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Roots of complex numbers

Lessons 41 · 1 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 10, Parts 1 and 2, 1st edition. EXTREMUM PRESS, Tashkent, 2017
41

Extracting the square root of a complex number

Textbook, Part 2: pp. 84–85
GoalExtract square roots (and cube and fourth roots) of complex numbers using the trigonometric form.
New words
square root · kvadrat ildizmodulus of a root · ildiz modulinth root · n-darajali ildizregular polygon · muntazam ko‘pburchak
Explanation

A square root of z = r(cos φ + i sin φ) is a complex number w = ρ(cos ψ + i sin ψ) whose square is z. From w² = z we get ρ² = r and 2ψ = φ + 2πn, so ρ = √r and ψ = (φ + 2πn)/2. For n = 0 and n = 1 we obtain two different roots, w₀ = √r(cos φ/2 + i sin φ/2) and w₁ = √r(cos(φ/2 + π) + i sin(φ/2 + π)) = −w₀; for n = 2, 3, … the roots repeat. So every non-zero complex number has two square roots, and they are opposite numbers. In the same way an nth root has modulus ⁿ√r and arguments (φ + 2πk)/n, k = 0, 1, …, n − 1: there are n roots and they lie at the vertices of a regular n-gon inscribed in the circle with centre at the origin and radius ⁿ√r. It helps to check the result by w² = z.

Worked examples
z = 9(cos 60° + i sin 60°). ρ = 3; ψ = 30° + 180°n. n = 0: 3(cos 30° + i sin 30°) = 3(√3/2 + i/2); n = 1: 3(cos 210° + i sin 210°) = −3(√3/2 + i/2).
z = −4: r = 4, φ = 180°. ρ = 2; ψ = 90° + 180°n. The roots are 2(cos 90° + i sin 90°) = 2i and 2(cos 270° + i sin 270°) = −2i. Check: (2i)² = −4.
Class activity

In pairs draw a circle and mark the three cube roots of z = 1 (arguments 0°, 120°, 240°); check with a protractor and ruler that they form a regular triangle.

Practice
1
How many square roots does a non-zero complex number have?
2
Write all values of √(−9).
3
What is the modulus of the square roots of z = 16(cos 80° + i sin 80°)?
4
Why are the two square roots opposite numbers?