Complex numbers in Cartesian and polar form: VCE Specialist Mathematics Unit 3
“Arithmetic and algebra of complex numbers in Cartesian form and polar form , the modulus and argument, conjugates, and representation of complex numbers and their operations on the Argand plane”
A complex number has real part and imaginary part , where . Its modulus is and its argument satisfies with the quadrant fixed by the signs of and . Polar form is . Add and subtract in Cartesian form; multiply and divide in polar form (multiply moduli, add arguments). The conjugate reflects across the real axis and gives .
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What this dot point is asking
VCAA wants you to work confidently with complex numbers in both Cartesian form and polar form , 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 where and is the imaginary unit with . We call the real part and the imaginary part (note the imaginary part is the real number , not ).
The Argand plane represents as the point , 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: , which is the parallelogram rule.
Modulus and argument
The modulus is the distance from the origin to the point :
The argument is the angle the position vector makes with the positive real axis, measured anticlockwise. It satisfies
Because repeats every , you must use the signs of and to place in the correct quadrant. The principal argument is the value in .
Conjugates
The conjugate of is , the reflection of across the real axis. Key properties:
The identity is the tool for division: to simplify , multiply numerator and denominator by so the denominator becomes the real number .
Polar (cis) form
Using and with , every nonzero complex number can be written
where is shorthand for . Polar form makes multiplication and division simple:
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.
Converting and dividing
Let and . Find in polar form.
Step 1: Convert 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: . Since and , the point is in the first quadrant, so . Thus .
Step 2: Convert to polar form
The negative imaginary part places the point in the fourth quadrant, so the argument is negative.
Modulus: . Since and , the point is in the fourth quadrant, so . Thus .
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.
Check the modulus by the rule , which matches. The argument lies in , 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 , for example, rotates a point anticlockwise without changing its distance from the origin. Choosing the right form for the operation is the single biggest efficiency gain on exam questions.
- Wrong quadrant for the argument
- on a calculator returns a value in , which is only correct when . If you must add or subtract to land in the correct quadrant. Always sketch the point first.
- Confusing the imaginary part with
- For , , not . The imaginary part is the real coefficient.
- Adding in polar form
- There is no simple rule for in polar form. Convert back to Cartesian, add, then reconvert if a polar answer is required.
- Dropping the sign in the conjugate
- negates only the imaginary part; the real part is unchanged.
Examples in context
Example 1. Square of a modulus. For , , confirming .
Example 2. Rationalising a quotient. .
Try this
Q1. Write in polar form. [2 marks]
- Cue. ; second quadrant so ; .
Q2. Simplify to Cartesian form. [2 marks]
- Cue. Multiply by : .
Q3. If and , find . [2 marks]
- Cue. Multiply moduli, add arguments: .
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 marksConsider 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 markLet 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
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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.