De Moivre’s theorem, n-th roots and square roots of complex numbers
Answers are for parents and teachers.
1
Find (√3 + i)⁹ in the form a + bi.
The modulus is 2 and the amplitude is 30°. So 2⁹(cos 270° + i sin 270°) = 512(0 − i) = −512i.
2
Use the imaginary part of (c + is)⁵ to prove sin 5θ = 16s⁵ − 20s³ + 5s. (Hint: c² = 1 − s².)
The imaginary terms of the expansion give sin 5θ = 5c⁴s − 10c²s³ + s⁵. With c² = 1 − s²: 5s(1 − s²)² − 10s³(1 − s²) + s⁵ = 5s − 10s³ + 5s⁵ − 10s³ + 10s⁵ + s⁵ = 16s⁵ − 20s³ + 5s.
3
Solve Z³ = −64i in algebraic form and say what shape the roots make.
−64i = 64(cos(−90°) + i sin(−90°)), modulus 4 and amplitudes (−90° + 360°k)/3 = −30°, 90°, 210°. The roots are 2√3 − 2i, 4i and −2√3 − 2i. They form an equilateral triangle.
4
Find the square roots of −24 + 10i.
x² − y² = −24 and x² + y² = 26, so x² = 1 and y² = 25; 2xy = 10 is positive, so the signs agree: ±(1 + 5i). Check: 1 − 25 + 10i.
5
Solve z² + (1 − 2i)z − 2i = 0.
The discriminant is (1 − 2i)² + 8i = −3 − 4i + 8i = −3 + 4i, with square roots ±(1 + 2i). Then z = (−1 + 2i ± (1 + 2i))/2: z = 2i or z = −1.