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Complex arithmetic and the Argand plane (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

“Complex numbers extend the reals with i where i squared equals minus one, and are represented as points or vectors on the Argand plane.”

8 min read

Quick answer

A complex number is z=a+biz=a+bi where a,ba,b are real and i2=−1i^2=-1. Add and subtract by combining real and imaginary parts; multiply by expanding and using i2=−1i^2=-1; divide by multiplying top and bottom by the conjugate zˉ=a−bi\bar z=a-bi. Each z=a+biz=a+bi is plotted as the point (a,b)(a,b) on the Argand plane.

Jump to a section
  1. What this dot point is asking
  2. Arithmetic in Cartesian form
  3. The Argand plane
  4. Powers of i and equality of complex numbers
  5. Loci on the Argand plane
  6. Common errors
  7. Why it matters
  8. Exam-style questions

What this dot point is asking

The equation x2=−1x^2=-1 has no real solution, so mathematicians define the imaginary unit ii with i2=−1i^2=-1. A complex number in Cartesian (rectangular) form is z=a+biz=a+bi, where a=Re⁡(z)a=\operatorname{Re}(z) is the real part and b=Im⁡(z)b=\operatorname{Im}(z) is the imaginary part.

Arithmetic in Cartesian form

Addition and subtraction act componentwise:

(a+bi)±(c+di)=(a±c)+(b±d)i.(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i.

Multiplication uses the distributive law and i2=−1i^2=-1:

(a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+(ad+bc)i.(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i.

Complex conjugate

The conjugate of z=a+biz=a+bi is zˉ=a−bi\bar z=a-bi. Conjugating reflects the point across the real axis. Crucially, zzˉ=(a+bi)(a−bi)=a2+b2z\bar z=(a+bi)(a-bi)=a^2+b^2, which is real and non-negative.

Division removes ii from the denominator by multiplying numerator and denominator by the conjugate of the denominator:

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

Dividing two complex numbers

Worked example

Express 3+2i1−4i\dfrac{3+2i}{1-4i} in the form a+bia+bi. To divide complex numbers in Cartesian form, multiply numerator and denominator by the conjugate of the denominator so that the denominator becomes real.

Step 1: Multiply by the conjugate of the denominator

The conjugate of 1−4i1-4i is 1+4i1+4i. Multiplying top and bottom by 1+4i1+4i leaves the value unchanged but clears the imaginary part from the denominator, because a complex number times its conjugate is always real.

3+2i1−4i⋅1+4i1+4i=(3+2i)(1+4i)(1−4i)(1+4i).\frac{3+2i}{1-4i}\cdot\frac{1+4i}{1+4i}=\frac{(3+2i)(1+4i)}{(1-4i)(1+4i)}.

Step 2: Expand the numerator

Expand the product in the numerator using the distributive law, and replace i2i^2 with −1-1 to collect real and imaginary parts.

Numerator: (3+2i)(1+4i)=3+12i+2i+8i2=3+14i−8=−5+14i.(3+2i)(1+4i)=3+12i+2i+8i^2=3+14i-8=-5+14i.

Step 3: Simplify the denominator and write the final form

The denominator is the product of conjugates, which equals the sum of squares of the real and imaginary parts and is therefore real.

Denominator: (1−4i)(1+4i)=12+42=17.(1-4i)(1+4i)=1^2+4^2=17.

So

3+2i1−4i=−5+14i17=−517+1417i.\frac{3+2i}{1-4i}=\frac{-5+14i}{17}=-\frac{5}{17}+\frac{14}{17}i.

Final answer: −517+1417i-\dfrac{5}{17}+\dfrac{14}{17}i.

The Argand plane

The Argand plane represents each complex number z=a+biz=a+bi as the point (a,b)(a,b), with the horizontal axis real and the vertical axis imaginary. Equivalently zz is the position vector from the origin to that point.

This geometric picture makes operations visual:

  • Addition of z1z_1 and z2z_2 follows the parallelogram (tip-to-tail vector) rule.
  • The conjugate zˉ\bar z is the reflection of zz in the real axis.
  • The modulus ∣z∣=a2+b2|z|=\sqrt{a^2+b^2} is the distance from the origin to the point, the length of the vector. Note ∣z∣2=zzˉ|z|^2=z\bar z.
Modulus and conjugate identities

∣z∣=a2+b2,zzˉ=∣z∣2,z1z2‾=zˉ1 zˉ2,∣z1z2∣=∣z1∣∣z2∣|z|=\sqrt{a^2+b^2},\qquad z\bar z=|z|^2,\qquad \overline{z_1z_2}=\bar z_1\,\bar z_2,\qquad |z_1z_2|=|z_1||z_2|

Modulus of a difference

Worked example

Find ∣z1−z2∣|z_1-z_2| where z1=4+iz_1=4+i and z2=1+5iz_2=1+5i.

Step 1: Subtract the complex numbers

Subtract real and imaginary parts separately to find the complex number z1−z2z_1-z_2, which represents the vector from z2z_2 to z1z_1 on the Argand plane.

z1−z2=(4−1)+(1−5)i=3−4i.z_1-z_2=(4-1)+(1-5)i=3-4i.

