Same length, equally spaced: A complex number has n different nth-roots. They sit on a circle, forming a regular n-sided polygon.
Where the roots sit
- They all have the same modulus R1/n, so they sit on a circle.
- They are equally spaced, 2π/n apart.
To get them all: Take the obvious root, then add 2π/n to the angle, n times.
The three cube roots of 1: equally spaced 120° apart on the unit circle (a triangle).
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IB-style question — cube roots of 1
Find the three cube roots of 1.
Step by step
- Write 1 in polar form. Its modulus is 1, argument 0.
- Cube roots: modulus 11/3 = 1; angles (0 + 2πk)/3 for k = 0, 1, 2.
- Write each (convert to a + bi).
Final answer
1, −½ + (√3/2)i and −½ − (√3/2)i.
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Root the modulus, spread the angle: 1. Put the number in polar form R cisφ. 2. Modulus of each root: R1/n. 3. Angles: (φ + 2πk)/n for k = 0, 1, …, n − 1 (each one 2π/n further round).
The four fourth-roots of −16: on a circle of radius 2, 90° apart (a square).
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IB-style question — fourth roots of −16
Find the four fourth-roots of −16, in polar form.
Step by step
- Polar form: −16 has modulus 16, argument π.
- Modulus of each root: 161/4 = 2.
- Angles: (π + 2πk)/4 for k = 0, 1, 2, 3.
- Write the four roots.
Final answer
2 cis(π/4), 2 cis(3π/4), 2 cis(5π/4), 2 cis(7π/4).