De Moivre’s formula

Section 1.3 De Moivre’s formula

The polar form description of complex multiplication given in Theorem 1.2.13 leads naturally to a similar description of integer powers of complex numbers, called de Moivre’s formula.

Subsection De Moivre’s formula

Theorem 1.3.1. De Moivre’s formula.

Let \(z=r(\cos\theta+i\sin\theta)\text{,}\) where \(r\geq 0\) and \(\theta\in \R\text{.}\)

Nonnegative integer powers.

For all nonnegative integers \(n\text{,}\) we have

\begin{equation} z^n=r^n(\cos(n\theta)+i\sin(n\theta))\text{.}\tag{1.11} \end{equation}
Negative integer powers.

Assume \(z\ne 0\text{.}\) For all positive integers \(n\) we have

\begin{equation} z^{-n}=r^{-n}(\cos(-n\theta)+i\sin(-n\theta))\text{.}\tag{1.12} \end{equation}

Proof.

First observe that statement (2) follows from (1) since for \(z\ne 0\) and positive integer \(n\) we have

\begin{align*} z^{-n} \amp= (z^{-1})^n \\ \amp = (r^{-1}(\cos(-\theta)+i\sin(-\theta))^n \amp (\knowl{./knowl/xref/th_polar_mult.html}{\text{1.2.16}})\\ \amp = (r^{-1})^n(\cos(-n\theta)+i\sin(-n\theta))\\ \amp = r^{-n}(\cos(-n\theta)+i\sin(-n\theta)) \amp \text{(by (1))}\text{.} \end{align*}

The proof of (1) is by induction on \(n\geq 0\text{.}\) (See Mathematical induction.)

Base case: \(n=0\text{.}\) We have

\begin{align*} r^0(\cos (0\theta)+i\sin(0\theta)) \amp = 1(\cos 0+i\sin 0) \\ \amp = 1\\ \amp = z^0 \amp (\knowl{./knowl/xref/d_integer_powers.html}{\text{1.1.17}})\text{,} \end{align*}

as desired.

Induction step: we assume the statement is true for \(n\text{,}\) and show it is true for \(n+1\text{.}\) Thus we assume that

\begin{equation*} z^n=r^n(\cos(n\theta)+i\sin(n\theta))\text{.} \end{equation*}

It follows that

\begin{align*} z^{n+1} \amp = z\, z^n\\ \amp = z\cdot r^n(\cos(n\theta)+i\sin(n\theta)) \amp (\text{ind. hyp.})\\ \amp = r(\cos \theta+i\sin\theta)\cdot r^n(\cos(n\theta)+i\sin(n\theta)) \\ \amp = r\, r^{n+1}(\cos(\theta+n\theta)+i\sin(\theta+n\theta)) \amp \text{(polar mult.)}\\ \amp =r^{n+1}(\cos((n+1)\theta)+i\sin((n+1)\theta))\text{,} \end{align*}

as desired.

Example 1.3.2. De Moivre’s formula.

Compute \((1+i)^{14}\text{.}\) Express you answer in both polar form and in the form \(a+ib\text{.}\)

Solution.

First write \(1+i\) in polar form as \(1+i=\sqrt{2}(\cos \pi/4+i\sin\pi/4)\text{.}\) Using de Moivre’s formula, we have

\begin{align*} (1+i)^{14} \amp = (\sqrt{2})^{14}(\cos (14\pi/4)+i\sin(14\pi/ 4))\\ \amp = 2^{14/2}(\cos 7\pi/2+i\sin 7\pi/2) \amp (\sqrt{2}=2^{1/2})\\ \amp 128(0-i) \amp (7\pi/2=3\pi/2+2\pi)\\ \amp =-128i\text{.} \end{align*}

Example 1.3.3. Double-angle formulas.

Use de Moivre’s formula to prove the double-angle formulas:

\begin{align*} \cos 2\theta \amp =\cos^2\theta-\sin^2\theta\\ \sin 2\theta \amp = 2\sin\theta\cos\theta\text{.} \end{align*}

Solution.

Let \(z=\cos\theta+i\sin\theta\text{.}\) We may compute \(z^2=z\, z\) using the algebraic definition of complex multiplication, or by using de Moivre’s formula. Equating the two resulting expressions yields

\begin{equation*} \cos 2\theta+i\sin 2\theta =\cos^2\theta-\sin^2\theta+ 2\sin\theta\cos\theta i\text{.} \end{equation*}

From the definition of complex number equality, it follows that

\begin{align*} \cos 2\theta \amp =\cos^2\theta-\sin^2\theta\\ \sin 2\theta \amp = 2\sin\theta\cos\theta\text{,} \end{align*}

as desired.

