Polar Representation of Complex Numbers
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Direct answer
Multiplication in polar form is addition of angles: z = r(cos θ + i sin θ) carries modulus r = √(x² + y²) and argument θ fixed by the quadrant, and z1z2 = r1r2[cos(θ1 + θ2) + i sin(θ1 + θ2)] — the property that turns powers into De Moivre's theorem and rotations into a single multiplication. The principal argument lives in (−π, π], one value per nonzero z. Conversions are exam bread: 1 + i has r = √2, θ = π/4; i alone has θ = π/2; a negative real has θ = π. Multiplying by cos α + i sin α rotates the plane about the origin by α without changing any length.
What you must remember
- Modulus: r = √(x² + y²), the distance from the origin; |z1z2| = |z1||z2| and |z1/z2| = |z1|/|z2| follow.
- Argument by quadrant: tan⁻¹(y/x) is only the first quadrant answer; quadrant two adds π, quadrant three subtracts π (principal value −π + tan⁻¹(y/x)), quadrant four uses tan⁻¹(y/x) negative.
- Principal range: θ ∈ (−π, π]; Arg z versus the multivalued arg z = Arg z + 2kπ distinction appears in Advanced statements.
- Product and quotient: arguments add for products, subtract in order for quotients — θ1 − θ2, and the reverse order is a planted error.
- Rotation as multiplication: multiplying by e^{iα} rotates by α about the origin; rotation about a point a is z → a + (z − a)e^{iα}.
- Standard arguments to know cold: 1 + i → π/4; −1 + i → 3π/4; −i → −π/2; the cube roots of unity sit at 0, ±2π/3.
- Conjugate in polar: z̄ has the same modulus and the negated argument — reflection in the real axis.
Powers and rotations
Compute (1 + i)^8 in polar. The modulus is √2 and the argument π/4, so (1 + i)^8 = (√2)^8 [cos(8 × π/4) + i sin(8 × π/4)] = 16[cos 2π + i sin 2π] = 16 — a one-line answer where binomial expansion would demand nine terms. Now rotation: turn the point 3 + 4i by 90° anticlockwise about the origin. Multiplying by i = e^{iπ/2} gives i(3 + 4i) = −4 + 3i, and the modulus 5 is preserved — visible in the arithmetic, guaranteed by the theory. Rotation about a different centre: rotate z about the point a by angle α by first translating a to the origin (z − a), rotating through multiplication by e^{iα}, then translating back: the composite a + (z − a)e^{iα}. This three-move pattern — translate, rotate, translate back — is the entire content of every "image under rotation" question, and recognising multiplication by i as the 90° shortcut saves a full polar conversion whenever the angle is a multiple of 90°.
Quadrant discipline
JEE Main asks conversions and small powers, and the argument is where marks die: for z = −2 − 2i, tan⁻¹(y/x) = tan⁻¹(1) = π/4 belongs to the first quadrant, while the number lives in the third, whose principal argument is −3π/4 (equivalently 5π/4 outside the principal range). Options always carry π/4 and 3π/4 beside the correct −3π/4. JEE Advanced prefers composite geometry: find the locus of z satisfying arg((z − 1)/(z + 1)) = π/4 — an arc of a circle through the points 1 and −1, because the quotient's argument is the angle subtended at z by the segment joining them. Rotation questions about arbitrary centres, and conjugation composed with rotation (reflect then rotate), extend the same arithmetic. The defensive routine: draw the point roughly before computing any argument — ten seconds of sketch eliminates the entire family of quadrant errors — and for products, add arguments only after each has been correctly placed, never before.
Frequently asked questions
What are the modulus and argument of z = x + iy?
r = √(x² + y²) and θ = atan by quadrant (tan θ = y/x, adjusted by π or −π outside the first and fourth quadrants).
What happens to modulus and argument under multiplication?
Moduli multiply and arguments add: z1z2 has modulus r1r2 and argument θ1 + θ2.
How does multiplication rotate the plane?
Multiplying by cos α + i sin α rotates every point about the origin by α with lengths unchanged — the polar form's geometric signature.
What is the principal argument's range?
(−π, π]; every other argument differs from it by a multiple of 2π.
How do you rotate z about a point a instead of the origin?
Compute a + (z − a)e^{iα}: translate a to the origin, rotate by multiplication, translate back.