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Complex numbers in Cartesian and polar form: VCE Specialist Mathematics Unit 3

Syllabus dot point

“Arithmetic and algebra of complex numbers in Cartesian form z=a+biz = a + bi and polar form z=r cis θz = r\,\mathrm{cis}\,\theta, the modulus and argument, conjugates, and representation of complex numbers and their operations on the Argand plane”

6 min read

Quick answer

A complex number z=a+biz = a + bi has real part aa and imaginary part bb, where i2=−1i^2 = -1. Its modulus is ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2} and its argument θ\theta satisfies tan⁡θ=b/a\tan\theta = b/a with the quadrant fixed by the signs of aa and bb. Polar form is z=r cis θ=r(cos⁡θ+isin⁡θ)z = r\,\mathrm{cis}\,\theta = r(\cos\theta + i\sin\theta). Add and subtract in Cartesian form; multiply and divide in polar form (multiply moduli, add arguments). The conjugate zˉ=a−bi\bar z = a - bi reflects zz across the real axis and gives zzˉ=∣z∣2z\bar z = |z|^2.

Jump to a section
  1. What this dot point is asking
  2. Cartesian form and the Argand plane
  3. Modulus and argument
  4. Conjugates
  5. Polar (cis) form
  6. Why both forms matter
  7. Examples in context
  8. Try this
  9. Exam-style questions

What this dot point is asking

VCAA wants you to work confidently with complex numbers in both Cartesian form z=a+biz = a + bi and polar form z=r cis θz = r\,\mathrm{cis}\,\theta, to compute the modulus and argument, to use the conjugate, and to plot complex numbers and visualise their arithmetic on the Argand plane. This material underpins De Moivre's theorem and the factorisation of polynomials over the complex numbers, so fluency here pays off across the whole complex-numbers strand.

Cartesian form and the Argand plane

A complex number is written z=a+biz = a + bi where a,b∈Ra, b \in \mathbb{R} and ii is the imaginary unit with i2=−1i^2 = -1. We call a=Re(z)a = \mathrm{Re}(z) the real part and b=Im(z)b = \mathrm{Im}(z) the imaginary part (note the imaginary part is the real number bb, not bibi).

The Argand plane represents z=a+biz = a + bi as the point (a,b)(a, b), with the horizontal axis the real axis and the vertical axis the imaginary axis. Addition of complex numbers corresponds to vector addition of the position vectors: (a1+b1i)+(a2+b2i)=(a1+a2)+(b1+b2)i(a_1 + b_1 i) + (a_2 + b_2 i) = (a_1 + a_2) + (b_1 + b_2)i, which is the parallelogram rule.

Modulus and argument

The modulus ∣z∣|z| is the distance from the origin to the point (a,b)(a, b):

∣z∣=a2+b2.|z| = \sqrt{a^2 + b^2}.

The argument arg⁡(z)=θ\arg(z) = \theta is the angle the position vector makes with the positive real axis, measured anticlockwise. It satisfies

cos⁡θ=a∣z∣,sin⁡θ=b∣z∣,tan⁡θ=ba.\cos\theta = \frac{a}{|z|}, \qquad \sin\theta = \frac{b}{|z|}, \qquad \tan\theta = \frac{b}{a}.

Because tan⁡\tan repeats every π\pi, you must use the signs of aa and bb to place θ\theta in the correct quadrant. The principal argument Arg(z)\mathrm{Arg}(z) is the value in (−π,π](-\pi, \pi].

Conjugates

The conjugate of z=a+biz = a + bi is zˉ=a−bi\bar z = a - bi, the reflection of zz across the real axis. Key properties:

z+zˉ=2a=2 Re(z),z−zˉ=2bi,zzˉ=a2+b2=∣z∣2.z + \bar z = 2a = 2\,\mathrm{Re}(z), \qquad z - \bar z = 2bi, \qquad z\bar z = a^2 + b^2 = |z|^2.

The identity zzˉ=∣z∣2z\bar z = |z|^2 is the tool for division: to simplify z1z2\dfrac{z_1}{z_2}, multiply numerator and denominator by zˉ2\bar z_2 so the denominator becomes the real number ∣z2∣2|z_2|^2.

