Power the size, multiply the angle: Multiplying in polar form adds the arguments. So multiplying z by itself n times adds θ to itself n times and multiplies r by itself n times. That's De Moivre's theorem:
to find zⁿ, raise the modulus to the power n and multiply the argument by n.
This turns a horrible bracket-expansion like (1 + i)⁸ into a two-line calculation. The same picture as before — each multiplication is a stretch-and-turn — repeated n times.
IB-style question — a power via De Moivre
Use De Moivre's theorem to evaluate (1 + i)⁸, giving your answer as a real number.
Step by step
- Put 1 + i in polar form: modulus and argument.
- Apply De Moivre: power the modulus, multiply the angle by 8.
- cis(2π) = cos 2π + i sin 2π = 1, so the result is real.
Final answer
(1 + i)⁸ = 16. Polar form turns an eighth power into one clean line.
IB-style question — least power that is real
z = 2 cis(π/6). Find the smallest positive integer n for which zⁿ is a positive real number.
Step by step
- By De Moivre, the argument of zⁿ is n × π/6.
- zⁿ is a positive real number when its argument is a whole multiple of 2π (points along the positive real axis).
- The smallest positive n takes k = 1.
Final answer
n = 12. (zⁿ is real — possibly negative — whenever nπ/6 is a multiple of π, i.e. n = 6; it is positive real first at n = 12.)
Why engineers love complex numbers: Two ideas run AI HL's complex-number applications.
AC circuits. Resistance and reactance combine into one complex impedance Z = R + iX. Its modulus |Z| is the total opposition to current; its argument is the phase angle between voltage and current. Impedances in series simply add.
Adding sinusoids. A wave A cos(ωt + φ) can be carried by the phasor A eiφ (its amplitude and phase). To add two waves of the same frequency, just add their phasors — then read off the new amplitude (modulus) and phase (argument). Adding arrows is far easier than wrangling trig.
IB-style question — impedance of an AC circuit
An AC circuit has resistance R = 3 Ω and reactance X = 4 Ω, giving impedance Z = 3 + 4i ohms.
Find the magnitude of the impedance and the phase angle, and comment on what they mean.
Step by step
- Magnitude is the modulus.
- Phase angle is the argument (quadrant 1, both parts positive).
Final answer
|Z| = 5 Ω and the phase angle is about 53.1°. Interpretation: the circuit opposes current with an effective 5 Ω, and the voltage leads the current by roughly 53° because of the reactance.
IB-style question — combine two sinusoids
Two voltages, v₁ = 5 cos(ωt) and v₂ = 12 sin(ωt), are added.
Write the sum in the single form R cos(ωt − φ), giving R and φ.
Step by step
- Use phasors: cos(ωt) carries phase 0 → 5, and sin(ωt) = cos(ωt − 90°) carries phase −90° → 12 e−i90° = −12i. Add the phasors.
- The amplitude R is the modulus of the resultant phasor.
- The phase −φ is the argument; the phasor 5 − 12i is in quadrant 4.
Final answer
v₁ + v₂ ≈ 13 cos(ωt − 67.4°). Adding two waves becomes adding two arrows: the resultant amplitude is 13 V.