1.14 De Moivre's Theorem (HL)

1. Foundational Propositions

PROPOSITION 1

Let $z = r\text{cis}\theta$. Then $z^{-1} = r^{-1}\text{cis}(-\theta)$, which implies $\frac{1}{z} = \frac{1}{r}\text{cis}(-\theta)$.

Proof: Utilizing the property $z\overline{z} = |z|^2 = r^2$ and $\overline{z} = r\text{cis}(-\theta)$.

$z^{-1} = \frac{1}{z} = \frac{\overline{z}}{z\overline{z}} = \frac{r\text{cis}(-\theta)}{r^2} = \frac{1}{r}\text{cis}(-\theta)$.

PROPOSITION 2

Let $z_1 = r_1\text{cis}\theta_1$ and $z_2 = r_2\text{cis}\theta_2$. Then $z_1 z_2 = r_1 r_2\text{cis}(\theta_1 + \theta_2)$.

Proof:

$z_1 z_2 = r_1\text{cis}\theta_1 \cdot r_2\text{cis}\theta_2$
$= r_1 r_2 (\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 + i\sin\theta_2)$
$= r_1 r_2 [\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2 + i(\sin\theta_1\cos\theta_2 + \sin\theta_2\cos\theta_1)]$
$= r_1 r_2 [\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)]$
$= r_1 r_2\text{cis}(\theta_1 + \theta_2)$.

PROPOSITION 3

Let $z_1 = r_1\text{cis}\theta_1$ and $z_2 = r_2\text{cis}\theta_2$. Then $\frac{z_1}{z_2} = \frac{r_1}{r_2}\text{cis}(\theta_1 - \theta_2)$.

Proof:

$\frac{z_1}{z_2} = z_1 \cdot z_2^{-1}$
$= r_1\text{cis}\theta_1 \cdot r_2^{-1}\text{cis}(-\theta_2)$
$= r_1 r_2^{-1}\text{cis}(\theta_1 - \theta_2)$
$= \frac{r_1}{r_2}\text{cis}(\theta_1 - \theta_2)$.

Notice on Modulus and Argument

The modulus $|z|$ strictly follows standard arithmetic operations, while the argument $\arg(z)$ behaves analogously to logarithms.

  • $|z_1 z_2| = |z_1||z_2|$ and $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$.
  • $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$ and $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$.

EXAMPLE 1

Let $z = 2\text{cis}\frac{\pi}{6}$ (which expands to $\sqrt{3} + i$) and $w = \text{cis}\frac{\pi}{3}$ (which expands to $\frac{1}{2} + \frac{\sqrt{3}}{2}i$). Operations yield:

$zw = 2 \cdot 1 \text{cis}(\frac{\pi}{6} + \frac{\pi}{3}) = 2\text{cis}(\frac{\pi}{2}) = 2(0 + i) = \mathbf{2i}$.
$\frac{z}{w} = \frac{2}{1}\text{cis}(\frac{\pi}{6} - \frac{\pi}{3}) = 2\text{cis}(-\frac{\pi}{6}) = 2(\frac{\sqrt{3}}{2} - \frac{1}{2}i) = \mathbf{\sqrt{3} - i}$.
$z^2 = z \cdot z = 2 \cdot 2\text{cis}(\frac{\pi}{6} + \frac{\pi}{6}) = 4\text{cis}(\frac{\pi}{3}) = \mathbf{2 + 2\sqrt{3}i}$.

2. De Moivre's Theorem

Let $z = r\text{cis}\theta$. For any positive integer $n \in \mathbb{Z}^+$, the theorem establishes the following relationship:

$z^n = r^n\text{cis}(n\theta)$

Proof by Mathematical Induction

Step 1: For $n=1$, the statement is trivially true since $z^1 = r^1\text{cis}(1\theta)$.
Step 2: Assume the statement is true for $n=k$, establishing the hypothesis $z^k = r^k\text{cis}(k\theta)$.
Step 3: Prove the statement holds for $n=k+1$, requiring $z^{k+1} = r^{k+1}\text{cis}((k+1)\theta)$.
$z^{k+1} = z^k \cdot z = r^k\text{cis}(k\theta) \cdot r\text{cis}\theta$
$= r^k r [\cos(k\theta) + i\sin(k\theta)][\cos\theta + i\sin\theta]$
$= r^{k+1} [\cos(k\theta + \theta) + i\sin(k\theta + \theta)]$ (Applying Proposition 2)
$= r^{k+1}\text{cis}((k+1)\theta)$.
Therefore, the statement holds universally by induction for any $n \in \mathbb{Z}^+$.

