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 $\dfrac{1}{z} = \dfrac{1}{r}\text{cis}(-\theta)$.
Proof: Utilizing the property $z\overline{z} = |z|^2 = r^2$ and $\overline{z} = r\text{cis}(-\theta)$.
$$
z^{-1} = \dfrac{1}{z} = \dfrac{\overline{z}}{z\overline{z}}
= \dfrac{r\text{cis}(-\theta)}{r^2}
= \dfrac{1}{r}\text{cis}(-\theta)
$$
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:
$$\begin{aligned}
z_1 z_2 &= r_1\text{cis}\theta_1 \cdot r_2\text{cis}\theta_2 \\
&= r_1 r_2 (\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 + i\sin\theta_2) \\
&= r_1 r_2 [\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2 + i(\sin\theta_1\cos\theta_2 + \sin\theta_2\cos\theta_1)] \\
&= r_1 r_2 [\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)] \\
&= r_1 r_2\text{cis}(\theta_1 + \theta_2)
\end{aligned}$$
PROPOSITION 3
Let $z_1 = r_1\text{cis}\theta_1$ and $z_2 = r_2\text{cis}\theta_2$. Then $\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}\text{cis}(\theta_1 - \theta_2)$.
Proof:
$$\begin{aligned}
\dfrac{z_1}{z_2} &= z_1 \cdot z_2^{-1} \\
&= r_1\text{cis}\theta_1 \cdot r_2^{-1}\text{cis}(-\theta_2) \\
&= r_1 r_2^{-1}\text{cis}(\theta_1 - \theta_2) \quad \text{[by Proposition 2]} \\
&= \dfrac{r_1}{r_2}\text{cis}(\theta_1 - \theta_2)
\end{aligned}$$
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 $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$.
- $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$ and $\arg\left(\dfrac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$.
EXAMPLE 1
Let $z = 2\text{cis}\dfrac{\pi}{6}$ (which expands to $\sqrt{3} + i$) and $w = \text{cis}\dfrac{\pi}{3}$ (which expands to $\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$). Operations yield:
$$\begin{aligned}
zw &= 2 \cdot 1 \text{cis}\left(\dfrac{\pi}{6} + \dfrac{\pi}{3}\right) = 2\text{cis}\left(\dfrac{\pi}{2}\right) = 2(0 + i) = \mathbf{2i} \\
\dfrac{z}{w} &= \dfrac{2}{1}\text{cis}\left(\dfrac{\pi}{6} - \dfrac{\pi}{3}\right) = 2\text{cis}\left(-\dfrac{\pi}{6}\right) = 2\left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i\right) = \mathbf{\sqrt{3} - i} \\
z^2 &= z \cdot z = 2 \cdot 2\text{cis}\left(\dfrac{\pi}{6} + \dfrac{\pi}{6}\right) = 4\text{cis}\left(\dfrac{\pi}{3}\right) = \mathbf{2 + 2\sqrt{3}i}
\end{aligned}$$
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:
$$z^n = r^n\text{cis}(n\theta)$$
NOTICE: Remember Euler’s notation $z = re^{i\theta} = r(\cos\theta + i\sin\theta)$. De Moivre’s Theorem is in strict accordance with this notation. Indeed:
$$z^n = (re^{i\theta})^n = r^n e^{in\theta} = r^n[\cos(n\theta) + i\sin(n\theta)]$$
which perfectly aligns with De Moivre’s statement.
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: We assume that the statement is true for $n=k$, establishing the hypothesis $z^k = r^k\text{cis}(k\theta)$.
