← Complex numbers

6 Roots of complex numbers

The n nth-roots z^{1/n}=r^{1/n}\operatorname{cis}\!\left(\frac{\theta+2\pi k}{n}\right),\quad k=0,1,\dots,n-1 There are exactly n roots, equally spaced by 2π/n around a circle of radius r^(1/n).

Roots & polynomial equations

• The n nth-roots are equally spaced around a circle: sketch them on an Argand diagram.
• A polynomial with real coefficients has complex roots in conjugate pairs: if a+bi is a root, so is a−bi.
• Use this to factorise: (z − (a+bi))(z − (a−bi)) = z² − 2az + (a²+b²), a real quadratic factor.

Worked example Cube roots of 8 Find all cube roots of 8. 8 = 8 cis(0). Roots: 8^(1/3) cis((0 + 2πk)/3) = 2 cis(2πk/3), k = 0,1,2. k=0: 2 cis 0 = 2. k=1: 2 cis(2π/3) = 2(−½ + i√3/2) = −1 + i√3. k=2: 2 cis(4π/3) = −1 − i√3. ✓ z = 2, −1 + i√3, −1 − i√3 (equally spaced 120° apart).

Practise just this