx 0.927$ , so $θ \approx 2.214$ rad $\approx 127°$ , which is in quadrant II ✓, and $5 cos (2.214) \approx - 3$ , $5 sin (2.214) \approx 4$ ✓.

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Example 7.22 (Convert the polar point $(2, \frac{3 π}{4})$ to Cartesian $(x, y)$ ).

x = r cos θ = 2 cos \frac{3 π}{4} = 2 (- \frac{2}{2}) = - 2

Solution.

$y = r sin θ = 2 sin \frac{3 π}{4} = 2 (\frac{2}{2}) = 2$ .
So the point is $(- 2, 2)$ .
Sanity check: $r = (- 2)^{2} + (2)^{2} = 2 + 2 = 2$ ✓, and the point is in quadrant II, matching $θ = \frac{3 π}{4} = 135°$ ✓.

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Common Mistakes

r negative in polar plots: Forgetting that negative r means go opposite direction produces points in the wrong half-plane.
Polar to Cartesian conversion sign: cos( quadrant II ) is negative; students often drop the sign when converting back.
θ in degrees vs rad: atan2 usually radians; convert if problem in degrees.
Forgetting r² = x² + y² for conversion: Always.
atan(y/x) instead of atan2: Loses quadrant info.
Assuming r≥0 always: Negative r valid and useful.

Exercises for §7.5 →

7.6Parametric equation

Parametric equations describe a curve by letting a parameter t vary, giving x and y as functions of t. They carry direction and timing information and can trace non-functions.

Parametric Equations

Definition 7.23 (Parametric equations). A curve can be described by a pair of functions

x = f (t), y = g (t),

where $t$ is a parameter (often time or angle).

Advantages:

Easy to plot direction of travel as t increases.
Can represent curves that fail the vertical line test (circles, loops).
Derivatives dx/dt, dy/dt give velocity components.

Eliminating the parameter $t$ sometimes recovers a Cartesian relation, but the parametric form keeps the timing information (where you are when, and in which direction you travel).

Classic examples: circle $x = cos t$ , $y = sin t$ ; line $x = t$ , $y = m t + b$ ; cycloid $x = r (t - sin t)$ , $y = r (1 - cos t)$ ; helix in 3D, etc.

Intuition. Parametric equations let the point "walk" according to a clock t. You describe where it is horizontally and vertically at each moment, without needing a single equation relating x and y directly. It is the difference between a photo of a road (the Cartesian curve) and a GPS track that also knows when you were at each spot and which way you were headed.

Example 7.24 (Parametrize the line segment from (1,2) to (4,8)). Use the "start plus $t$ times displacement" template with $t$ going from $0$ to $1$ . The displacement is $(4 - 1, 8 - 2) = (3, 6)$ .

Solution.

$x = 1 + 3 t$ (at $t = 0$ gives $1$ , at $t = 1$ gives $4$ ).
$y = 2 + 6 t$ (at $t = 0$ gives $2$ , at $t = 1$ gives $8$ ).
Check the endpoints: $t = 0 \Rightarrow (1, 2)$ ✓ and $t = 1 \Rightarrow (4, 8)$ ✓. The direction vector $(3, 6) = 3 (1, 2)$ , confirming the path stays on the line through the two points.

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Example 7.25 (Eliminate the parameter from $x = cos t$ , $y = sin t$ ( $0 \leq t < 2 π$ )). Solve each for the trig value: $cos t = x$ , $sin t = y$ .

Solution.

Use the Pythagorean identity $cos^{2} t + sin^{2} t = 1$ : substitute to get $x^{2} + y^{2} = 1$ .
So the Cartesian curve is the unit circle. Note what was lost: the Cartesian equation does not record that the point starts at $(1, 0)$ and travels counterclockwise once — that direction and timing live only in the parametrization.
Sanity check a point: at $t = \frac{π}{2}$ , $(x, y) = (0, 1)$ , and indeed $0^{2} + 1^{2} = 1$ ✓.

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Common Mistakes

Parametric vs implicit: A parametric curve may trace only part of the algebraic curve obtained by eliminating t, or may trace it multiple times.
Forgetting the range of t: The interval for t determines which portion of the curve is drawn.
Eliminating t incorrectly: Solve for t from one eq and sub; may introduce extraneous or lose parts.
Velocity vs speed: (dx/dt, dy/dt) is velocity vector; speed is its magnitude.
Assuming t must be time: Can be angle or any parameter.
Not checking orientation: As t increases, may go clockwise or counter.

Exercises for §7.6 →

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter