UNSW Sydney · FACULTY OF MATHEMATICS

MATH1131 Chap.4 Polar Form, Complex Roots and Polynomials

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Chapter 4 of 12 · MATH1131

Polar Form, Complex Roots and Polynomials

Polar form writes a non-zero complex number as r times (cos theta + i sin theta), or r e^{i theta} using Euler's formula, where r is the modulus and theta an argument. The course fixes the principal argument in the interval from -pi (excluded) to pi (included), and finding it correctly depends on locating the quadrant first.

In polar form multiplication multiplies moduli and adds arguments, and De Moivre's theorem gives integer powers immediately. Expanding De Moivre's identity also produces multiple-angle formulas such as cos 3 theta in terms of cos theta, and Euler's formula rewrites powers of cosine and sine as sums of multiple angles.

Every non-zero complex number has exactly n distinct nth roots, sharing one modulus and spaced 2 pi/n apart, so they form a regular polygon in the Argand diagram. The chapter ends with polynomials: roots and factors, the factor theorem, and the rule that non-real roots of real polynomials occur in conjugate pairs, which lets any real polynomial be factorised into real linear and quadratic factors.

In this chapter

What this chapter covers

  • 01

    Modulus, argument and principal argument

  • 02

    Euler's formula and exponential polar form

  • 03

    Multiplication and division in polar form

  • 04

    De Moivre's theorem and powers

  • 05

    Multiple-angle trigonometric identities

  • 06

    Roots of unity and nth roots

  • 07

    Roots, factors and conjugate pairs of polynomials

Worked example · free

A tenth power using polar form

Q [4 marks]. Evaluate (1 + i)^10 in the form a + bi. The marks shown are a study allocation for practice, not an official University marking scheme.
The cube roots of 8i.The cube roots of 8i. Since 8i = 8e iπ/2 , each root has modulus 2 and arguments π/6, 5π/6 and 3π/2; the three points form an equilateral triangle on the circle |w| = 2.ReIm−22−22√3 + i−√3 + i−2i|w| = 2
The cube roots of 8i. Since 8i = 8eiπ/2, each root has modulus 2 and arguments π/6, 5π/6 and 3π/2; the three points form an equilateral triangle on the circle |w| = 2.
  • 1Polar form: |1 + i| = sqrt(2) and the point is in the first quadrant with argument pi/4, so 1 + i = sqrt(2) e^{i pi/4}.
  • 1De Moivre: (sqrt(2))^10 e^{10 i pi/4} = 32 e^{5 i pi/2}.
  • 1Reduce the argument: 5 pi/2 - 2 pi = pi/2.
  • 1Convert: 32(cos(pi/2) + i sin(pi/2)) = 32i.
(1 + i)^10 = 32i. As a check, (1 + i)^2 = 2i, so (1 + i)^10 = (2i)^5 = 32 i^5 = 32i.
Sia tip — Plot the base before reading off its argument, because the ratio b/a cannot distinguish opposite quadrants.
Glossary

Key terms

Principal argument
The unique argument of a non-zero complex number lying in the interval from minus pi, excluded, to pi, included.
Euler's formula
The identity e to the i theta equals cos theta plus i sin theta for real theta.
De Moivre's theorem
The rule that raising r e to the i theta to an integer power n gives r to the n times e to the i n theta.
Roots of unity
The complex solutions of z to the n equal to 1, spaced evenly on the unit circle.
Conjugate pair
Two non-real roots a plus bi and a minus bi that occur together for polynomials with real coefficients.
Factor theorem
The result that z minus alpha divides a polynomial exactly when alpha is a root of it.
Polar form
The expression of a non-zero complex number by its modulus and an argument, r times cos theta plus i sin theta.
FAQ

Polar Form, Complex Roots and Polynomials FAQ

How do I find the argument of a complex number in the third quadrant?

Find the reference angle from the absolute values of the parts, then use the quadrant: for a point with both parts negative the principal argument is the reference angle minus pi, which lies between minus pi and minus pi over 2.

Why does z^n = w have exactly n solutions?

Writing w with every argument phi plus 2k pi and taking nth roots gives a new root for each k, but increasing k by n returns to the same root. That leaves exactly n distinct values for k from 0 to n minus 1.

When can I use the conjugate-pair rule for polynomial roots?

Only when every coefficient of the polynomial is real. Then a non-real root forces its conjugate to be a root too, and the pair combines into a real quadratic factor. State that the coefficients are real before using it.

How is De Moivre's theorem used to prove trigonometric identities?

Expand (cos theta + i sin theta) to the nth power using the binomial theorem, then compare real and imaginary parts with cos n theta + i sin n theta. This expresses cos n theta and sin n theta as polynomials in cos theta and sin theta.

What do all the nth roots of unity add up to?

For n of at least two they add to zero. They are the vertices of a regular polygon centred at the origin, so their vectors balance, which also follows from the sum of a geometric series.

Study strategy

Exam move

Drill conversion between Cartesian and polar form with the standard angles until the quadrant check is automatic. Then practise each use of polar form separately: powers through De Moivre, roots through the general root formula, and identities through binomial expansion. For root questions, always draw the regular polygon of roots; it catches missing or duplicated answers instantly.

For polynomials, practise the full chain of finding one root, pairing its conjugate, dividing out the quadratic factor and checking by expansion, since exam questions often ask for all three steps. Check every polar answer by converting it back to Cartesian form at least once per problem set, since a wrong quadrant shows up immediately as a sign error in one coordinate.

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