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Raising 1 + i to the eighth power by multiplying it out takes seven multiplications and invites sign errors. With one formula it takes two lines. That formula carries the name of Abraham de Moivre, a mathematician who never held a university post and spent his life teaching in London coffee houses. Here is his story, the theorem, and how to use it without the usual mistakes.
Who was Abraham de Moivre?
De Moivre was born on 26 May 1667 in Vitry-le-François, France, and died in London on 27 November 1754, according to his MacTutor biography. His parents were Protestants. After Louis XIV revoked the Edict of Nantes in 1685, Protestants faced serious persecution, and de Moivre was imprisoned for his beliefs. By around 1688 he had reached England.
In London he worked as a private tutor, walking between pupils' homes and teaching in coffee houses. MacTutor records that when he first saw Newton's *Principia*, he bought a copy, cut the pages apart and carried a few at a time to read between lessons. He became friends with Halley and Newton and was elected a Fellow of the Royal Society in 1697. Even so, as a foreigner he never obtained a university chair, and MacTutor notes that he died in poverty.
What else did de Moivre discover?
Most of his reputation comes from probability. His book *The Doctrine of Chances* appeared in 1718, with expanded editions in 1738 and 1756, and MacTutor notes that it contains the definition of statistical independence. In a Latin pamphlet dated 13 November 1733 he showed that the binomial distribution, for a large number of trials, can be approximated by the curve we now call the normal distribution. He also wrote *Annuities on Lives* (1724), one of the early mathematical treatments of life insurance. Our post on probability distributions of a random variable covers the binomial distribution he was approximating.
What does de Moivre's theorem say?
For any integer n:
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
Raising a complex number of modulus 1 to the nth power multiplies its angle by n. MacTutor notes that the formula appears in this form in a paper de Moivre published in 1722, with a closely related result in an earlier paper of 1707, and that it "took trigonometry into analysis".
For a general complex number z = r(cos θ + i sin θ), the theorem gives zⁿ = rⁿ(cos nθ + i sin nθ). Powers multiply the angle and raise the modulus to the same power.
Worked example 1: (1 + i)⁸
Modulus. r = √(1² + 1²) = √2.
Argument. The point (1, 1) is in the first quadrant, so θ = π/4.
Apply the theorem. (1 + i)⁸ = (√2)⁸ (cos 2π + i sin 2π) = 16 × (1 + 0i) = 16.
Check another way. (1 + i)² = 1 + 2i + i² = 2i. Then (2i)⁴ = 16 i⁴ = 16. The answers agree.
Worked example 2: (√3 + i)⁶
r = √(3 + 1) = 2 and θ = π/6, since tan θ = 1/√3 with the point in the first quadrant.
(√3 + i)⁶ = 2⁶ (cos π + i sin π) = 64 × (−1) = −64.
Worked example 3: a triple-angle formula for free
Expand (cos θ + i sin θ)³ with the binomial theorem and compare real parts with cos 3θ:
cos 3θ = cos³θ − 3 cos θ sin²θ = cos³θ − 3 cos θ (1 − cos²θ) = 4cos³θ − 3cos θ.
That is the identity in your trigonometry chapter, derived in three lines.
Common mistakes and how to avoid them
Wrong quadrant for the argument. For −1 + i, tan θ = −1, but the point is in the second quadrant, so θ = 3π/4, not −π/4. Always sketch the point first.
Forgetting the modulus. (1 + i)⁸ is not cos 2π + i sin 2π = 1. The (√2)⁸ = 16 factor matters.
Mixing degrees and radians mid-problem. Pick one and stay with it.
Not checking small cases. Squaring by hand, as in step 4 above, catches most slips.
On the syllabus: the JEE (Main) 2026 syllabus lists modulus, argument and the Argand diagram, and the current NCERT Class 11 chapter on complex numbers covers the Argand plane. Neither names de Moivre's theorem, so treat it as a tool for checking and simplifying rather than a topic to memorise, and check the bulletin for your exam year. Our post on complex numbers for JEE and the Argand plane shows the geometry behind it.
If you keep getting the argument wrong, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can see where the quadrant decision is made.
What can a student learn from de Moivre?
De Moivre studied the hardest book of his time in scraps, between paid lessons, and still mastered it. He also built on others openly: he extended Huygens and Montmort on chance and improved Jacob Bernoulli's law of large numbers. Steady reading in small pieces, and building on work you have understood, still works.
Practice routine:
Find (1 − i)⁶, (−1 + √3 i)³ and (1 + i)¹⁰ using the theorem, then check one by hand.
Derive sin 3θ = 3 sin θ − 4 sin³θ from the imaginary part of (cos θ + i sin θ)³.
Write the argument of eight points, two in each quadrant, before computing anything.
Next step: do the three powers in step 1 tonight and check your answers against a direct expansion.