Notice: De Moivre's theorem is valid for any integer exponent $n \in \mathbb{Z}$. For example, evaluating $z^{-5}$ produces $r^{-5}\text{cis}(-5\theta)$, which is computationally equal to $\frac{1}{r^5}(\cos 5\theta - i\sin 5\theta)$.

3. Applications and Examples

EXAMPLE 2

Given $z = 2\text{cis}\frac{\pi}{6}$ (which corresponds to $\sqrt{3} + i$):

$z^2 = 2^2\text{cis}(\frac{2\pi}{6}) = 4\text{cis}(\frac{\pi}{3}) = \mathbf{2 + 2\sqrt{3}i}$.
$z^{60} = 2^{60}\text{cis}(\frac{60\pi}{6}) = 2^{60}\text{cis}(10\pi) = 2^{60}(\cos 10\pi + i\sin 10\pi) = \mathbf{2^{60}}$.

Expanding an exponent like 60 using the binomial theorem would be mathematically prohibitive, demonstrating the necessity of De Moivre's Theorem.

EXAMPLE 3

Evaluate the expression $(1+i)^{10}$.

Solution: The complex number must first be converted into polar form: $1+i = \sqrt{2}\text{cis}\frac{\pi}{4}$.

$(1+i)^{10} = (\sqrt{2})^{10}\text{cis}(\frac{10\pi}{4}) = 2^5\text{cis}(\frac{10\pi}{4})$.
Because $\frac{10\pi}{4} = \frac{5\pi}{2} = 2\pi + \frac{\pi}{2}$, the principal argument reduces to $\frac{\pi}{2}$.
The final evaluation is $32\text{cis}\frac{\pi}{2} = 32(0 + i) = \mathbf{32i}$.

EXAMPLE 4 (Extracting Trigonometric Identities)

Calculate $z^3$ for the unit complex number $z = \cos\theta + i\sin\theta$ using two distinct mathematical methods to formulate identities.

Method 1 (De Moivre's Theorem): $z^3 = \cos 3\theta + i\sin 3\theta$.
Method 2 (Binomial Theorem): $z^3 = (\cos\theta + i\sin\theta)^3$
$= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$.
Grouping the real and imaginary components yields: $(\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta)$.
Equating the real parts establishes the identity: $\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$.
Substituting the Pythagorean identity $\sin^2\theta = 1 - \cos^2\theta$ produces the final form: $\cos 3\theta = \mathbf{4\cos^3\theta - 3\cos\theta}$.
Equating the imaginary parts similarly formulates: $\sin 3\theta = \mathbf{3\sin\theta - 4\sin^3\theta}$.

EXAMPLE 5

Utilizing the established properties $z^n + z^{-n} = 2\cos n\theta$ and $z^n - z^{-n} = 2i\sin n\theta$, expand the expression $(z+z^{-1})^3$.

Method 1 (Direct Substitution): $(z+z^{-1})^3 = (2\cos\theta)^3 = 8\cos^3\theta$.
Method 2 (Binomial Expansion): $(z+z^{-1})^3 = z^3 + 3z + 3z^{-1} + z^{-3}$.
Grouping the paired terms provides: $(z^3 + z^{-3}) + 3(z + z^{-1})$.
Substituting the trigonometric identities back in yields: $2\cos 3\theta + 6\cos\theta$.
Equating both methodologies determines that: $8\cos^3\theta = 2\cos 3\theta + 6\cos\theta$, which simplifies algebraically to $\cos^3\theta = \mathbf{\frac{1}{4}\cos 3\theta + \frac{3}{4}\cos\theta}$.

