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

Before we start, one shared picture. Everything below lives on the Argand plane: the horizontal axis measures the real part $a$ (how far right), the vertical axis measures the imaginary part $b$ (how far up), and every number $z = a + bi$ is an arrow from the origin $O$ to the point $(a, b)$ .

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

Two labels on that arrow: its length is the modulus $r = a^{2} + b^{2}$ (always $\geq 0$ ), and the angle it makes with the positive real axis, measured anticlockwise, is the argument $θ$ . See Modulus and Argument and Argand Diagram if either word feels unfamiliar.

Level 1 — Recognition

Recall Solution L1.1

$cis θ = cos θ + i sin θ$ . WHAT it is: a sum — cosine, plus $i$ times sine. WHY the name: read the letters "cos i sin". It is NOT a product $cos θ \cdot i \cdot sin θ$ .

Recall Solution L1.2

$θ = \frac{π}{2}$ is a quarter-turn anticlockwise from the positive real axis — straight up. With length $3$ , the point is at $(0, 3)$ , i.e. $z = 3 i$ . Check: $a = 3 cos \frac{π}{2} = 0$ , $b = 3 sin \frac{π}{2} = 3$ . So $z = 0 + 3 i = 3 i$ . ✓

Recall Solution L1.3

False. $r = a^{2} + b^{2}$ squares both parts before taking a (non-negative) square root, so $r \geq 0$ always. A number "pointing backwards" is captured by the angle $θ$ , never by a negative length.

Level 2 — Application

Recall Solution L2.1

Step 1 (length): $r = (3)^{2} + 1^{2} = 3 + 1 = 2$ . Why Pythagoras? legs $3$ and $1$ , hypotenuse $r$ . Step 2 (quadrant): $(3, 1)$ — both positive ⇒ Quadrant I, no adjustment needed. Step 3 (reference angle): $α = tan^{- 1} (\frac{1}{3}) = \frac{π}{6}$ . Result: $z = 2 cis \frac{π}{6}$ . Check: $2 cos \frac{π}{6} = 2 \cdot \frac{3}{2} = 3$ ✓, $2 sin \frac{π}{6} = 2 \cdot \frac{1}{2} = 1$ ✓.

Recall Solution L2.2

Step 1: $a = 6 cos \frac{π}{3} = 6 \cdot \frac{1}{2} = 3$ . Step 2: $b = 6 sin \frac{π}{3} = 6 \cdot \frac{3}{2} = 33$ . Result: $z = 3 + 33 i$ . Back-check: $3^{2} + (33)^{2} = 9 + 27 = 36 = 6$ ✓ — modulus restored.

Recall Solution L2.3

Step 1 (length): $r = 1^{2} + (- 3)^{2} = 1 + 3 = 2$ . Step 2 (quadrant): $(1, - 3)$ — real part positive, imaginary part negative ⇒ Quadrant IV (right and down). Step 3 (reference angle): use magnitudes, $α = tan^{- 1} \frac{- 3}{1} = tan^{- 1} (3) = \frac{π}{3}$ . Step 4 (fix the quadrant): In Q IV the arrow dips below the positive real axis, so the principal argument is $θ = - α = - \frac{π}{3}$ . Why negative? We turn clockwise (downward) from the positive real axis, and clockwise angles are negative. Result: $z = 2 cis (- \frac{π}{3})$ . Check: $2 cos (- \frac{π}{3}) = 2 \cdot \frac{1}{2} = 1$ ✓, $2 sin (- \frac{π}{3}) = 2 \cdot (- \frac{3}{2}) = - 3$ ✓.

Recall Solution L2.4

$r = 5^{2} + (- 12)^{2} = 25 + 144 = 169 = 13$ . Why the sign of $- 12$ vanishes: squaring kills it. The negative only tells us the arrow points down, which will matter for the argument, not the length.

