IB Maths AA HLTopic 1 — Number & AlgebraPaper 1 & 2HL only~10 min read
Modulus-Argument (Polar) Form
Cartesian form is the natural way to write a complex number when you’re adding or subtracting. But once multiplication, division, or powers come into play, a different form makes life much easier — modulus-argument form, also called polar form. Instead of “go right x, then up y“, polar form says “stand at the origin, point in direction θ, then walk a distance r“. The same complex number, just described differently. The beauty? Multiplication becomes a one-line calculation: moduli multiply, arguments add. No more expanding brackets, no more i-squared substitutions. This note is the gateway to that simpler world.
📘 What you need to know
Polar form:z = r(cos θ + i sin θ), sometimes written compactly as z = r cis θ.
r = |z| is the modulus, and θ = arg(z) is the argument. The formula is in the formula booklet.
To convert from Cartesian to polar: compute r with Pythagoras and θ using the quadrant + reference angle.
To convert from polar to Cartesian: evaluate cos θ and sin θ, then expand to get x + yi.
The conjugate in polar form is z* = r(cos(−θ) + i sin(−θ)) = r cis(−θ). Same modulus, negated argument.
If a new argument falls outside the required range, add or subtract 2π to bring it back into range.
What is polar form?
Every complex number can be described by two numbers: how far it is from the origin (the modulus), and what direction it points in (the argument). Polar form combines these two numbers into a single expression.
Modulus-argument (polar) formz = r(cos θ + i sin θ) = r cis θ
✓ in the formula booklet
Where r = |z| (the modulus) and θ = arg(z) (the argument). The shorthand “cis θ” is read as “cos plus isin θ” — neat little mnemonic.
A complex number in polar form: distance r and direction θ
Read off the diagram: the horizontal component is r cos θ and the vertical component is r sin θ. So the Cartesian form x + yi can always be rewritten as r cos θ + i·r sin θ = r(cos θ + i sin θ). That’s where the polar form comes from.
A few examples to anchor it. z = 1 + i√3 has modulus 2 and argument π/3, so in polar form it’s 2(cos(π/3) + i sin(π/3)) or 2 cis(π/3). And z = 4(cos(π/4) + i sin(π/4)) converts to Cartesian form by computing 4·(√2/2) + 4·(√2/2)·i = 2√2 + 2√2 i.
Negative arguments must stay visible. If θ = −π/3, write z = r(cos(−π/3) + i sin(−π/3)). Don’t try to “simplify” by pulling the minus out of the sin — that would change the form into something that’s no longer valid polar form.
The complex conjugate in polar form
For z = r(cos θ + i sin θ), the conjugate is:
Conjugate in polar formz* = r(cos(−θ) + i sin(−θ)) = r cis(−θ)
Same modulus, negated argument. This matches what you already know geometrically: the conjugate is a reflection in the real axis, which doesn’t change the length but flips the direction angle.
🤔 Why does the formula work?
Because cos is an even function and sin is an odd function: cos(−θ) = cos θ and sin(−θ) = −sin θ. So r(cos(−θ) + i sin(−θ)) = r(cos θ − i sin θ) = x − yi, which is exactly z*.
Common trap:z* in polar form is notr(cos θ − i sin θ). That’s a Cartesian-style rewrite of the right answer — but it’s not in standard polar form, because the form requires a “+” between cos and i sin. Always write the conjugate as r(cos(−θ) + i sin(−θ)).
Multiplication in polar form
Here’s the payoff. Multiplying complex numbers in Cartesian form means expanding brackets and collecting i-squared terms. In polar form, it’s a single line.
Multiplication in polar formz1z2 = r1r2 · cis(θ1 + θ2)
Moduli
multiply
|z1z2| = |z1|·|z2|
Arguments
add
arg(z1z2) = arg(z1) + arg(z2)
The same rule extends to powers: since z2 = z · z, you get z2 = r2 cis(2θ). And z3 = r3 cis(3θ). This pattern is the heart of De Moivre’s theorem, which you’ll meet shortly.
Division in polar form
Division reverses both operations. Moduli divide; arguments subtract:
Division in polar formz1z2 = r1r2 · cis(θ1 − θ2)
Compare this to dividing in Cartesian form, where you’d have to multiply top and bottom by the conjugate of the denominator. The polar version skips all that — just divide the moduli and subtract the arguments. If multiplication and division are the main job in a question, convert to polar first.
Adjusting the argument’s range
After multiplication or division, the new argument might fall outside the standard range −π < θ ≤ π. When that happens, add or subtract 2π to bring it back into range — without changing the actual angle on the diagram.
🧭 Recipe — fixing an out-of-range argument
Compute the new argument by adding (multiplication) or subtracting (division) the two existing arguments.
Check whether it lies in the required range — usually −π < θ ≤ π or 0 ≤ θ < 2π.
If it’s too big, subtract 2π. If it’s too small (more negative than the lower bound), add 2π.
Check the new value is in range. If not, repeat.
Quick example: if you compute a new argument of 7π/6, that’s outside −π < θ ≤ π (since π = 6π/6 and 7π/6 > 6π/6). Subtract 2π = 12π/6 to get 7π/6 − 12π/6 = −5π/6. Now it’s in range, and represents the same angle on the diagram.
Worked examples
WE 1
Convert from Cartesian to polar form
Express z = √3 + i in modulus-argument (polar) form, with the argument in the range −π < θ ≤ π.