1.15 Roots of $z^n = a$ (HL)

1. Concept of Equality

An underlying mathematical observation is that an equality established in complex numbers functions as a powerful relation, naturally yielding a system of two independent equations.

For Cartesian forms: $x+yi = a+bi \Leftrightarrow \begin{cases} x = a \\ y = b \end{cases}$
For Polar forms: $r\text{cis}\theta = \rho\text{cis}\phi \Leftrightarrow \begin{cases} r = \rho \\ \theta = \phi + 2k\pi \end{cases}$

2. The n-th Roots of 1 (Roots of Unity)

The generalized solutions to the equation $z^n = 1$, which correspond directly to the roots of the polynomial $z^n - 1$, are classified as the $n$-th roots of 1. By defining an arbitrary root as $z = r\text{cis}\theta$, the equation translates into polar form as $r^n\text{cis}(n\theta) = 1\text{cis}(0)$.

This generates the following functional system:

$r^n = 1 \Rightarrow r = 1$
$n\theta = 0 + 2k\pi \Rightarrow \theta = \frac{2k\pi}{n}$

Consequently, the $n$ distinct roots of 1 are calculated iteratively using the integers $k = 0, 1, 2, \dots, n-1$:

$z_k = \text{cis}\left(\frac{2k\pi}{n}\right) = \cos\left(\frac{2k\pi}{n}\right) + i\sin\left(\frac{2k\pi}{n}\right)$

EXAMPLE 1

The 3rd roots of 1 (representing the solutions to $z^3 = 1$) are determined by evaluating $z_k = \text{cis}\left(\frac{2k\pi}{3}\right)$ across $k=0,1,2$.

  • $z_0 = \text{cis}(0) = \mathbf{1}$
  • $z_1 = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}$
  • $z_2 = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}$

Geometric interpretation: The modulus of every calculated root is precisely 1, anchoring them strictly to the perimeter of the unit circle. With defined arguments of $0$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$, these three roots partition the unit circle into three perfectly equal arcs.

EXAMPLE 2

The 4th roots of 1 (representing the solutions to $z^4 = 1$) are computed via $z_k = \text{cis}\left(\frac{2k\pi}{4}\right)$ for $k=0,1,2,3$.

  • $z_0 = \mathbf{1}$
  • $z_1 = \text{cis}\frac{\pi}{2} = \mathbf{i}$
  • $z_2 = \text{cis}\pi = \mathbf{-1}$
  • $z_3 = \text{cis}\frac{3\pi}{2} = \mathbf{-i}$

Geometrically, the four solutions $1, i, -1, -i$ sector the unit circle into four identical quadrants. Universally, the $n$-th roots of 1 geometrically divide the unit circle into exactly $n$ equal portions.

Notice on Root Sequences

If the primary non-real root $z_1$ is designated as $w = \text{cis}\left(\frac{2\pi}{n}\right)$, applying De Moivre's theorem reveals that $w^2 = z_2$, $w^3 = z_3$, and so forth. This establishes that the $n$-th roots of 1 can be represented holistically as a geometric sequence: $1, w, w^2, \dots, w^{n-1}$.

A crucial mathematical result arises from summing these roots as a finite geometric series:

$1 + w + w^2 + \dots + w^{n-1} = \frac{1 - w^n}{1 - w} = 0$

This evaluates to 0 because $w^n = 1$. Therefore, the sum of all $n$-th roots of 1 is universally exactly 0.

EXAMPLE 3 (Factoring $z^n - 1$)

Determine the 5th roots of 1, factorize the polynomial $z^5 - 1$, and utilize the root sum property to demonstrate that $\cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = -\frac{1}{2}$.