Step 3: We claim that the statement holds for $n=k+1$, requiring $z^{k+1} = r^{k+1}\text{cis}((k+1)\theta)$.
$$\begin{aligned}
z^{k+1} &= z^k \cdot z \\
&= r^k\text{cis}(k\theta) \cdot r\text{cis}\theta \quad \text{[by hypothesis]} \\
&= r^k r [\cos(k\theta) + i\sin(k\theta)][\cos\theta + i\sin\theta] \\
&= r^{k+1} [\cos(k\theta + \theta) + i\sin(k\theta + \theta)] \quad \text{[applying Proposition 2]} \\
&= r^{k+1}\text{cis}((k+1)\theta)
\end{aligned}$$
Therefore, by mathematical induction, the statement is true for any $n \in \mathbb{Z}^+$. (Q.E.D.)
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 $\dfrac{1}{r^5}(\cos 5\theta - i\sin 5\theta)$.
3. Applications and Examples
EXAMPLE 2
Given $z = 2\text{cis}\dfrac{\pi}{6}$ (which corresponds to $\sqrt{3} + i$):
$$\begin{aligned}
z^2 &= 2^2\text{cis}\left(\dfrac{2\pi}{6}\right) = 4\text{cis}\left(\dfrac{\pi}{3}\right) = \mathbf{2 + 2\sqrt{3}i} \\
z^{60} &= 2^{60}\text{cis}\left(\dfrac{60\pi}{6}\right) = 2^{60}\text{cis}(10\pi) = 2^{60}(\cos 10\pi + i\sin 10\pi) = \mathbf{2^{60}}
\end{aligned}$$
Expanding an exponent like 60 using the binomial theorem would be mathematically prohibitive, demonstrating the sheer 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: $1+i = \sqrt{2}\text{cis}\dfrac{\pi}{4}$.
$$
(1+i)^{10} = (\sqrt{2})^{10}\text{cis}\left(\dfrac{10\pi}{4}\right)
= 2^5\text{cis}\left(\dfrac{5\pi}{2}\right)
$$
Because $\dfrac{5\pi}{2} = 2\pi + \dfrac{\pi}{2}$, the principal argument reduces to $\dfrac{\pi}{2}$.
$$32\text{cis}\dfrac{\pi}{2} = 32(0 + i) = \mathbf{32i}$$
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):
$$\begin{aligned}
z^3 &= (\cos\theta + i\sin\theta)^3 \\
&= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta
\end{aligned}$$
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):
$$\begin{aligned}
(z+z^{-1})^3 &= z^3 + 3z + 3z^{-1} + z^{-3} \\
&= (z^3 + z^{-3}) + 3(z + z^{-1})
\end{aligned}$$
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:
$$\cos^3\theta = \mathbf{\dfrac{1}{4}\cos 3\theta + \dfrac{3}{4}\cos\theta}$$
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: $\quad x+yi = a+bi \iff \begin{cases} x = a \\ y = b \end{cases}$
For Polar forms: $\quad r\text{cis}\theta = \rho\text{cis}\phi \iff \begin{cases} r = \rho \\ \theta = \phi + 2k\pi \end{cases}$
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:
$$\begin{cases}
r^n = 1 \implies r = 1 \\
n\theta = 0 + 2k\pi \implies \theta = \dfrac{2k\pi}{n}
\end{cases}$$
Consequently, the $n$ distinct roots of 1 are calculated iteratively using the integers $k = 0, 1, 2, \dots, n-1$:
$$z_k = \text{cis}\left(\dfrac{2k\pi}{n}\right) = \cos\left(\dfrac{2k\pi}{n}\right) + i\sin\left(\dfrac{2k\pi}{n}\right)$$
EXAMPLE 1 (3rd Roots of Unity)
The 3rd roots of 1 (representing the solutions to $z^3 = 1$) are determined by evaluating $z_k = \text{cis}\left(\dfrac{2k\pi}{3}\right)$ across $k=0,1,2$.
$$\begin{aligned}
z_0 &= \text{cis}(0) = \mathbf{1} \\
z_1 &= \cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3} = \mathbf{-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i} \\
z_2 &= \cos\dfrac{4\pi}{3} + i\sin\dfrac{4\pi}{3} = \mathbf{-\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i}
\end{aligned}$$
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 $0$, $\dfrac{2\pi}{3}$, and $\dfrac{4\pi}{3}$, these three roots partition the unit circle into three perfectly equal arcs.
