How does writing a complex number by its modulus and angle make multiplication, division and powers easy?
Polar form expresses a complex number by its modulus and argument, and De Moivre's theorem raises it to integer powers.
Converting between Cartesian and polar form, the modulus and argument, multiplication and division in polar form, and using De Moivre's theorem to compute integer powers of complex numbers.
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What this dot point is asking
Cartesian form is convenient for adding, but clumsy for multiplying and taking powers. Polar form describes the same number by how far it is from the origin and in what direction.
Modulus and argument
For the modulus is and the argument is the angle the vector makes with the positive real axis, measured anticlockwise. Then and , so
often abbreviated .
Multiplication and division in polar form
If and , then
So multiplication is a scaling and rotation: the new vector is times as long and rotated by . This follows from the compound-angle formulae applied to and .
De Moivre's theorem
Applying the multiplication rule times gives De Moivre's theorem: for any integer ,
This makes large powers, which would be hopeless to expand in Cartesian form, immediate.
The theorem also derives multiple-angle identities. Expanding by the binomial theorem and equating real parts with gives .
Common errors
Why it matters
Polar form turns multiplication into rotation and division into the reverse, and De Moivre's theorem is the gateway to finding the roots of complex numbers. The conversion technique builds directly on the modulus and conjugate ideas from complex arithmetic.
Exam-style practice questions
Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2018 SACE Stage 22 marksLet z = cis(theta) be a complex number with modulus 1. Using de Moivre's theorem, show that z^n + z^(-n) = 2 cos(n theta).Show worked answer →
Since z = cis(theta) = cos(theta) + i sin(theta), de Moivre's theorem gives z^n = cos(n theta) + i sin(n theta). [1 mark]
The negative power: z^(-n) = cos(-n theta) + i sin(-n theta) = cos(n theta) - i sin(n theta), using that cosine is even and sine is odd.
Adding the two: z^n + z^(-n) = [cos(n theta) + i sin(n theta)] + [cos(n theta) - i sin(n theta)] = 2 cos(n theta), since the imaginary parts cancel. [1 mark]
2018 SACE Stage 23 marksWith z = cis(theta), expand (z + z^(-1))^3 to show that cos^3(theta) = (1/4) cos(3 theta) + (3/4) cos(theta).Show worked answer →
From the previous result, z + z^(-1) = 2 cos(theta) and z^3 + z^(-3) = 2 cos(3 theta).
Expand using the binomial theorem: (z + z^(-1))^3 = z^3 + 3z^2(z^(-1)) + 3z(z^(-2)) + z^(-3) = z^3 + 3z + 3z^(-1) + z^(-3). [1 mark]
Group the symmetric pairs: = (z^3 + z^(-3)) + 3(z + z^(-1)) = 2 cos(3 theta) + 3(2 cos(theta)) = 2 cos(3 theta) + 6 cos(theta). [1 mark]
But (z + z^(-1))^3 = (2 cos(theta))^3 = 8 cos^3(theta). So 8 cos^3(theta) = 2 cos(3 theta) + 6 cos(theta). Dividing by 8 gives cos^3(theta) = (1/4) cos(3 theta) + (3/4) cos(theta). [1 mark]