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Complex number arithmetic and the Argand plane in HSC Mathematics Extension 2: Cartesian and polar form, modulus, argument, conjugate and geometric interpretation

Syllabus dot point

“Represent complex numbers in Cartesian and polar form, perform arithmetic, and interpret modulus, argument and conjugate geometrically on the Argand plane”

HSCMaths Extension 2Introduction to Complex Numbers (MEX-N1)14 min read

Quick answer

A complex number is z=x+iyz = x + iy where i2=−1i^2 = -1. On the Argand plane it is the point (x,y)(x, y). Its modulus is ∣z∣=x2+y2|z| = \sqrt{x^2 + y^2} (distance from the origin) and its argument is the angle θ\theta the vector to zz makes with the positive real axis. The conjugate zˉ=x−iy\bar z = x - iy is the reflection in the real axis. Polar form is z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta). Addition follows the parallelogram law; multiplication rotates and scales.

Jump to a section
  1. What this dot point is asking
  2. Why complex numbers exist
  3. Cartesian arithmetic
  4. The Argand plane
  5. Modulus and argument
  6. The conjugate
  7. Polar (modulus-argument) form
  8. Addition is the parallelogram law
  9. Multiplication is rotate and scale
  10. Geometry of conjugation, negation and multiplying by i
  11. Powers of ii and the imaginary unit
  12. Equality and solving for real unknowns
  13. How exam questions ask about complex arithmetic
  14. Exam-style questions

What this dot point is asking

NESA wants you to work fluently with complex numbers in both Cartesian form z=x+iyz = x + iy and polar (modulus-argument) form z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta). You must add, subtract, multiply and divide them, find the modulus, argument and conjugate, and read all of these geometrically on the Argand plane. This is the foundation dot point of the whole complex-numbers course: modulus-argument form, de Moivre's theorem, loci and complex polynomials are all written in this language, so the fluency built here is repaid in every later topic.

Why complex numbers exist

Real numbers run out when you try to solve x2=−1x^2 = -1, because no real number squares to a negative. Defining a single new number ii with i2=−1i^2 = -1 fixes this, and remarkably it fixes everything at once: once ii is allowed, every polynomial equation has a solution (the fundamental theorem of algebra). A complex number combines a real part and an imaginary part into z=x+iyz = x + iy, where x=Re⁡(z)x = \operatorname{Re}(z) and y=Im⁡(z)y = \operatorname{Im}(z) are both real. Note that Im⁡(z)=y\operatorname{Im}(z) = y is the real coefficient of ii, not iyiy itself: the imaginary part of 2+3i2 + 3i is 33, not 3i3i.

The real numbers sit inside the complex numbers as the special case y=0y = 0, and the purely imaginary numbers are the case x=0x = 0. This is why the two axes of the Argand plane are the real axis and the imaginary axis: a complex number is genuinely two-dimensional, carrying twice the information of a real number.

Cartesian arithmetic

For z=a+biz = a + bi and w=c+diw = c + di:

z+w=(a+c)+(b+d)i,z−w=(a−c)+(b−d)i.z + w = (a + c) + (b + d)i, \qquad z - w = (a - c) + (b - d)i.

Addition and subtraction work componentwise, exactly like adding vectors: combine the real parts, then combine the imaginary parts. Multiplication uses ordinary expansion together with i2=−1i^2 = -1:

zw=(a+bi)(c+di)=(ac−bd)+(ad+bc)i.zw = (a + bi)(c + di) = (ac - bd) + (ad + bc)i.

You never have to memorise this formula; just expand the brackets and replace i2i^2 with −1-1. Division is performed by multiplying numerator and denominator by the conjugate of the denominator, which "realises" the denominator:

zw=a+bic+di⋅c−dic−di=(ac+bd)+(bc−ad)ic2+d2.\frac{z}{w} = \frac{a + bi}{c + di} \cdot \frac{c - di}{c - di} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}.

The denominator c2+d2=∣w∣2c^2 + d^2 = |w|^2 is real and positive whenever w≠0w \neq 0, so the quotient lands cleanly in the form x+iyx + iy.

The Argand plane

The Argand plane represents z=x+iyz = x + iy as the point (x,y)(x, y) or, equivalently, the position vector (arrow) from the origin to that point. The horizontal axis is the real axis and the vertical axis the imaginary axis. Every fact about a complex number has a picture here: the modulus is a length, the argument is an angle, the conjugate is a reflection, addition is a parallelogram, and multiplication is a rotation combined with a scaling. Sketching the point before doing algebra is the single habit that prevents the most common Extension 2 mistakes, especially picking the wrong quadrant for an argument.

