Complex numbers - TCE Mathematics Specialised (Tasmania)
“Represent complex numbers in Cartesian and polar form, perform arithmetic, and apply De Moivre's theorem.”
A complex number is where . You must move fluently between Cartesian form and polar form , do arithmetic in whichever form is easier, and use De Moivre's theorem to find powers and roots.
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 .
Multiply the moduli and add the arguments; divide the moduli and subtract the arguments.
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 .
Find all solutions of .
Step 1: Write the right-hand side in polar form
Before applying De Moivre's theorem we need the right-hand side expressed as , because the root formula takes the modulus and argument of that number directly. Here lies on the negative real axis, so it has modulus and argument :
Step 2: Apply the -th root formula
The cube roots of all have modulus and arguments that step through equally spaced values. With and , the modulus of each root is and the three arguments are:
Step 3: Evaluate each root
Substituting in turn gives the three distinct roots.
For : , giving .
For : , giving .
For : , giving .
Final answer: The three cube roots are . As a check, the two complex roots are conjugates (expected since the coefficients are real), and they are spaced apart on the circle of radius .
A very common error is to compute as without checking the quadrant. For , a calculator gives , but is in the third quadrant, so the principal argument is . Always sketch on the Argand diagram first and read off the correct quadrant before quoting an argument.
Express in polar form, then use it to evaluate .
Step 1: Find the modulus and argument of
To convert to polar form we need both and . The modulus comes from Pythagoras and the argument requires us to identify the correct quadrant before applying arctangent.
The modulus is . The point is in the second quadrant. The reference angle is , so the principal argument is . Hence
Step 2: Write in polar form
We also convert the second factor, since multiplying in polar form is much simpler than expanding in Cartesian form: we just multiply the moduli and add the arguments.
Step 3: Multiply using the polar multiplication rule
Multiplying in polar form multiplies moduli and adds arguments, so there is no bracket expansion needed at all.
Final answer: . Converting back to Cartesian form (if required) uses and , but the polar answer is exact and compact, which is exactly why polar form is preferred for products.
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. The exponential form is the same information as polar form, and it makes the multiplication and division rules obvious because exponents add and subtract: . Switching at the right moment is what separates fast, clean solutions from long algebraic struggles, so practise converting both ways until it is automatic.
The fundamental theorem in practice
Allowing guarantees solutions to equations that have none over the reals. For instance has discriminant , so over the reals it has no solution. Using , the quadratic formula gives , a conjugate pair. This is the bridge to factorising polynomials over the complex field, where every quadratic factor with a negative discriminant splits into two complex linear factors.
Exam-style questions
Questions in the style of TASC 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.
Practice7 marks(a) Express in both Cartesian and polar form. (b) Hence, or otherwise, deduce an exact value for .Show worked answer →
(a) Cartesian: multiply top and bottom by the conjugate . The numerator is , and the denominator is . So . Polar: , and . So . (4 marks)
(b) Comparing the two forms, . Rationalise by multiplying top and bottom by : . So . (3 marks)
Practice4 marksIf , find the minimum possible value of .Show worked answer →
The condition describes a ray starting at the point , that is , pointing in the direction making angle with the positive real axis.
Here 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 to the origin points in the direction of angle and has length . The angle between this segment and the ray is . The perpendicular (shortest) distance is therefore .
Checking that the foot of the perpendicular lies on the ray (not its backward extension): the projection of the vector from the start point to the origin onto the ray direction is positive, so the foot is on the ray. Hence the minimum value of is .