1 · IB Maths
Roots of a complex number
The n roots of a complex number sit equally spaced on a circle, forming a regular polygon in the Argand plane. Change n, the modulus and the argument of z and watch the polygon shrink, grow and rotate.
Predict first
Start from the default settings (n = 5, R = 2.00, φ = 0°). Set the argument φ to 90°. What is the argument of the first root, k = 0?
The first root has argument φ ÷ n = 90° ÷ 5 = 18°. Each root's argument is a fifth of φ plus a multiple of 360° ÷ 5.
Check it in the model: Move the Argument of z, φ slider from 0° to 90° and read the First root, arg φ ÷ n line.
z the n roots (k = 0 is the first)
roots: R^(1/n) (cos θk + i sin θk)
θk = (φ + 360°k) ÷ n, k = 0 … n − 1
Try this
- Use "Roots of unity" and change n from 2 to 8. What shape is each polygon, and what happens to the angle between neighbouring roots?
- Set n = 3, R = 8 and φ = 0°. Check that one root is 2 and that the other two are complex conjugates of each other.
- Keep n = 6 and R = 2.0, then increase φ from 0° to 60°. By how much does the polygon turn? Then find the φ that brings the polygon back to its original position.
Exam tip: the roots all share the modulus R^(1/n), but the argument of the first root is φ ÷ n, not φ. Add 360° to φ before dividing to find each of the other roots.
More on 1: Arithmetic and geometric series · Compound interest and depreciation · Proof by induction · Permutations and combinations · De Moivre's theorem · Systems of linear equations · Complex numbers on the Argand plane · Complex loci · Matrix transformations · Sequences and series · Exponentials and logarithms · Binomial theorem