A complex number on the Argand plane with its modulus, argument and conjugate The point z = 3 + 2i is plotted on the Argand plane. The vector from the origin to z has length r (the modulus) and makes an angle theta (the argument) with the positive real axis. The conjugate 3 minus 2i is the reflection of z in the real axis. Re Im 1 2 3 2i -2i θ r z = 3 + 2i z̅ = 3 - 2i

Modulus and argument

The modulus is the distance from the origin:

∣z∣=x2+y2.|z| = \sqrt{x^2 + y^2}.

The argument arg⁡z=θ\arg z = \theta is the angle measured anticlockwise from the positive real axis to the vector representing zz. It satisfies tan⁡θ=yx\tan\theta = \dfrac{y}{x}, but you must choose the quadrant using the signs of xx and yy. The principal argument is taken in (−π,π](-\pi, \pi].

Key identities, each of which has a clean geometric reading:

∣z∣2=zzˉ,∣zw∣=∣z∣ ∣w∣,arg⁡(zw)=arg⁡z+arg⁡w.|z|^2 = z\bar z, \qquad |zw| = |z|\,|w|, \qquad \arg(zw) = \arg z + \arg w.

The first turns "modulus squared" into a product you can compute by multiplying out; the second and third are the algebra behind the rotate-and-scale picture of multiplication below.

The conjugate

The conjugate zˉ=x−iy\bar z = x - iy is the reflection of zz in the real axis. Useful properties include z+w‾=zˉ+wˉ\overline{z + w} = \bar z + \bar w, zw‾=zˉ wˉ\overline{zw} = \bar z\,\bar w, and z+zˉ=2Re⁡(z)z + \bar z = 2\operatorname{Re}(z), while z−zˉ=2iIm⁡(z)z - \bar z = 2i\operatorname{Im}(z). The single most useful fact is

zzˉ=(x+iy)(x−iy)=x2+y2=∣z∣2,z\bar z = (x + iy)(x - iy) = x^2 + y^2 = |z|^2,

a real number. This is exactly why multiplying by the conjugate realises a denominator: it converts a complex divisor into its squared modulus.

Polar (modulus-argument) form

Writing x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta where r=∣z∣r = |z| gives

z=r(cos⁡θ+isin⁡θ).z = r(\cos\theta + i\sin\theta).

In polar form, multiplication multiplies the moduli and adds the arguments, while division divides the moduli and subtracts the arguments. This makes polar form the natural setting for powers and roots, and it is developed fully in the modulus-argument-form and de Moivre dot points.

Addition is the parallelogram law

Because real and imaginary parts add separately, complex addition is vector addition: z+wz + w is found by the head-to-tail (triangle) rule or, equivalently, by completing the parallelogram on zz and ww. The sum is the diagonal of that parallelogram from the origin. The other diagonal, drawn from the head of ww to the head of zz, represents the difference z−wz - w. Recognising this turns many "find the fourth vertex" or "show this is a rhombus" geometry questions into one-line vector statements.

The figure below builds z+wz + w for z=3+iz = 3 + i and w=1+2iw = 1 + 2i one move at a time, finishing with the full parallelogram.

Stage 1, plot the two numbers as vectors. Draw z=3+iz = 3 + i and w=1+2iw = 1 + 2i as arrows from the origin OO. Nothing has been combined yet; these are the two numbers we want to add.

Adding complex numbers on the Argand plane, stage 1 Addition of z = 3 + i and w = 1 + 2i on the Argand plane shown as vectors; the sum 4 + 3i is the diagonal of the parallelogram. Re Im 1 2 3 4 1i 2i 3i z w O Plot z and w as vectors from the origin O.

Stage 2, slide w to the head of z. A complex number is a free vector, so move a copy of ww until its tail sits at the head of zz. Walking along zz and then along ww lands you at the sum.

Adding complex numbers on the Argand plane, stage 2 Addition of z = 3 + i and w = 1 + 2i on the Argand plane shown as vectors; the sum 4 + 3i is the diagonal of the parallelogram. Re Im 1 2 3 4 1i 2i 3i z w w O Slide a copy of w to the head of z (head to tail).

Stage 3, the sum is the diagonal. Completing the parallelogram, the sum z+wz + w is the diagonal from OO to the far corner. Reading off coordinates, z+w=(3+1)+(1+2)i=4+3iz + w = (3 + 1) + (1 + 2)i = 4 + 3i, which matches adding the components directly.

Adding complex numbers on the Argand plane, stage 3 Addition of z = 3 + i and w = 1 + 2i on the Argand plane shown as vectors; the sum 4 + 3i is the diagonal of the parallelogram. Re Im 1 2 3 4 1i 2i 3i z w w z + w = 4 + 3i O Sum = diagonal from O: add components (3+1, 1+2).