Level 3 — Analysis

Recall Solution L3.1

Step 1 (length): $r = (- 1)^{2} + (- 3)^{2} = 1 + 3 = 2$ . Step 2 (quadrant): $(- 1, - 3)$ — both negative ⇒ Quadrant III (down-left). Step 3 (reference angle): use magnitudes, $α = tan^{- 1} \frac{- 3}{- 1} = tan^{- 1} (3) = \frac{π}{3}$ . Step 4 (fix the quadrant): In Q III the principal argument is $θ = - (π - α) = - (π - \frac{π}{3}) = - \frac{2 π}{3}$ . Why negative? Measuring anticlockwise would give $π + \frac{π}{3} = \frac{4 π}{3}$ , but that exceeds $π$ ; the principal value swings the short way clockwise instead. Result: $z = 2 cis (- \frac{2 π}{3})$ . Check: $2 cos (- \frac{2 π}{3}) = 2 \cdot (- \frac{1}{2}) = - 1$ ✓, $2 sin (- \frac{2 π}{3}) = 2 \cdot (- \frac{3}{2}) = - 3$ ✓.

The figure below contrasts the two ways to name the Q III direction: the yellow arc is the reference angle $α = \frac{π}{3}$ measured off the negative real axis, while the green arc is the principal argument $θ = - \frac{2 π}{3}$ swung the short way clockwise from the positive real axis. Seeing both arcs on one picture is what makes "turn the short way" concrete — the green route is visibly shorter than going anticlockwise the long way round.

Recall Solution L3.2

The error: $(- 4, 4)$ lives in Quadrant II (left and up). The calculator's $- \frac{π}{4}$ points into Quadrant IV (right and down) — the opposite direction. They share the same $tan$ value because $tan$ repeats every $π$ ; the calculator picked the wrong one of the two. The fix: reference angle $α = tan^{- 1} \frac{4}{- 4} = tan^{- 1} (1) = \frac{π}{4}$ . In Q II, $θ = π - α = π - \frac{π}{4} = \frac{3 π}{4}$ . Result: $ar g (- 4 + 4 i) = \frac{3 π}{4}$ .

Recall Solution L3.3

Why $tan^{- 1}$ breaks: $a = 0$ means $b / a$ divides by zero — the reference-angle formula has nothing to feed on. This is a degenerate case; we read the direction straight off the picture. $z = - 7 i$ sits at $(0, - 7)$ — straight down the negative imaginary axis. The arrow points at $θ = - \frac{π}{2}$ (principal value). Result: $r = 7$ , $ar g z = - \frac{π}{2}$ , so $z = 7 cis (- \frac{π}{2})$ .

Level 4 — Synthesis

Recall Solution L4.1

Step 1 — convert each factor. $1 + i$ : $r_{1} = 2$ , Q I, $θ_{1} = \frac{π}{4}$ . So $1 + i = 2 cis \frac{π}{4}$ . $3 + i$ : $r_{2} = 2$ , Q I, $θ_{2} = \frac{π}{6}$ (from L2.1). So $3 + i = 2 cis \frac{π}{6}$ . Step 2 — multiply the polar forms. Rule (see Multiplication and Division of Complex Numbers): multiply moduli, add arguments. $r = 2 \cdot 2 = 22, θ = \frac{π}{4} + \frac{π}{6} = \frac{3 π}{12} + \frac{2 π}{12} = \frac{5 π}{12} .$ So the product is $22 cis \frac{5 π}{12}$ . Step 3 — the exact values of $cos \frac{5 π}{12}$ and $sin \frac{5 π}{12}$ . Note $\frac{5 π}{12} = \frac{π}{4} + \frac{π}{6}$ , so we use the angle-sum formulas $cos (A + B) = cos A cos B - sin A sin B$ and $sin (A + B) = sin A cos B + cos A sin B$ with $A = \frac{π}{4}$ , $B = \frac{π}{6}$ : $cos \frac{5 π}{12} = \frac{1}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{1}{2} = \frac{3 - 1}{2 2} = \frac{6 - 2}{4},$ $sin \frac{5 π}{12} = \frac{1}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3 + 1}{2 2} = \frac{6 + 2}{4} .$ Step 4 — back to Cartesian. $a = 22 \cdot \frac{6 - 2}{4} = \frac{2 ( 6 - 2 )}{2} = \frac{12 - 2}{2} = \frac{2 3 - 2}{2} = 3 - 1$ . $b = 22 \cdot \frac{6 + 2}{4} = \frac{12 + 2}{2} = 3 + 1$ . Result: $(3 - 1) + (3 + 1) i$ . Cross-check by direct expansion: $(1 + i) (3 + i) = 3 + i + 3 i + i^{2} = 3 - 1 + (3 + 1) i$ ✓.

