EULER’S FORMULA AND DE MOIVRE’S FORMULA FOR QUATERNIONS

Eungchun Cho

Abstract. Natural generalizations of Euler’s formula and De Moivre’s formula for quaternions are derived.

1. Introduction. A quaternion q is a linear combination a1 + bi + cj + dk, where a, b, c, and d are real numbers and

\begin{aligned}1 & = (1, 0, 0, 0), i = (0, 1, 0, 0), \\ j & = (0, 0, 1, 0), k = (0, 0, 0, 1). \end{aligned}

The sum of quaternions is the usual component–wise sum and the multiplication is defined so that (1, 0, 0, 0) is the identity and i, j, and k satisfy

i^{2} = j^{2} = k^{2} = ijk = −1.  (1)

It follows from (1) that

ij = k, jk = i, ki = j, and ij = −ji, jk = −kj, ki = −ik.

A quaternion is usually written as a + bi + cj + dk or as α + βj, where α and β are complex numbers. The complex numbers do not commute with j, but satisfy jβ = \overline{β}j. We can also write q = a + ω, where ω = bi + cj + dk, called the pure quaternion part of q. a is called the real part of q. The conjugate of q is \overline{q} = a − ω. We can view the pure quaternion part ω = bi + cj + dk as a vector in ℝ^{3}. A simple computation shows

ω_{1}ω_{2} = −ω_{1} · ω_{2} + ω_{1} × ω_{2},  (2)

where ω_{1} · ω_{2} is the dot product and ω_{1} × ω_{2} is the cross product in ℝ^{3}. It follows from (2) that ω_{2}ω_{1} = \overline{ω_{1}ω_{2}} for any pure quaternion ω_{1} and ω_{2}. Let a_{i} be real numbers and β_{i} be pure quaternions. Then

(a_{1} + β_{1})(a_{2} + β_{2}) = (a_{1}a_{2} − β_{1} · β_{2}) + a_{1}β_{2} + a_{2}β_{1} + β_{1} × β_{2}.  (3)

It follows from (3) that \overline{q_{1}q_{2}} = \overline{q_{2}} \overline{q_{1}} for any quaternion q_{i}. For more details on quaternions, we refer to [1].

2. Euler’s Formula and De Moivre’s Formula for Quaternions. We will use the notation

S^{3} = \{q : |q| = 1\} and S^{2} = \{ω : |ω| = 1, \overline{ω} = −ω\}.

S^{3} is the set of all unit quaternions and S^{2} is the set of all unit pure quaternions. S^{3} is a group under quaternion multiplication and is isomorphic to SU(2), the group of all 2 by 2 unitary matrices of determinant 1. The map

(a, b, c, d) ↦ ( \begin{matrix}a + bi & −c + di \\ c + di & a − bi \end{matrix} )

is a group isomorphism between S^{3} and SU(2).

Since ω·ω = 1 and ω×ω = 0 for any ω ∈ S^{2}, we have the following proposition.

\underline{Proposition 1}. ω^{2} = −1 for any ω ∈ S^{2}, hence, any ω ∈ S^{2} has order 4.

We can express any q = a + bi + cj + dk ∈ S^{3} as

q = cos θ + ω sin θ,  (4)

where cos θ = a and

ω = \frac{1}{\sqrt{b^{2} + c^{2} + d^{2}}} (bi + cj + dk) = \frac{1}{\sqrt{1 − a^{2}}} (bi + cj + dk).

This is similar to the polar coordinate expression of a complex number. We can view θ as the angle between the vector q ∈ ℝ^{4} and the real axis (the subspace of real numbers) and ω sin θ as the projection of q onto the subspace ℝ^{3} of pure quaternions. We will call (4) the polar expression of a unit quaternion q. Since ω^{2} = −1 for any ω ∈ S^{2}, we have a natural generalization of Euler’s formula for quaternions,

e^{ωθ} = 1 + ωθ − \frac{θ^{2}}{2!} − ω \frac{θ^{3}}{3!} + \frac{θ^{4}}{4!} + ⋯

= 1 − \frac{θ^{2}}{2!} + \frac{θ^{4}}{4!} − ⋯ + ω(θ − \frac{θ^{3}}{3!} + \frac{θ^{5}}{5!} − ⋯)

= cos θ + ω sin θ for any real θ. If the power series definition

cos x = 1 − \frac{x^{2}}{2!} + \frac{x^{4}}{4!} − ⋯ and sin x = x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − ⋯

is used for quaternion x, then

cos ω = cos i = cosh 1
and sin ω = −ωi sin i = iω sinh 1

for every ω ∈ S^{2}. We note the cosine function is constant on the set S^{2}. For more on Euler’s formula for complex numbers, we refer to [2].

A simple computation and the addition formula for cosine and sine, i.e.,

cos(θ + ψ) = cos θ cos ψ − sin θ sin ψ and sin(θ + ψ) = cos θ sin ψ + sin θ cos ψ

prove the following lemma.

\underline{Lemma}. For any ω ∈ S^{2}, we have
(cos θ + ω sin θ)(cos ψ + ω sin ψ) = cos(θ + ψ) + ω sin(θ + ψ).

\underline{Remark}. It follows from the lemma that K_{ω} = \{cos θ + ω sin θ : 0 ≤ θ < 2π\} is a subgroup of S^{3} and is isomorphic to S^{1}.

\underline{Proposition 2 (De Moivre’s formula)}. Let q = e^{ωθ} = cos θ+ω sin θ ∈ S^{3}, where

θ is a real and ω ∈ S^{2}. Then,

q^{n} = e^{ωnθ} = (cos θ + ω sin θ)^{n} = cos nθ + ω sin nθ  (5)

for every integer n.

\underline{Proof}. The proof is by induction on the nonnegative integers n.

q^{n+1} = (cos θ + ω sin θ)^{n+1}
= (cos nθ + ω sin nθ)(cos θ + ω sin θ)
= cos(n + 1)θ + ω sin(n + 1)θ.

The formulas holds for all integers n, since

q^{−1} = cos θ − ω sin θ
and q^{−n} = cos nθ − ω sin nθ = cos(−nθ) + ω sin(−nθ).

\underline{Corollary}. There are infinitely many unit quaternions satisfying x^{n} = 1.

\underline{Proof}. For every ω ∈ S^{2}, we have a quaternion q = cos 2π/n + ω sin 2π/n of order n.

\underline{Example}. \frac{1}{2}(1 + i + j + k) = cos \frac{π}{3} + sin \frac{π}{3} (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) is of order 6 and

\frac{1}{2}(−1 + i + j + k) = cos \frac{2π}{3} + sin \frac{2π}{3} (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) is of order 3.

\underline{References}

1. N. Jacobson, Basic Algebra, W. H. Freeman and Company, San Francisco, CA, 1974.

2. R. P. Boas, Invitation to Complex Analysis, Random House, New York, 1987.

Eungchun Cho Division of Mathematics and Science Kentucky State University Frankfort, KY 40601 email: eccho@gwmail.kysu.edu