الرياضيات البحتة (بالإنجليزية) الصف الثالث الثانوي، الأسبوع 6: The n-th roots of a complex number, and the roots as corners of a regular polygon on the Argand plane

De Moivre’s theorem, n-th roots and square roots of complex numbers

الأسبوع 6 · نظرية ديموافر والجذور النونية للعدد المركب والجذران التربيعيان والمعادلات التربيعية ذات المعاملات المركبة · 3 دروس

The n-th roots of a complex number, and the roots as corners of a regular polygon on the Argand plane

Pure Mathematics Sec 3 Term 1, Unit 2, Lesson 2

GoalFind all n roots of Zⁿ = a, write them in algebraic form, and place them on the Argand plane.

New words

n-th root · a number Z with Zⁿ = aregular polygon · a polygon with equal sides and equal angles; its corners lie on a circlestep angle · 360° ÷ n, the turn from one root to the next

Explanation

The equation Zⁿ = a, with a complex and n a positive whole number, has exactly n different solutions, called the n-th roots of a (in Arabic الجذور النونية). To find them, write a in trigonometric form r(cos θ + i sin θ). Adding any whole number of full turns to θ gives the same a, so a = r(cos(θ + 360°k) + i sin(θ + 360°k)). Take the n-th root with De Moivre’s theorem: Z = r^(1/n) (cos((θ + 360°k)/n) + i sin((θ + 360°k)/n)) for k = 0, 1, …, n − 1. After n values the angles repeat, so there are exactly n roots. In radians replace 360° by 2π. The geometry is neat. All roots have the same modulus r^(1/n), so they lie on one circle with centre O. Their amplitudes grow by the same step 360°/n, so the roots are the corners of a regular polygon with n sides. Find one root and you can find the others by turning it about O through the step angle again and again. The picture shows the six roots of Z⁶ = 64.

Worked examples

Solve Z³ = 27i. Write 27i = 27(cos 90° + i sin 90°). Every root has modulus 27^(1/3) = 3 and the amplitudes are (90° + 360°k)/3 = 30°, 150°, 270°. So the roots are 3(cos 30° + i sin 30°) = (3√3/2) + (3/2)i, 3(cos 150° + i sin 150°) = −(3√3/2) + (3/2)i and 3(cos 270° + i sin 270°) = −3i. In exponential form they are 3e^(iπ/6), 3e^(5iπ/6) and 3e^(−iπ/2). They are the corners of an equilateral triangle.

Solve Z⁴ = −8 + 8√3 i. The modulus of the right side is √(64 + 192) = 16 and its amplitude is 120° (second quadrant, tan = −√3). Each root has modulus 16^(1/4) = 2 and amplitude (120° + 360°k)/4 = 30°, 120°, 210°, 300°. In principal form the amplitudes are 30°, 120°, −150°, −60°, and the roots are √3 + i, −1 + √3 i, −√3 − i and 1 − √3 i. The step angle is 90°, so they make a square.

The six roots of Z⁶ = 64 in the picture above. The modulus is 64^(1/6) = 2 and the amplitude of 64 is 0°, so the amplitudes are 0°, 60°, 120°, 180°, 240°, 300°. The roots are 2, 1 + √3 i, −1 + √3 i, −2, −1 − √3 i, 1 − √3 i. They are the corners of a regular hexagon whose side equals the radius 2.

Class activity

Turn a root. Draw a circle of radius 2 cm with a compass and mark the eight corners of a regular octagon on it, starting on the positive x-axis, 45° apart (use a protractor). These are the eighth roots of 256 = 2⁸. Which four corners are the numbers 2, 2i, −2 and −2i? Which corner is √2(1 + i), and why does it have modulus 2?

Practice

Solve Z⁴ = 81 and write the roots in the form a + bi.

Find the three cube roots of 64 in algebraic form.

Show that the roots of Z⁴ = −4 are ±1 ± i. What shape do they form?

The roots of Z⁸ = 256 are the corners of a regular polygon. Name it, give the radius of its circle, and the angle between neighbouring roots.

One fifth root of a number a is 2(cos 24° + i sin 24°). Write the other four roots, and find a.