Recall Solution L4.2

First, read the moduli. Remember $cis θ = cos θ + i sin θ$ has modulus $cos^{2} θ + sin^{2} θ = 1 = 1$ for every $θ$ — a lone $cis$ is always a unit-length arrow. So the numerator has modulus $2$ and the denominator has modulus $1$ . Rule for division: divide the moduli, subtract the arguments. $r = \frac{2}{1} = 2, θ = \frac{3 π}{4} - \frac{π}{4} = \frac{2 π}{4} = \frac{π}{2} .$ So the quotient is $2 cis \frac{π}{2}$ . Cartesian: $a = 2 cos \frac{π}{2} = 0$ , $b = 2 sin \frac{π}{2} = 2$ . Result: $2 i$ . Why this is nicer than Cartesian division: no conjugate, no rationalising — subtracting angles does all the work.

Recall Solution L4.3

Reasoning: $z^{3} = z \cdot z \cdot z$ . Each multiplication by $cis \frac{π}{5}$ adds $\frac{π}{5}$ to the angle and multiplies the modulus by $1$ (since $∣ z ∣ = 1$ ). $z^{3} = cis (\frac{π}{5} + \frac{π}{5} + \frac{π}{5}) = cis \frac{3 π}{5} .$ Argument: $\frac{3 π}{5}$ , modulus still $1$ . This is exactly $(cis θ)^{n} = cis n θ$ with $n = 3$ .

Level 5 — Mastery

Recall Solution L5.1

The set: fixed modulus $2$ and a fixed direction $\frac{π}{3}$ pin down exactly one arrow. $z = 2 cis \frac{π}{3} = 2 cos \frac{π}{3} + 2 i sin \frac{π}{3} = 1 + i 3$ . Why $2 π k$ is harmless: adding a whole turn ( $2 π$ ) to the angle returns the arrow to the same direction, because $cos$ and $sin$ have period $2 π$ . So $cis (θ + 2 π k) = cis θ$ for every integer $k$ — all those "different" arguments name the same point. That is why we single out one principal argument in $(- π, π]$ . Result: the unique number $1 + i 3$ .

Recall Solution L5.2

Modulus of every root is $2$ . Arguments: $0, \frac{π}{2}, π, \frac{3 π}{2}$ (equivalently $- \frac{π}{2}$ ).

$k = 0$ : $2 cis 0 = 2$ .
$k = 1$ : $2 cis \frac{π}{2} = 2 i$ .
$k = 2$ : $2 cis π = - 2$ .
$k = 3$ : $2 cis \frac{3 π}{2} = - 2 i$ . Result: ${2, 2 i, - 2, - 2 i}$ — four points evenly spaced $\frac{π}{2}$ apart on a circle of radius $2$ (see Roots of Complex Numbers). Check one: $(2 i)^{4} = 2^{4} i^{4} = 16 \cdot 1 = 16$ ✓.

Recall Solution L5.3

Step 1 — polar form of the base. $1 + i = 2 cis \frac{π}{4}$ , so $\frac{1 + i}{2} = \frac{2}{2} cis \frac{π}{4} = 1 \cdot cis \frac{π}{4}$ . Modulus $1$ , argument $\frac{π}{4}$ . Step 2 — raise to the 8th (De Moivre). Modulus $1^{8} = 1$ ; argument $8 \cdot \frac{π}{4} = 2 π$ . $(cis \frac{π}{4})^{8} = cis 2 π = cos 2 π + i sin 2 π = 1 + 0 i = 1.$ Why it lands on $1$ : raising to the 8th power rotates the unit arrow eight quarter-turns of $\frac{π}{4}$ each $= 2 π$ , one full loop, back to the start. Result: $1$ . ∎

Connections

Modulus and Argument — the two numbers every problem here extracts.
Argand Diagram — the plane every arrow lives in.
Multiplication and Division of Complex Numbers — the MMAA rule used in L4.
De Moivre's Theorem — powers in L4.3, L5.3.
Roots of Complex Numbers — L5.2.
Euler's Formula — the reason angles add.