Step 1: Sketch — z is in Quadrant 1 (real > 0, imag > 0)argument will be positive acuteStep 2: Modulus by Pythagoras|z| = √((√3)² + 1²) = √(3 + 1) = √4 = 2Step 3: Argument using arctanarg(z) = arctan(1/√3) = π/6Step 4: Write in polar formz = 2(cos(π/6) + i sin(π/6))z = 2 cis(π/6)Q1 means the argument equals the reference angle directly — no adjustment needed
WE 2
Conjugate in polar form
Given z = 5(cos(2π/5) + i sin(2π/5)), write the complex conjugate z* in modulus-argument (polar) form.
Step 1: Same modulus, negated argumentr stays the same: r = 5θ flips sign: 2π/5 → −2π/5Step 2: Write in polar form (keep the +, not −, between cos and i sin)z* = 5(cos(−2π/5) + i sin(−2π/5))z* = 5 cis(−2π/5)don’t simplify to 5(cos(2π/5) − i sin(2π/5)) — that’s not standard polar form even though it equals z*
WE 3
Multiplication in polar form
Let z1 = 3 cis(π/4) and z2 = 2 cis(π/6). Find z1z2 in polar form.
Step 1: Moduli multiply|z₁z₂| = 3 · 2 = 6Step 2: Arguments addarg(z₁z₂) = π/4 + π/6common denominator: 3π/12 + 2π/12 = 5π/12Step 3: Write the answerz₁z₂ = 6 cis(5π/12)5π/12 lies in (−π, π], so no range adjustment needed
WE 4
Division in polar form
Let z1 = 12(cos(5π/6) + i sin(5π/6)) and z2 = 4(cos(π/3) + i sin(π/3)). Find z1/z2 in polar form, with the argument in the range −π < θ ≤ π.
Step 1: Moduli divide|z₁/z₂| = 12 / 4 = 3Step 2: Arguments subtractarg(z₁/z₂) = 5π/6 − π/3common denominator: 5π/6 − 2π/6 = 3π/6 = π/2Step 3: Write the answerz₁/z₂ = 3(cos(π/2) + i sin(π/2)) = 3 cis(π/2)π/2 is in range, no adjustment needed — and notice this is just 3i (since cis(π/2) = i)
WE 5
Range adjustment after multiplication
Let z1 = 2 cis(3π/4) and z2 = 5 cis(2π/3). Find z1z2 with the argument in the range −π < θ ≤ π.
Step 1: Moduli multiply|z₁z₂| = 2 · 5 = 10Step 2: Arguments addarg(z₁z₂) = 3π/4 + 2π/3common denominator: 9π/12 + 8π/12 = 17π/12Step 3: Check range — 17π/12 vs π = 12π/1217π/12 > π, so out of range — subtract 2π = 24π/1217π/12 − 24π/12 = −7π/12Step 4: Write the answer in rangez₁z₂ = 10 cis(−7π/12)always check the range after adding two arguments — going past π happens often
WE 6
Combined multiplication and division
Let z1 = 6 cis(π/3), z2 = 2 cis(π/4) and z3 = 3 cis(π/6). Find (z1 · z2) / z3 in polar form.
Step 1: Moduli — multiply two, then divide by third(6 · 2) / 3 = 12 / 3 = 4Step 2: Arguments — add two, then subtract third(π/3 + π/4) − π/6common denominator 12: (4π/12 + 3π/12) − 2π/12= 7π/12 − 2π/12 = 5π/12Step 3: Check range and write answer5π/12 is in (−π, π] ✓(z₁ · z₂) / z₃ = 4 cis(5π/12)in polar form, combined operations are just arithmetic on the moduli and arguments — way faster than expanding brackets
💡 Top tips
Use polar form whenever you see multiplication, division, or powers. Cartesian form is for adding and subtracting; polar is for everything else.
Always sketch first when computing arguments. The sketch decides the quadrant, which decides the sign and size of the argument.
Keep the “+” between cos and i sin. The standard form requires it. If you have a negative argument, write cos(−θ) + i sin(−θ), not cos θ − i sin θ.
cis is just a shorthand for cos + i sin. “cis θ” means “cos θ + i sin θ”. Use whichever notation you prefer (or whichever the question uses).
After multiplying or dividing, check the range. If the new argument falls outside the required range, add or subtract 2π.
Recognise common arguments: 0, π/6, π/4, π/3, π/2, 2π/3, 3π/4, 5π/6, π. These plus their negatives cover most exam questions.
Powers extend the multiplication rule:z2 = r2 cis(2θ), z3 = r3 cis(3θ), and so on. This is De Moivre’s theorem in disguise.
⚠ Common mistakes
Writing the conjugate as r(cos θ − i sin θ). That’s the Cartesian-style version. Polar form needs cos(−θ) + i sin(−θ) — keep the plus between cos and i sin.
Multiplying the arguments instead of adding. arg(z1·z2) = arg(z1) + arg(z2), not the product.
Adding the moduli instead of multiplying. |z1·z2| = |z1|·|z2|, not the sum.
Forgetting to check the range after computing a new argument. Going beyond π or below −π is common and needs a 2π adjustment.
Mixing up “add 2π” and “subtract 2π”. If the argument is too big (above π), subtract. If too small (below −π), add.
Using degrees instead of radians. In IB, complex number arguments are always in radians — always check your calculator’s mode.
Forgetting that “r” must be positive in polar form. If you’d compute a negative r, you’ve made an error — the modulus is never negative.
Polar form is the cleaner of the two main complex-number representations when products, quotients, and powers are involved. Once you internalise the moduli-multiply-and-arguments-add rule, you’ll find polar form is the form you reach for in most HL exam questions on complex numbers. The next note introduces an even more elegant version of the same idea: exponential (Euler’s) form, which uses the surprising fact that eiθ = cos θ + i sin θ. Same content, different notation — and somehow even slicker.
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