Polar (cis) form

Using a=rcos⁡θa = r\cos\theta and b=rsin⁡θb = r\sin\theta with r=∣z∣r = |z|, every nonzero complex number can be written

z=r(cos⁡θ+isin⁡θ)=r cis θ,z = r(\cos\theta + i\sin\theta) = r\,\mathrm{cis}\,\theta,

where cis θ\mathrm{cis}\,\theta is shorthand for cos⁡θ+isin⁡θ\cos\theta + i\sin\theta. Polar form makes multiplication and division simple:

z1z2=r1r2 cis(θ1+θ2),z1z2=r1r2 cis(θ1−θ2).z_1 z_2 = r_1 r_2\,\mathrm{cis}(\theta_1 + \theta_2), \qquad \frac{z_1}{z_2} = \frac{r_1}{r_2}\,\mathrm{cis}(\theta_1 - \theta_2).

In words: to multiply, multiply the moduli and add the arguments; to divide, divide the moduli and subtract the arguments. This is why polar form is preferred for powers and roots, while Cartesian form is preferred for addition and subtraction.

Worked example

Converting and dividing

Let z1=1+iz_1 = 1 + i and z2=3−iz_2 = \sqrt{3} - i. Find z1z2\dfrac{z_1}{z_2} in polar form.

Step 1: Convert z1z_1 to polar form

The modulus is found using Pythagoras, and the argument is placed in the correct quadrant by checking the signs of the real and imaginary parts.

Modulus: ∣z1∣=12+12=2|z_1| = \sqrt{1^2 + 1^2} = \sqrt{2}. Since a=1>0a = 1 > 0 and b=1>0b = 1 > 0, the point is in the first quadrant, so θ1=tan⁡−1(1/1)=π4\theta_1 = \tan^{-1}(1/1) = \dfrac{\pi}{4}. Thus z1=2 cis π4z_1 = \sqrt{2}\,\mathrm{cis}\,\dfrac{\pi}{4}.

Step 2: Convert z2z_2 to polar form

The negative imaginary part places the point in the fourth quadrant, so the argument is negative.

Modulus: ∣z2∣=(3)2+(−1)2=3+1=2|z_2| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2. Since a=3>0a = \sqrt{3} > 0 and b=−1<0b = -1 < 0, the point is in the fourth quadrant, so θ2=tan⁡−1 ⁣(−13)=−π6\theta_2 = \tan^{-1}\!\left(\dfrac{-1}{\sqrt{3}}\right) = -\dfrac{\pi}{6}. Thus z2=2 cis ⁣(−π6)z_2 = 2\,\mathrm{cis}\!\left(-\dfrac{\pi}{6}\right).

Step 3: Divide using the polar rule

Dividing in polar form means dividing the moduli and subtracting the arguments, which is far simpler than multiplying out in Cartesian form.

z1z2=22 cis ⁣(π4−(−π6))=22 cis ⁣(3π12+2π12)=22 cis 5π12.\frac{z_1}{z_2} = \frac{\sqrt{2}}{2}\,\mathrm{cis}\!\left(\frac{\pi}{4} - \left(-\frac{\pi}{6}\right)\right) = \frac{\sqrt{2}}{2}\,\mathrm{cis}\!\left(\frac{3\pi}{12} + \frac{2\pi}{12}\right) = \frac{\sqrt{2}}{2}\,\mathrm{cis}\,\frac{5\pi}{12}.

Check the modulus by the rule ∣z1/z2∣=∣z1∣/∣z2∣=2/2|z_1/z_2| = |z_1|/|z_2| = \sqrt{2}/2, which matches. The argument 5π12=75∘\dfrac{5\pi}{12} = 75^\circ lies in (−π,π](-\pi, \pi], so it is already the principal argument.

Why both forms matter

Cartesian form makes the real and imaginary parts explicit and is the natural setting for adding, subtracting, and reading off coordinates on the Argand plane. Polar form encodes the geometry of multiplication: every multiplication is a rotation (by the argument) combined with a scaling (by the modulus). Multiplying by i=cis π2i = \mathrm{cis}\,\frac{\pi}{2}, for example, rotates a point 90∘90^\circ anticlockwise without changing its distance from the origin. Choosing the right form for the operation is the single biggest efficiency gain on exam questions.