Multiplication is rotate and scale

Multiplication has a strikingly different picture from addition. Since ∣zw∣=∣z∣∣w∣|zw| = |z||w| and arg⁡(zw)=arg⁡z+arg⁡w\arg(zw) = \arg z + \arg w, multiplying zz by ww scales the length of zz by the factor ∣w∣|w| and rotates it anticlockwise by the angle arg⁡w\arg w. Multiplication does not slide points around as addition does; it spins and stretches them about the origin.

Below, z=2+iz = 2 + i is multiplied by w=1+iw = 1 + i. Since ∣w∣=2|w| = \sqrt2 and arg⁡w=π4\arg w = \tfrac{\pi}{4} (45∘45^\circ), the result zwzw is zz rotated 45∘45^\circ anticlockwise and stretched by 2\sqrt2. Expanding confirms it: (2+i)(1+i)=2+2i+i+i2=1+3i(2 + i)(1 + i) = 2 + 2i + i + i^2 = 1 + 3i.

Multiplication as rotation and scaling on the Argand plane Multiplying z = 2 + i by w = 1 + i rotates the vector anticlockwise by the argument of w (45 degrees) and scales its length by the modulus of w (root 2), giving zw = 1 + 3i. Re Im 1 2 3 1i 2i 3i z zw = 1 + 3i rotate 45°, scale ×√2 O

Geometry of conjugation, negation and multiplying by i

Three transformations of z=x+iyz = x + iy appear constantly, and each is a rigid rotation or reflection about the origin:

  • Conjugation z↦zˉz \mapsto \bar z reflects the point in the real axis (it negates the imaginary part, and so negates the argument).
  • Negation z↦−zz \mapsto -z rotates it by 180∘180^\circ about the origin.
  • Multiplying by ii rotates the point 90∘90^\circ anticlockwise without changing its distance from the origin, because ∣i∣=1|i| = 1 and arg⁡i=π2\arg i = \tfrac{\pi}{2}.

Reading these geometrically is often faster than algebra: iziz is always perpendicular to zz, which is why zz and iziz form two adjacent sides of a square in many locus problems.

Conjugation, negation and multiplication by i on the Argand plane For z = 3 + 1.6i: the conjugate reflects z in the real axis, negation rotates z by 180 degrees about the origin, and multiplying by i rotates z by 90 degrees anticlockwise. Re Im z z̅ -z iz z̅ reflects, -z turns 180°, iz turns 90° anticlockwise.

Powers of ii and the imaginary unit

The imaginary unit ii satisfies i2=−1i^2 = -1, and its powers cycle with period four: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1, then the pattern repeats. To simplify a high power such as i23i^{23}, divide the exponent by 44 and keep the remainder: 23=4(5)+323 = 4(5) + 3, so i23=i3=−ii^{23} = i^3 = -i. This cyclic behaviour is just the rotate-by-90∘90^\circ picture above repeated: i=cos⁡π2+isin⁡π2i = \cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}, so each power turns the point a further quarter turn anticlockwise about the origin, returning to the start after four steps.

Equality and solving for real unknowns

Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal. This lets you turn one complex equation into two real simultaneous equations. For example, if (a+bi)(2−i)=5+5i(a + bi)(2 - i) = 5 + 5i with a,ba, b real, expand the left side to (2a+b)+(2b−a)i(2a + b) + (2b - a)i, then equate parts: 2a+b=52a + b = 5 and 2b−a=52b - a = 5. Solving gives a=1a = 1, b=3b = 3. Splitting into real and imaginary parts is the standard route for finding square roots of a complex number and for any problem that supplies a Cartesian target.

Key fact

For z=x+iyz = x + iy: the modulus is ∣z∣=x2+y2|z| = \sqrt{x^2 + y^2} (distance from OO), the argument θ=arg⁡z\theta = \arg z is the anticlockwise angle from the positive real axis, and the conjugate is zˉ=x−iy\bar z = x - iy (reflection in the real axis), with zzˉ=∣z∣2z\bar z = |z|^2. Addition adds components (parallelogram law); multiplication multiplies moduli and adds arguments (rotate and scale). Divide by realising the denominator: multiply top and bottom by the conjugate of the denominator.

