IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 HL only ~10 min read

Modulus & Argument

When you draw a complex number on an Argand diagram, you can describe it in two new ways. The modulus is the length of the arrow from the origin — how far the number is from zero. The argument is the angle the arrow makes with the positive real axis — the direction it points. Together, these two numbers give you a completely new “address” for any complex number, replacing real-and-imaginary parts with size-and-angle. This is the foundation of polar form, exponential form, and pretty much everything HL does with complex numbers from this point on. The skill to nail down here is computing the argument correctly in every quadrant — that’s where most marks are won or lost.

📘 What you need to know

The modulus of z = x + yi is its distance from the origin on an Argand diagram. Written |z| or r.
By Pythagoras: |z| = √(x² + y²). The modulus is never negative.
The argument of z is the angle from the positive real axis to the vector representing z, measured counter-clockwise in radians. Written arg(z) or θ.
Default range: −π < arg(z) ≤ π. Some questions use 0 ≤ arg(z) < 2π instead — read the question.
To find the argument, sketch first to identify the quadrant, then use right-angled trigonometry with tan to find the reference angle.
Modulus & conjugate: z·z* = |z|² — the multiplication of a complex number by its conjugate gives the squared modulus.
Under multiplication of two complex numbers, moduli multiply and arguments add. Under division, moduli divide and arguments subtract.

The modulus — how far from the origin?

Picture z = x + yi as a point or vector on an Argand diagram. The modulus is just the distance from the origin to that point. Drop a perpendicular to the real axis and you’ve got a right-angled triangle with horizontal side x and vertical side y. Pythagoras finishes the job.

Modulus of a complex number |z| = √(x² + y²) for z = x + yi ✓ in the formula booklet

The modulus is just the hypotenuse

So if z = 5 + 12i, then |z| = √(5² + 12²) = √(25 + 144) = √169 = 13. If z = −3 − 4i, then |z| = √(9 + 16) = 5. Negative parts make no difference because they get squared.

The modulus connects to the conjugate: multiplying a complex number by its conjugate gives the squared modulus. z·z* = (a + bi)(a − bi) = a² + b² = |z|².

A small warning: the moduli of two complex numbers do not add up to the modulus of their sum. So |z₁ + z₂| ≠ |z₁| + |z₂| in general. For example, |3 + 4i| = 5 and |−3 + 4i| = 5, but their sum is 8i with modulus 8 — not 10.

The argument — what direction does it point?

The argument measures the angle the vector makes with the positive real axis, going counter-clockwise. It’s basically the “direction” of the complex number, given as an angle in radians.

Argument — the direction angle arg(z) = θ measured counter-clockwise from positive real axis
default range: −π < θ ≤ π

The sign of the argument depends entirely on which quadrant z sits in. Here’s a complete map — memorise it:

Argument by quadrant — sign and size

Quadrant 1

+ acute

arg(z) is positive and between 0 and π/2

Quadrant 2

+ obtuse

arg(z) is positive and between π/2 and π

Quadrant 3

− obtuse

arg(z) is negative, between −π and −π/2

Quadrant 4

− acute

arg(z) is negative, between −π/2 and 0

How to compute the argument

The trick is always to use right-angled trigonometry on the triangle formed by the vector and the real axis. The angle in that triangle gives you the reference angle, written α (alpha). Then you adjust the sign and add or subtract from π depending on the quadrant.

The reference angle α = arctan(|y| / |x|) (use absolute values — this is always positive acute)

🧭 Recipe — finding arg(z)

Sketch z on an Argand diagram and identify the quadrant.
Compute the reference angle α = arctan(|y| / |x|) using the absolute values.
Adjust to get the actual argument based on the quadrant:
Q1 → arg = +α Q2 → arg = π − α
Q3 → arg = −(π − α) Q4 → arg = −α
Check the range: −π < arg ≤ π. If you slipped outside, add or subtract 2π.

Don’t blindly use arctan(y/x) on your calculator. The arctan function only returns values in (−π/2, π/2) — it can’t tell Q2 from Q4 by itself. The sketch is what fixes the sign and the quadrant. Skip the sketch and you’ll lose marks for arguments in Q2 or Q3 nearly every time.

🤔 Why does Q2 use π − α?

Because the reference angle α sits inside the triangle below the vector, between the vector and the negative real axis. The full argument from the positive real axis goes all the way to the negative side (which is π) and then backs off by α to reach the vector. So arg = π − α. The same logic gives the other three formulas.

Modulus & argument under multiplication and division

Here’s the rule that makes complex numbers genuinely beautiful. When you multiply two complex numbers, their moduli multiply and their arguments add. When you divide, moduli divide and arguments subtract.

Multiplication |z₁·z₂| = |z₁| · |z₂| AND arg(z₁·z₂) = arg(z₁) + arg(z₂)

Division |z₁/z₂| = |z₁| / |z₂| AND arg(z₁/z₂) = arg(z₁) − arg(z₂)

🤔 Why is this so neat?

Because in the modulus-argument picture, multiplying complex numbers becomes “scale and rotate”. The lengths multiply (scaling) and the angles add (rotating). This single fact is why polar form, exponential form, and De Moivre’s theorem all work — they’re all consequences of this rule. You’ll meet it in a much more powerful form in the Further Complex Numbers notes.

Worked examples

WE 1

Find the modulus

Find the modulus of z = 8 − 6i, giving an exact answer.

Step 1: Identify x and y x = 8, y = −6 Step 2: Apply the formula |z| = √(x² + y²) |z| = √(8² + (−6)²) = √(64 + 36) = √100 = 10 |z| = 10 negative imaginary part squares to a positive — the modulus is always a non-negative real number

WE 2

Argument in Quadrant 1

Find the modulus and argument of z = 5 + 5i.