Example 1.3.4. Triple-angle formula.

Use de Moivre’s formula to derive triple-angle formulas for \(\cos\theta\) and \(\sin\theta\text{.}\) In more detail, find polynomials \(f(x,y)\) and \(g(x,y)\) such that

\begin{align*} \cos 3\theta \amp = f(\cos\theta,\sin\theta)\\ \sin 3\theta \amp = g(\cos\theta,\sin\theta)\text{.} \end{align*}

You may use the fact that the binomial theorem holds for complex numbers: i.e., given any \(z,w\in \C\) and positive integer \(n\text{,}\) we have

\begin{equation*} (z+w)^n=\sum_{k=0}^n {n \choose k} z^{n-k}w^k=z^n+{n\choose 1} z^{n-1}w+\cdots +{n \choose n-1}zw^{n-1}+w^n\text{,} \end{equation*}

where

\begin{equation*} {n\choose k}=\frac{n!}{k!\, (n-k)!}\text{.} \end{equation*}

Solution.

Let as an exercise.

Subsection \(n\)-th roots

One powerful application of de Moivre’s formula is that it allows us to compute all \(n\)-th roots of a given complex number \(w\text{.}\)

Definition 1.3.5. \(n\)-th roots.

Let \(n\) be a positive integer and let \(w\in \C\text{.}\) An \(n\)-th root of \(w\) is a complex number \(z\in \C\) satisfying \(z^n=w\text{.}\) Equivalently, an \(n\)-th root of \(w\) is a complex root (or zero) of the polynomial

\begin{equation*} f(z)=z^n - w\text{.} \end{equation*}

An \(n\)-th root of unity is an \(n\)-th root of \(1\text{.}\)

Example 1.3.6. Cube-roots of \(8i\).

Find all cube-roots of \(8i\) and sketch these as points in the complex plane. You may express your answer in polar form.

Solution.

First write \(8i\) in polar form as \(8i=8(\cos \pi/2+i\sin \pi/2)\text{.}\) If \(z=r(\cos\theta+i\sin\theta)\) satisfies \(z^3=8i\text{,}\) then using de Moivre’s theorem we would have

\begin{equation*} r^3(\cos 3\theta+i\sin 3\theta)=8(\cos \pi/2+i\sin \pi/2)\text{.} \end{equation*}

Using Theorem 1.2.13 we conclude that we must have

\begin{align*} r^3 \amp =8\\ 3\theta \amp =\pi/2+2\pi k \text{ for some } k\in \Z\text{,} \end{align*}

or equivalently,

\begin{align*} r \amp =2\\ \theta \amp = \pi/6+2\pi k/3 \text{ for some } k\in \Z \text{.} \end{align*}

Taking \(k=0,1,2\text{,}\) we see that

\begin{align*} z_0 \amp = 2(\cos(\pi/6)+i\sin(\pi/6))=\sqrt{3}+i\\ z_1 \amp = 2(\cos(5\pi/6)+i\sin(5\pi/6))=-\sqrt{3}+i\\ z_2 \amp = 2(\cos(3\pi/2)+i\sin(3\pi/2))=-2i \end{align*}

are cube-roots of \(8i\text{.}\) We claim further that these are all of the cube-roots of \(8i\text{.}\) This follows from the fact that any angle of the form \(\pi/6+2\pi k/n\) is obtained from one of the three angles \(pi/6, 5\pi/6, 3\pi/2\) by adding a multiple of \(2\pi\text{.}\)

The reasoning in Example 1.3.6 can easily be generalized to produce a procedure for computing \(n\)-th roots of an arbitrary complex number.

Theorem 1.3.8. \(n\)-th roots.

Let \(n\) be a positive integer. A nonzero complex number \(w\) has \(n\) distinct \(n\)-th roots. In more detail, writing \(w\) in polar form as \(w=r(\cos\theta+i\sin\theta)\text{,}\) the \(n\)-th roots of \(w\) are \(z_0,z_1,\dots, z_{n-1}\text{,}\) where for all \(0\leq k\leq n-1\text{,}\) we have

\begin{equation} z_k=\sqrt[n]{r}(\cos(\theta/n+2\pi k/n)+i\sin(\theta/n+2\pi k/n))\text{.}\tag{1.13} \end{equation}

Proof.