How exam questions ask about complex arithmetic

The wording varies, but each phrasing maps to one routine:

  • "Express in the form x+iyx + iy" with a fraction: realise the denominator (multiply by the conjugate of the denominator), then simplify.
  • "Simplify z2z^2" or a product: expand the brackets and replace every i2i^2 with −1-1.
  • "Find ∣z∣|z|" or "the modulus": compute x2+y2\sqrt{x^2 + y^2}; for a product or quotient, use ∣zw∣=∣z∣∣w∣|zw| = |z||w| or ∣z/w∣=∣z∣/∣w∣|z/w| = |z|/|w| as a shortcut.
  • "Find arg⁡z\arg z": sketch the point, find the reference angle from tan⁡−1∣y/x∣\tan^{-1}|y/x|, then fix the quadrant from the signs of xx and yy; give the principal value in (−π,π](-\pi, \pi].
  • "Find the conjugate" or a symbol zˉ\bar z: change the sign of the imaginary part.
  • "On an Argand diagram, plot / show that ...": read addition as the parallelogram and multiplication by a unit number as a rotation; "multiply by ii" means rotate 90∘90^\circ.
  • "Find real aa and bb such that ...": equate real and imaginary parts to get two simultaneous equations.
Worked example

Express 3+i1−2i\dfrac{3 + i}{1 - 2i} in Cartesian form and find its modulus

Express 3+i1−2i\dfrac{3 + i}{1 - 2i} in Cartesian form, then find its modulus.

Realise the denominator by multiplying by the conjugate: Division is complete only when the denominator is real. The conjugate of 1−2i1 - 2i is 1+2i1 + 2i, and multiplying by 1+2i1+2i\frac{1+2i}{1+2i} (which equals 11) turns the denominator into a real number without changing the value:

3+i1−2i⋅1+2i1+2i=(3+i)(1+2i)12+22.\frac{3 + i}{1 - 2i} \cdot \frac{1 + 2i}{1 + 2i} = \frac{(3 + i)(1 + 2i)}{1^2 + 2^2}.

Expand the numerator using i2=−1i^2 = -1: Multiply out (3+i)(1+2i)(3 + i)(1 + 2i) using the distributive law, remembering to replace i2i^2 with −1-1:

(3+i)(1+2i)=3+6i+i+2i2=3+7i−2=1+7i.(3 + i)(1 + 2i) = 3 + 6i + i + 2i^2 = 3 + 7i - 2 = 1 + 7i.

The denominator is 12+22=51^2 + 2^2 = 5, so:

1+7i5=15+75i.\frac{1 + 7i}{5} = \frac{1}{5} + \frac{7}{5}i.

Verify using the modulus quotient rule: The identity ∣zw∣=∣z∣∣w∣\left|\frac{z}{w}\right| = \frac{|z|}{|w|} offers a quick check. Compute the two moduli separately:

∣3+i∣=9+1=10,∣1−2i∣=1+4=5.|3 + i| = \sqrt{9 + 1} = \sqrt{10}, \qquad |1 - 2i| = \sqrt{1 + 4} = \sqrt{5}.

∣3+i1−2i∣=105=2.\left|\frac{3+i}{1-2i}\right| = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}.

Computing directly from the Cartesian form 15+75i\frac{1}{5} + \frac{7}{5}i gives (1/5)2+(7/5)2=50/25=2\sqrt{(1/5)^2 + (7/5)^2} = \sqrt{50/25} = \sqrt{2}. Both methods agree.

Final answer: 3+i1−2i=15+75i\dfrac{3+i}{1-2i} = \dfrac{1}{5} + \dfrac{7}{5}i, with modulus 2\sqrt{2}.

Worked example: square roots from real and imaginary parts

Find the square roots of 3+4i3 + 4i

Find the two square roots of 3+4i3 + 4i in the form a+iba + ib with a,ba, b real.

Set up the defining equation: A square root a+iba + ib must satisfy (a+ib)2=3+4i(a + ib)^2 = 3 + 4i. Expand the left side, using i2=−1i^2 = -1, to separate real and imaginary parts:

(a+ib)2=a2−b2+2ab i.(a + ib)^2 = a^2 - b^2 + 2ab\,i.

Equate real and imaginary parts: Matching the real parts and the imaginary parts of a2−b2+2ab i=3+4ia^2 - b^2 + 2ab\,i = 3 + 4i gives two real equations:

a2−b2=3,2ab=4.a^2 - b^2 = 3, \qquad 2ab = 4.

Add the modulus equation to solve: Taking the modulus of both sides of (a+ib)2=3+4i(a + ib)^2 = 3 + 4i gives a third, very convenient equation, since ∣a+ib∣2=a2+b2|a + ib|^2 = a^2 + b^2 and ∣3+4i∣=9+16=5|3 + 4i| = \sqrt{9 + 16} = 5:

a2+b2=5.a^2 + b^2 = 5.