Step 2: Compute the modulus using the distance formula

The modulus of a complex number a+bia+bi is a2+b2\sqrt{a^2+b^2}, the straight-line distance from the origin to that point. Applying this to 3−4i3-4i gives the distance between the original two points.

∣z1−z2∣=32+(−4)2=9+16=25=5.|z_1-z_2|=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5.

Geometrically this is the distance between the two points on the Argand plane.

Final answer: ∣z1−z2∣=5|z_1-z_2|=5.

Powers of i and equality of complex numbers

The powers of ii cycle with period four: i1=ii^1=i, i2=−1i^2=-1, i3=−ii^3=-i, i4=1i^4=1, and then the pattern repeats. To simplify a high power, divide the exponent by 44 and use the remainder; for example i27=i24⋅i3=1⋅(−i)=−ii^{27}=i^{24}\cdot i^3=1\cdot(-i)=-i. Two complex numbers are equal only when both their real parts and their imaginary parts match, so an equation between complex numbers gives two simultaneous real equations - the technique behind the first worked exam question above and behind solving for unknowns in complex equations generally.

Loci on the Argand plane

Because ∣z−w∣|z-w| is the distance between the points zz and ww, modulus equations describe geometric shapes. ∣z−c∣=r|z-c|=r is a circle of radius rr centred at cc; ∣z−a∣=∣z−b∣|z-a|=|z-b| is the perpendicular bisector of the segment joining aa and bb, since it is the set of points equidistant from both. Recognising a modulus condition as a distance lets you translate between the algebra of complex numbers and the geometry of the plane, which is exactly what the circle question above tests.

Common errors

Treating i2i^2 as +1+1

By definition i2=−1i^2=-1, so i3=−ii^3=-i and i4=1i^4=1. A frequent slip is dropping the sign change when an i2i^2 appears in a product, for example writing (2i)(3i)=6(2i)(3i)=6 instead of 6i2=−66i^2=-6. Always replace i2i^2 with −1-1 immediately.

Forgetting to rationalise when dividing

You cannot leave ii in a denominator. To simplify 1c+di\dfrac{1}{c+di} you must multiply by c−dic−di\dfrac{c-di}{c-di}; multiplying by anything else, or by c+dic+di\dfrac{c+di}{c+di}, will not clear the imaginary part.

Why it matters

Cartesian arithmetic and the Argand plane are the foundation for the rest of Topic 2. Multiplication and division become far simpler once you convert to polar form and apply De Moivre's theorem, and the geometric view is essential when finding roots of complex numbers.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2022-style2 marks
Calculator-free. Let z=x+iyz = x + iy be a complex number. Find all solutions of the equation z=zˉz = \bar{z}, where zˉ\bar{z} is the conjugate of zz.
Show worked answer →

Write z=x+iyz = x + iy, so zˉ=x−iy\bar{z} = x - iy.

The equation z=zˉz = \bar{z} becomes x+iy=x−iyx + iy = x - iy. [1 mark]

Equating imaginary parts: iy=−iyiy = -iy, so 2iy=02iy = 0, hence y=0y = 0; the real part xx is unrestricted. [1 mark]

Therefore the solutions are all z=xz = x with xx real - exactly the points on the real axis of the Argand plane.

SACE 2021-style2 marks
Calculator-free. A circle in the complex plane has centre AA representing 10+25i10 + 25i and radius 2020. Write an equation, in terms of zz, describing all points on the circumference.
Show worked answer →

A circle is the set of points zz whose distance from a fixed centre equals the radius, and that distance is the modulus ∣z−centre∣|z - \text{centre}|. [1 mark]

With centre 10+25i10 + 25i and radius 2020, the circumference is ∣z−(10+25i)∣=20|z - (10 + 25i)| = 20. [1 mark]

Equivalently, with z=x+iyz = x + iy, this is (x−10)2+(y−25)2=400(x - 10)^2 + (y - 25)^2 = 400. Either form earns full marks.

SACE 2023-style3 marks
Calculator-free. Given z1=2+3iz_1 = 2 + 3i and z2=1−iz_2 = 1 - i, express z1z2\dfrac{z_1}{z_2} in the form a+bia + bi.
Show worked answer →

Multiply numerator and denominator by the conjugate of the denominator, 1+i1 + i:

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

Numerator: (2+3i)(1+i)=2+2i+3i+3i2=2+5i−3=−1+5i(2 + 3i)(1 + i) = 2 + 2i + 3i + 3i^2 = 2 + 5i - 3 = -1 + 5i. Denominator: 12+12=21^2 + 1^2 = 2.

So z1z2=−1+5i2=−12+52i\dfrac{z_1}{z_2} = \dfrac{-1 + 5i}{2} = -\dfrac{1}{2} + \dfrac{5}{2}i.

Marks: one for multiplying by the conjugate, one for the numerator, one for the simplified form.

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