Worked examples · Sheet 3

Powers with De Moivre's formula

Theory used: De Moivre's formula Trigonometric form and argument Euler's formula and exponential form

Exercise 1

Compute

(1 + i)^{8} .

1 + i = 2 (cos \frac{π}{4} + i sin \frac{π}{4})

(1 + i)^{8} = (2)^{8} (cos \frac{8 π}{4} + i sin \frac{8 π}{4}) = 16 (cos 2 π + i sin 2 π) = 16

The number $1 + i$ has modulus $1 + 1 = 2$ and positive real part, so its argument is $arctan 1 = \frac{π}{4}$ . By De Moivre’s formula, in the eighth power the modulus is raised to the eighth power and the argument is multiplied by $8$ : the modulus becomes $(2)^{8} = 2^{4} = 16$ and the argument becomes $2 π$ , which identifies the same point as the argument $0$ . Therefore the power is the real number $16$ .

Exercise 2

Compute

(3 + i)^{6} .

3 + i = 2 (cos \frac{π}{6} + i sin \frac{π}{6})

(3 + i)^{6} = 2^{6} (cos \frac{6 π}{6} + i sin \frac{6 π}{6}) = 64 (cos π + i sin π) = - 64

The modulus of $3 + i$ is $3 + 1 = 2$ ; the real part is positive, so the argument is $arctan \frac{1}{3} = \frac{π}{6}$ . Applying De Moivre’s formula with $n = 6$ , the modulus becomes $2^{6} = 64$ and the argument becomes $6 \cdot \frac{π}{6} = π$ . Since $cos π = - 1$ and $sin π = 0$ , the result is the real number $- 64$ .

Exercise 3

Compute

(1 - i 3)^{5} .

1 - i 3 = 2 (cos (- \frac{π}{3}) + i sin (- \frac{π}{3}))

(1 - i 3)^{5} = 2^{5} (cos (- \frac{5 π}{3}) + i sin (- \frac{5 π}{3}))

- \frac{5 π}{3} + 2 π = \frac{π}{3} ⟹ (1 - i 3)^{5} = 32 (cos \frac{π}{3} + i sin \frac{π}{3}) = 16 + 163 i

The modulus of $1 - i 3$ is $1 + 3 = 2$ ; the real part is positive, so the argument is $arctan (- 3) = - \frac{π}{3}$ , an angle in the fourth quadrant. By De Moivre’s formula the fifth power has modulus $2^{5} = 32$ and argument $- \frac{5 π}{3}$ .

The argument $- \frac{5 π}{3}$ does not belong to the interval $(- π, π]$ , so $2 π$ is added and one obtains the principal argument $\frac{π}{3}$ , which identifies the same point. Since $cos \frac{π}{3} = \frac{1}{2}$ and $sin \frac{π}{3} = \frac{3}{2}$ , the result in algebraic form is $32 \cdot \frac{1}{2} + 32 \cdot \frac{3}{2} i = 16 + 163 i$ .