Step 1: Sketch — z is in Q1 (real > 0, imag > 0) argument will be positive acute Step 2: Modulus |z| = √(5² + 5²) = √50 = 5√2 Step 3: Reference angle α = arctan(|y|/|x|) α = arctan(5/5) = arctan(1) = π/4 Step 4: Q1 → arg = +α arg(z) = π/4 |z| = 5√2, arg(z) = π/4 Q1 is the easiest case — the reference angle IS the argument, no adjustments needed

WE 3

Argument in Quadrant 2

Find the modulus and argument of z = −√3 + i.

Step 1: Sketch — z is in Q2 (real < 0, imag > 0) argument will be positive obtuse, between π/2 and π Step 2: Modulus |z| = √((−√3)² + 1²) = √(3 + 1) = √4 = 2 Step 3: Reference angle α = arctan(|y|/|x|) α = arctan(1/√3) = π/6 Step 4: Q2 → arg = π − α arg(z) = π − π/6 = 5π/6 |z| = 2, arg(z) = 5π/6 if you’d just typed arctan(1/(−√3)) into the calculator, you’d have got −π/6 — wrong quadrant. The sketch saves you.

WE 4

Argument in Quadrant 3

Find the modulus and argument of z = −2 − 2i.

Step 1: Sketch — z is in Q3 (real < 0, imag < 0) argument will be negative obtuse, between −π and −π/2 Step 2: Modulus |z| = √((−2)² + (−2)²) = √8 = 2√2 Step 3: Reference angle α α = arctan(|−2|/|−2|) = arctan(1) = π/4 Step 4: Q3 → arg = −(π − α) arg(z) = −(π − π/4) = −3π/4 |z| = 2√2, arg(z) = −3π/4 Q3 → measure clockwise (negative), and start from the negative real axis to get the obtuse angle

WE 5

Modulus and argument under multiplication

Two complex numbers have |z₁| = 4 with arg(z₁) = π/3 and |z₂| = 3 with arg(z₂) = π/6. Find |z₁z₂| and arg(z₁z₂).

Step 1: Moduli multiply |z₁z₂| = |z₁| · |z₂| = 4 · 3 = 12 Step 2: Arguments add arg(z₁z₂) = arg(z₁) + arg(z₂) = π/3 + π/6 = 2π/6 + π/6 = 3π/6 = π/2 |z₁z₂| = 12, arg(z₁z₂) = π/2 no need to actually multiply (a + bi)(c + di) here — modulus-argument form makes it a one-line calculation

WE 6

Argument in the [0, 2π) range

Find the argument of z = −1 − i in the range 0 ≤ arg(z) < 2π.

Step 1: Sketch — z in Q3 in the standard range, arg would be negative obtuse Step 2: Find the standard argument first α = arctan(1/1) = π/4 Q3 → arg(z) = −(π − π/4) = −3π/4 (in standard range) Step 3: Convert to [0, 2π) range — add 2π arg(z) = −3π/4 + 2π = −3π/4 + 8π/4 = 5π/4 arg(z) = 5π/4 (in [0, 2π) range) if a question gives a range that excludes negatives, just add 2π to any negative argument to land in [0, 2π)

💡 Top tips

Always sketch first. The sketch decides the quadrant, which decides the sign and size of the argument. Skip the sketch and you’ll get arguments wrong.
Memorise the four quadrant rules. Q1 → +α, Q2 → π−α, Q3 → −(π−α), Q4 → −α. These four lines are the whole technique.
Use absolute values inside arctan. The reference angle is always positive acute — get it from arctan(|y|/|x|), then use the quadrant to set the actual sign.
Modulus is always positive. If you ever get a negative modulus, you’ve made a mistake — recheck your squaring.
Memorise common reference angles. arctan(1) = π/4, arctan(√3) = π/3, arctan(1/√3) = π/6. These come up in nearly every exam question.
Read the range carefully. Default is −π < arg ≤ π. Some questions ask for [0, 2π). If your answer falls outside, just add 2π to bring it into range.
For products and quotients, use the modulus-argument rule instead of multiplying out brackets. Moduli multiply/divide; arguments add/subtract. Way faster.

⚠ Common mistakes

Trusting arctan(y/x) on the calculator. Calculator arctan only returns angles in (−π/2, π/2) — it can’t distinguish Q2 from Q4. Always sketch and adjust.
Forgetting the minus sign in Q3 and Q4. Both quadrants give negative arguments in the standard range. If you get a positive answer for a Q3 number, you’ve slipped.
Giving the argument in degrees instead of radians. The IB always expects radians for complex numbers. Watch your calculator’s MODE setting.
Adding moduli of two complex numbers. |z₁ + z₂| ≠ |z₁| + |z₂|. Sums of complex numbers don’t have additive moduli.
Computing arctan with raw signed values. Use absolute values for the reference angle, then adjust — this is the safe pattern.
Forgetting to check the range. If your final answer is outside −π < arg ≤ π (or [0, 2π) when specified), add or subtract 2π.
Trying to find arg(0). The complex number 0 has no argument — there’s no direction associated with the origin. If you see this in a question, the answer is “undefined”.

Modulus and argument are the building blocks for everything that comes next in HL complex numbers — polar form, exponential form, De Moivre’s theorem, and the geometry of multiplication. Get really comfortable with the four-quadrant argument rule before moving on. Spend an evening doing 10–12 mixed problems where you compute modulus and argument from scratch — by the end of it, the technique will feel automatic, and the rest of the topic will fall into place much faster.

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