We seek \(z=s(\cos\psi+i\sin\psi)\) (\(s\geq 0\text{,}\) \(\psi\in \R\)) satisfying \(z^n=w\text{.}\) By de Moivre’s formula, this is equivalent to

\begin{equation*} s^n(\cos(n\psi)+i\sin(n\psi))=r(\cos\theta+i\sin\theta)\text{.} \end{equation*}

By Theorem 1.2.13, this is true if and only if

\begin{align*} s^n \amp = r\\ n\psi \amp = \theta+2\pi k \text{ for some } k\in \Z\text{.} \end{align*}

Solving for \(s\) and \(\psi\) we conclude that

\begin{align*} s \amp = \sqrt[n]{r}\\ \psi \amp = \theta/n +2\pi k/n \text{ for some } k\in \Z\text{.} \end{align*}

It follows that the complex numbers

\begin{equation*} z_k=\sqrt[n]{r}(\cos(\theta/n+2\pi k/n)+i\sin(\theta/n+2\pi k/n)), 0\leq k\leq n-1 \end{equation*}

are \(n\)-th roots of \(w\text{.}\) That they are distinct follows again from Theorem 1.2.13 and the fact that

\begin{equation*} (\theta/n+2\pi k/n)-(\theta/n+2\pi j/n) = 2\pi(k-j)/n < 2\pi \end{equation*}

for all \(0\leq j< k\leq n-1\text{.}\) Lastly, since the polynomial \(f(z)=z^n-w\) has at most \(n\) distinct roots (see Theorem 1.3.11), the \(n\) roots \(z_k\) are the only \(n\)-th roots of \(w\text{.}\)

Remark 1.3.9. Visualizing \(n\)-th roots.

Let \(z_0, z_1, \dots, z_{n-1}\) be the distinct \(n\)-th roots of the nonzero complex number \(w\text{,}\) as described in (1.13). We make some geometric observations about the \(z_k\text{:}\)

Each \(z_k\) has modulus \(\sqrt[n]{r}\text{,}\) and thus lives on the circle of radius \(\sqrt[n]{r}\) centered at the origin.
Looking at the arguments of the \(z_k\text{,}\) we see that for each \(k\geq 1\text{,}\) \(z_k\) is obtained from \(z_{k-1}\) by a rotation of \(2\pi/n\) about the origin. It follows that starting with \(z_0\text{,}\) we can obtain the other roots by successive rotation by \(2\pi/n\text{.}\)
From the previous observations, it follows that the \(n\) distinct \(n\)-th roots of \(w\) are the vertices of a regular \(n\)-gon inscribed in the circle of radius \(\sqrt[n]{r}\) centered at the origin. (See Figure 1.3.7.)

Subsection Polynomials

We can view formula (1.13) as providing explicit roots to the polynomial \(f(z)=z^n-w\text{,}\) which in turn allows us to factor \(f\) into linear terms. To make proper sense of this, we introduce some terminology and elementary facts around complex polynomials.

Definition 1.3.10. Polynomials.

A (complex) polynomial is a function \(f\colon \C\rightarrow \C\) of the form

\begin{equation} f(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots +a_1z+a_0\text{,}\tag{1.14} \end{equation}

where \(n\) is a nonnegative integer and \(a_k\in \C\) for all \(0\leq k\leq n\text{.}\)

We call \(a_kx^k\) the \(k\)-th term of \(f\text{,}\) and \(a_k\) the \(k\)-th coefficient; \(a_0\) is called the constant term of \(f\text{.}\)

If \(a_n\ne 0\text{,}\) then \(a_nx^{n}\) is called the leading term of \(f\text{,}\) \(a_{n}\) its leading coefficient, and \(n\) the degree, denoted \(\deg f=n\text{.}\)

Lastly, a root (or zero) of \(f\) is a a complex number \(w\) satisfying \(f(w)=0\text{.}\)

Theorem 1.3.11. Elementary properties of polynomials.

If \(f\) and \(g\) are polynomials, then so are \(f+g\text{,}\) \(fg\text{,}\) and \(cf\) for any \(c\in \C\text{.}\)
If \(f\) is a polynomial and \(w_0\in \C\) is a root of \(f\text{,}\) then there is a polynomial \(g\) such that

\begin{equation} f(z)=(z-w_0)g(z)\text{.}\tag{1.15} \end{equation}
If \(f\) is a polynomial of degree \(n\text{,}\) then \(f\) has at most \(n\) distinct roots.