(a) Root Determination: The 5th roots are $z_0=1$, $z_1=\text{cis}\frac{2\pi}{5}$, $z_2=\text{cis}\frac{4\pi}{5}$, $z_3=\text{cis}\frac{6\pi}{5}$, and $z_4=\text{cis}\frac{8\pi}{5}$. Note that $z_3$ can be expressed as $\text{cis}(-\frac{4\pi}{5})$, acting strictly as the complex conjugate to $z_2$. Likewise, $z_4$ is the conjugate to $z_1$.
(b) Factorization: The polynomial possesses one linear factor $(z-1)$ and two irreducible quadratics formulated by multiplying the conjugate root pairs.
The first quadratic is evaluated as: $(z - z_1)(z - z_4) = (z - \cos\frac{2\pi}{5} - i\sin\frac{2\pi}{5})(z - \cos\frac{2\pi}{5} + i\sin\frac{2\pi}{5}) = z^2 - 2z\cos\frac{2\pi}{5} + 1$.
The second quadratic is similarly evaluated as $z^2 - 2z\cos\frac{4\pi}{5} + 1$.
Thus, the final factored form over the real numbers is $z^5 - 1 = (z - 1)(z^2 - 2z\cos\frac{2\pi}{5} + 1)(z^2 - 2z\cos\frac{4\pi}{5} + 1)$.
(c) Sum Property Execution: Given the total sum of the roots is 0, the aggregated sum of their real parts is simultaneously 0.
$1 + \cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} + \cos\frac{6\pi}{5} + \cos\frac{8\pi}{5} = 0$.
Because the cosines of conjugate angles match perfectly ($\cos(x) = \cos(-x)$), this relation reduces to $1 + 2\cos\frac{2\pi}{5} + 2\cos\frac{4\pi}{5} = 0$, algebraically leading to $\cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = \mathbf{-\frac{1}{2}}$.

Remark on the General Factorization of $z^n - 1$

Following the methodology established in the previous example, larger polynomials behave similarly based on their degree's parity:

  • If $n$ is an odd integer: Factoring $z^7 - 1$ produces one linear root and three quadratics: $(z-1)(z^2 - 2z\cos\frac{2\pi}{7} + 1)(z^2 - 2z\cos\frac{4\pi}{7} + 1)(z^2 - 2z\cos\frac{6\pi}{7} + 1)$.
  • If $n$ is an even integer: Factoring $z^6 - 1$ introduces two real roots ($\pm 1$), giving the factored form $(z-1)(z+1)(z^2 - z + 1)(z^2 + z + 1)$.

3. The n-th Roots of a Complex Number $a$

To compute the solutions for an equation in the format $z^n = a$, the complex constant $a$ must first be converted into its polar form $\rho\text{cis}\phi$. By assuming a generic root $z = r\text{cis}\theta$, the equation maps to $r^n\text{cis}(n\theta) = \rho\text{cis}\phi$.

This establishes the necessary mathematical system:

$r^n = \rho \Rightarrow r = \sqrt[n]{\rho}$
$n\theta = \phi + 2k\pi \Rightarrow \theta = \frac{\phi + 2k\pi}{n}$

For integer values $k = 0, 1, 2, \dots, n-1$, the $n$ distinct roots are defined formally as:

$z_k = \sqrt[n]{\rho} \text{cis}\left(\frac{\phi + 2k\pi}{n}\right)$

EXAMPLE 4

Solve the specific equation $z^3 = 8i$.

Solution: The constant $8i$ is translated to the polar form $8\text{cis}\frac{\pi}{2}$. Equating $r^3\text{cis}(3\theta) = 8\text{cis}\frac{\pi}{2}$ yields the radius $r = 2$ and the angle equation $3\theta = \frac{\pi}{2} + 2k\pi$.

Solving for the argument $\theta$ produces $\theta = \frac{\pi}{6} + \frac{2k\pi}{3} = \frac{\pi + 4k\pi}{6}$. Computing for $k=0, 1, 2$ establishes the three distinct roots:
$z_0 = \mathbf{2\text{cis}\frac{\pi}{6}}$
$z_1 = \mathbf{2\text{cis}\frac{5\pi}{6}}$
$z_2 = 2\text{cis}\frac{9\pi}{6} = \mathbf{2\text{cis}\frac{3\pi}{2}}$

Geometric interpretation: All three computed roots sit on a circle defined by a radius of 2. They divide the circle into three perfectly equal arcs, but the initial root $z_0$ begins at an offset angle of $\frac{\pi}{6}$ rather than at $0$.