How do complex numbers extend the real number system and let us solve every polynomial?
Represent complex numbers in Cartesian and polar form, perform arithmetic, and apply De Moivre's theorem.
Cartesian and polar forms of complex numbers, modulus-argument arithmetic, De Moivre's theorem and roots, for TCE Mathematics Specialised Unit 3.
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What this dot point is asking
Complex numbers extend the real numbers by introducing the imaginary unit , defined by . Once we allow , every polynomial equation has a solution, which is the content of the Fundamental Theorem of Algebra. This dot point is about being completely comfortable with the two standard representations and knowing when to use each.
Cartesian form
In Cartesian (rectangular) form we write , where is the real part and is the imaginary part. Addition and subtraction are done component by component:
Multiplication uses the distributive law together with :
The conjugate of is . It is the key tool for division, because is always a non-negative real number. To divide, multiply top and bottom by the conjugate of the denominator:
Modulus and argument
The modulus is , the distance from the origin to on the Argand diagram. The argument is the angle measured anticlockwise from the positive real axis. The principal argument is taken in . Be careful: only directly when is in the first or fourth quadrant; otherwise adjust by using the signs of and .
Polar and exponential form
In polar form , often abbreviated , and in exponential form via Euler's formula .
De Moivre's theorem
For any integer ,
This makes powers trivial in polar form and is the route to finding roots. The solutions of where are
These roots lie equally spaced on a circle of radius .
Choosing the right form
Use Cartesian form for addition and subtraction, and for stating exact real and imaginary parts. Use polar or exponential form for multiplication, division, powers and roots. Switching at the right moment is what separates fast, clean solutions from long algebraic struggles, so practise converting both ways until it is automatic.
Exam-style practice questions
Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2021 TASC7 marks(a) Express w = (1 + i)/(sqrt(3) + i) in both Cartesian and polar forms. (b) Hence, or otherwise, deduce an exact value for tan(pi/12).Show worked answer →
(a) Cartesian: multiply top and bottom by the conjugate sqrt(3) - i. Numerator (1 + i)(sqrt(3) - i) = sqrt(3) - i + sqrt(3) i - i^2 = (sqrt(3) + 1) + (sqrt(3) - 1)i. Denominator (sqrt(3) + i)(sqrt(3) - i) = 3 + 1 = 4. So w = [(sqrt(3) + 1) + (sqrt(3) - 1)i] / 4. Polar: |w| = |1 + i| / |sqrt(3) + i| = sqrt(2)/2, and arg(w) = arg(1 + i) - arg(sqrt(3) + i) = pi/4 - pi/6 = pi/12. So w = (sqrt(2)/2)(cos(pi/12) + i sin(pi/12)). (4 marks)
(b) Comparing the two forms, tan(pi/12) = (imaginary part)/(real part) of w = (sqrt(3) - 1)/(sqrt(3) + 1). Rationalise by multiplying top and bottom by (sqrt(3) - 1): = (3 - 2sqrt(3) + 1)/(3 - 1) = (4 - 2sqrt(3))/2 = 2 - sqrt(3). So tan(pi/12) = 2 - sqrt(3). (3 marks)
2024 TASC4 marksIf arg(z - 5) = 2pi/3 find the minimum possible value for |z|.Show worked answer →
The condition arg(z - 5) = 2pi/3 describes a ray starting at the point 5 (i.e. (5, 0)) and pointing in the direction making angle 2pi/3 with the positive real axis.
|z| is the distance from the origin to a point on this ray, and the minimum distance is the perpendicular distance from the origin to the line carrying the ray.
The segment from (5, 0) to the origin points in the direction of angle pi (length 5). The angle between this segment and the ray is pi - 2pi/3 = pi/3. The perpendicular (shortest) distance is therefore 5 sin(pi/3) = 5 sqrt(3)/2.
Checking that the foot of the perpendicular lies on the ray (not its backward extension): the projection of (origin - start) onto the ray direction is positive, so the foot is on the ray. Hence the minimum value of |z| is 5 sqrt(3)/2.