Common exam traps
Wrong quadrant for the argument
tan⁡−1(b/a)\tan^{-1}(b/a) on a calculator returns a value in (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), which is only correct when a>0a > 0. If a<0a < 0 you must add or subtract π\pi to land in the correct quadrant. Always sketch the point first.
Confusing the imaginary part with bibi
For z=3−4iz = 3 - 4i, Im(z)=−4\mathrm{Im}(z) = -4, not −4i-4i. The imaginary part is the real coefficient.
Adding in polar form
There is no simple rule for z1+z2z_1 + z_2 in polar form. Convert back to Cartesian, add, then reconvert if a polar answer is required.
Dropping the sign in the conjugate
a+bi‾=a−bi\overline{a + bi} = a - bi negates only the imaginary part; the real part is unchanged.

Examples in context

Example 1. Square of a modulus. For z=3+4iz = 3 + 4i, zzˉ=(3+4i)(3−4i)=9−12i+12i−16i2=9+16=25=∣z∣2z\bar z = (3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i^2 = 9 + 16 = 25 = |z|^2, confirming ∣z∣=5|z| = 5.

Example 2. Rationalising a quotient. 2+i1−i=(2+i)(1+i)(1−i)(1+i)=2+2i+i+i21+1=1+3i2=12+32i\dfrac{2 + i}{1 - i} = \dfrac{(2 + i)(1 + i)}{(1 - i)(1 + i)} = \dfrac{2 + 2i + i + i^2}{1 + 1} = \dfrac{1 + 3i}{2} = \tfrac{1}{2} + \tfrac{3}{2}i.

Try this

Q1. Write z=−1+3 iz = -1 + \sqrt{3}\,i in polar form. [2 marks]

  • Cue. ∣z∣=1+3=2|z| = \sqrt{1 + 3} = 2; second quadrant so θ=π−π3=2π3\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}; z=2 cis 2π3z = 2\,\mathrm{cis}\,\frac{2\pi}{3}.

Q2. Simplify 43+i\dfrac{4}{3 + i} to Cartesian form. [2 marks]

  • Cue. Multiply by 3−i3−i\frac{3 - i}{3 - i}: 4(3−i)10=12−4i10=65−25i\frac{4(3 - i)}{10} = \frac{12 - 4i}{10} = \frac{6}{5} - \frac{2}{5}i.

Q3. If z1=2 cis π3z_1 = 2\,\mathrm{cis}\,\frac{\pi}{3} and z2=3 cis π6z_2 = 3\,\mathrm{cis}\,\frac{\pi}{6}, find z1z2z_1 z_2. [2 marks]

  • Cue. Multiply moduli, add arguments: 6 cis π2=6i6\,\mathrm{cis}\,\frac{\pi}{2} = 6i.

Exam-style questions

Questions in the style of VCAA 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.

2023 VCAA-style3 marks
Consider the complex number z = (b - i)^3, where b is in R+. Find b given that arg(z) = -pi/2.
Show worked answer →

Work in polar form, since taking a power multiplies the argument.

The base is b - i, which corresponds to the point (b, -1) in the fourth quadrant (b > 0). Its argument is arg(b - i) = -arctan(1/b).

By De Moivre's theorem the argument of the cube is multiplied by 3:
arg(z) = 3 arg(b - i) = -3 arctan(1/b).

Set this equal to the required value: -3 arctan(1/b) = -pi/2, so arctan(1/b) = pi/6.

Therefore 1/b = tan(pi/6) = 1/sqrt(3), giving b = sqrt(3).

Check: for b = sqrt(3), the point (sqrt(3), -1) has argument -pi/6, and 3 times -pi/6 = -pi/2, as required.

2025 VCAA-style1 mark
Let z be in C. Given that |z| = 1 and z is not 1, Re(1 / (1 - z)) is A. -1/2 B. 0 C. 1/2 D. sqrt(3)/2
Show worked answer →

The answer is C.

Since |z| = 1, write z = cos(t) + i sin(t). Then 1 - z = (1 - cos(t)) - i sin(t).

Rationalise by multiplying numerator and denominator by the conjugate of the denominator:
1 / (1 - z) = ((1 - cos(t)) + i sin(t)) / ((1 - cos(t))^2 + sin^2(t)).

The denominator is 1 - 2cos(t) + cos^2(t) + sin^2(t) = 2 - 2cos(t) = 2(1 - cos(t)).

So the real part is (1 - cos(t)) / (2(1 - cos(t))) = 1/2 (valid because z is not 1, so 1 - cos(t) is not 0).

The real part is constant at 1/2 for every such z, which is option C.

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Sources & how we know this

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