Given a nonzero \(w\in \C\) and positive integer \(n\text{,}\) let \(z_0,z_1,\dots, z_{n-1}\) be the \(n\) distinct \(n\)-th roots of \(w\text{.}\) Since \(z_k\) is a root of \(f(z)=z^n-w\) for all \(0\leq k\leq n-1\text{,}\) it follows that we can factor \(f\) into linear factors as

\begin{equation*} f(z)=z^n-w =(z-w_0)(z-w_1)\cdots (z-w_{n-1})\text{.} \end{equation*}

In particular, we now know how to completely factor any real polynomial of the form \(f(z)=z^n-a\text{,}\) where \(a\in \R\text{.}\) The next theorem delves into factorization details of the polynomial \(f(z)=z^n-1\text{.}\)

Theorem 1.3.12. Roots of unity.

Fix a positive integer \(n\) and let \(f(z)=z^n-1\text{.}\)

The \(n\) distinct roots of \(f\) are

\begin{equation*} \zeta_k=\cos(2\pi k/n)+i\sin(2\pi k/n)\text{,} \end{equation*}

where \(0\leq k\leq n-1\text{.}\) Note that \(\zeta_0=1\text{.}\)
\(f(z)\) factors as

\begin{align} z^n-1\amp=\prod_{k=0}^{n-1}(z-\zeta_k) \tag{1.16}\\ \amp =(z-1)\prod_{k=1}^{n-1}(z-\zeta_k)\text{.}\tag{1.17} \end{align}
We have

\begin{equation} z^n-1=(z-1)(z^{n-1}+z^{n-2}+\cdots +z+1)\tag{1.18} \end{equation}

and thus, letting \(g(z)=z^{n-1}+z^{n-2}+\cdots +z+1\text{,}\)

\begin{equation*} g(z)=\prod_{k=1}^{n-1}(z-\zeta_k)\text{.} \end{equation*}

Example 1.3.13. Factoring \(f(x)=x^6-1\).

Give two distinct factorizations of \(f(x)=x^6-1\text{:}\)

as a product of irreducible real polynomials;
as a product of linear terms with complex coefficients.

Solution.

Using the difference of squares factoring identity, we have

\begin{equation*} x^6-1=(x^3-1)(x^3+1)\text{.} \end{equation*}

Next, since \(1\) is a root of \(x^3-1\text{,}\) and \(-1\) is a root of \(x^3+1\text{,}\) we can use polynomial division to factor

\begin{align*} x^3-1 \amp = (x-1)(x^2+x+1) \amp (x^3+1)\amp=(x+1)(x^2-x+1)\text{.} \end{align*}

It is now easy to see, using the quadratic formula, that \(x^2+x+1\) and \(x^2-x+1\) are irreducible over \(\R\text{,}\) since they have no real roots. Thus

\begin{equation*} x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x-1) \end{equation*}

is a factorization of \(x^6-1\) into irreducible real polynomials.
Using Theorem 1.3.12, we have

\begin{align*} x^6-1 \amp =\prod_{k=0}^{5}(x-\zeta_k)\text{,} \end{align*}

where

\begin{equation*} \zeta_k=\cos(2\pi k/6+i\sin 2\pi k /6)=\cos(\pi k/3)+i\sin(\pi k/3)\text{.} \end{equation*}

Take a moment to survey these results: thanks to complex numbers, we now know how to completely factor the following polynomials:

\begin{align*} f(z) \amp = z^n-w \amp \\ f(z)\amp =z^n-1 \\ f(z)\amp =z^{n-1}+z^{n-2}+\cdots +z+1 \text{.} \end{align*}

Pretty impressive! As it turns out, every complex polynomial can be factored in this manner. This is the content of the fundamental theorem of algebra, a full proof of which we will be able to provide by the end of this course.

Theorem 1.3.14. Fundamental theorem of algebra.

Let \(f\colon \C\rightarrow \C\) be a nonconstant polynomial: i.e., \(\deg f \geq 1\text{.}\)

\(f\) has a complex root.
There is a \(w\in \C\) such that \(f(w)=0\text{.}\)
\(f\) spits completely.

There are complex numbers \(w_1,w_2,\dots, w_n\) such that

\begin{equation*} f(z)=c\prod_{k=1}^n(z-w_k)\text{.} \end{equation*}

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