Adding a2−b2=3a^2 - b^2 = 3 and a2+b2=5a^2 + b^2 = 5 gives 2a2=82a^2 = 8, so a2=4a^2 = 4 and a=±2a = \pm 2. Then b2=1b^2 = 1, so b=±1b = \pm 1. From 2ab=4>02ab = 4 > 0, aa and bb have the same sign.

State both roots and check: The two square roots are therefore 2+i2 + i and −(2+i)=−2−i-(2 + i) = -2 - i. Check the first: (2+i)2=4+4i+i2=3+4i(2 + i)^2 = 4 + 4i + i^2 = 3 + 4i. Correct.

Final answer: the square roots of 3+4i3 + 4i are ±(2+i)\pm(2 + i).

Common traps
Forgetting i2=−1i^2 = -1
When expanding products the i2i^2 term flips sign and merges with the real part. Leaving it as +i2+i^2 loses marks.
Wrong quadrant for the argument
Using arctan⁡(y/x)\arctan(y/x) blindly puts every argument in (−π/2,π/2)(-\pi/2, \pi/2). Sketch the point and check the signs of xx and yy to place the angle in the correct quadrant.
Not realising the denominator
Division is only complete once the denominator is a real number. Multiply by the conjugate, do not stop at a complex denominator.
Confusing modulus with argument
The modulus is a length; the argument is an angle. Multiplying complex numbers adds the arguments but multiplies the moduli.
Calling the imaginary part iyiy
Im⁡(z)\operatorname{Im}(z) is the real number yy, not iyiy. The imaginary part of 2+3i2 + 3i is 33.
Exam technique

Sketch an Argand diagram before answering anything about modulus or argument; it is the cheapest insurance against a wrong quadrant. Lay division out as "multiply by the conjugate over itself", show the realised denominator c2+d2c^2 + d^2 explicitly, and only then split into x+iyx + iy. Keep moduli as exact surds (2\sqrt{2}, 252\sqrt{5}) unless a decimal is asked for, and quote the principal argument in (−π,π](-\pi, \pi]. When a question gives an Argand diagram and asks you to mark z+wz + w, iziz or zˉ\bar z, use the geometry (parallelogram, 90∘90^\circ turn, reflection) rather than recomputing coordinates; it is faster and it is what the marks reward.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2022 HSC-style2 marks
Express (3 - i)/(2 + i) in the form x + iy, where x and y are real numbers.
Show worked answer →

Realise the denominator by multiplying top and bottom by the conjugate 2 - i.

(3 - i)/(2 + i) = (3 - i)(2 - i) / ((2 + i)(2 - i)).

Numerator: (3 - i)(2 - i) = 6 - 3i - 2i + i^2 = 6 - 5i - 1 = 5 - 5i.

Denominator: (2 + i)(2 - i) = 4 - i^2 = 4 + 1 = 5.

So the expression equals (5 - 5i)/5 = 1 - i, giving x = 1 and y = -1.

Mark notes: 1 mark for multiplying by the conjugate to get a real denominator, 1 mark for the correct final form 1 - i.

2021 HSC-style2 marks
The complex numbers z = 5 + i and w = 2 - 4i are given. Find (conjugate of z)/w, giving your answer in Cartesian form.
Show worked answer →

The conjugate of z = 5 + i is z-bar = 5 - i. We need (5 - i)/(2 - 4i).

Multiply numerator and denominator by the conjugate of the denominator, 2 + 4i.

(5 - i)(2 + 4i) = 10 + 20i - 2i - 4i^2 = 10 + 18i + 4 = 14 + 18i.

(2 - 4i)(2 + 4i) = 4 - 16i^2 = 4 + 16 = 20.

So (conjugate of z)/w = (14 + 18i)/20 = 7/10 + (9/10)i.

Mark notes: 1 mark for forming the conjugate and multiplying out, 1 mark for the simplified Cartesian answer 7/10 + (9/10)i.

2024 HSC-style2 marks
Let z = 2 + 3i and w = 1 - 5i. (i) Find z + (conjugate of w). (ii) Find z^2.
Show worked answer →

Part (i). The conjugate of w = 1 - 5i is w-bar = 1 + 5i.

z + w-bar = (2 + 3i) + (1 + 5i) = 3 + 8i.

Part (ii). Square z directly, using i^2 = -1.

z^2 = (2 + 3i)^2 = 4 + 12i + 9i^2 = 4 + 12i - 9 = -5 + 12i.

Mark notes: this was set as two 1 mark parts. 1 mark for z + (conjugate of w) = 3 + 8i, and 1 mark for z^2 = -5 + 12i. Add the real and imaginary parts separately and remember i^2 = -1 when expanding the square.

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