EXAMPLE 2 (4th Roots of Unity)
The 4th roots of 1 (representing the solutions to $z^4 = 1$) are computed via $z_k = \text{cis}\left(\dfrac{2k\pi}{4}\right)$ for $k=0,1,2,3$.
$$\begin{aligned}
z_0 &= \mathbf{1} \\
z_1 &= \text{cis}\dfrac{\pi}{2} = \mathbf{i} \\
z_2 &= \text{cis}\pi = \mathbf{-1} \\
z_3 &= \text{cis}\dfrac{3\pi}{2} = \mathbf{-i}
\end{aligned}$$
Geometric interpretation: 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 $w = \text{cis}\left(\dfrac{2\pi}{n}\right)$, 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:
$$1 + w + w^2 + \dots + w^{n-1} = \dfrac{1 - w^n}{1 - w} = 0$$
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 $\cos\dfrac{2\pi}{5} + \cos\dfrac{4\pi}{5} = -\dfrac{1}{2}$.
(a) Root Determination: The 5th roots are $z_0=1$, $z_1=\text{cis}\dfrac{2\pi}{5}$, $z_2=\text{cis}\dfrac{4\pi}{5}$, $z_3=\text{cis}\dfrac{6\pi}{5}$, and $z_4=\text{cis}\dfrac{8\pi}{5}$. Note that $z_3$ can be expressed as $\text{cis}\left(-\dfrac{4\pi}{5}\right)$, 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:
$$\begin{aligned}
(z - z_1)(z - z_4) &= \left(z - \cos\dfrac{2\pi}{5} - i\sin\dfrac{2\pi}{5}\right)\left(z - \cos\dfrac{2\pi}{5} + i\sin\dfrac{2\pi}{5}\right) \\
&= z^2 - 2z\cos\dfrac{2\pi}{5} + 1
\end{aligned}$$
The second quadratic is similarly evaluated as $z^2 - 2z\cos\dfrac{4\pi}{5} + 1$.
Thus, the final factored form over the real numbers is:
$$z^5 - 1 = (z - 1)\left(z^2 - 2z\cos\dfrac{2\pi}{5} + 1\right)\left(z^2 - 2z\cos\dfrac{4\pi}{5} + 1\right)$$
(c) Sum Property Execution: Given the total sum of the roots is 0, the aggregated sum of their real parts is simultaneously 0.
$$1 + \cos\dfrac{2\pi}{5} + \cos\dfrac{4\pi}{5} + \cos\dfrac{6\pi}{5} + \cos\dfrac{8\pi}{5} = 0$$
Because the cosines of conjugate angles match perfectly ($\cos(x) = \cos(-x)$), this relation reduces to:
$$\begin{aligned}
1 + 2\cos\dfrac{2\pi}{5} + 2\cos\dfrac{4\pi}{5} &= 0 \\
2\left(\cos\dfrac{2\pi}{5} + \cos\dfrac{4\pi}{5}\right) &= -1 \\
\implies \cos\dfrac{2\pi}{5} + \cos\dfrac{4\pi}{5} &= \mathbf{-\dfrac{1}{2}}
\end{aligned}$$
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: For example, factoring $z^7 - 1$ produces one linear root and three quadratics:
$$(z-1)\left(z^2 - 2z\cos\dfrac{2\pi}{7} + 1\right)\left(z^2 - 2z\cos\dfrac{4\pi}{7} + 1\right)\left(z^2 - 2z\cos\dfrac{6\pi}{7} + 1\right)$$
By using the sum of the 7th roots of 1, we can identically show that
$$
\cos\dfrac{2\pi}{7} + \cos\dfrac{4\pi}{7} + \cos\dfrac{6\pi}{7} = -\dfrac{1}{2}
$$
Can you guess a similar formula by using the 9th roots of 1?
-
If $n$ is an even integer: For example, factoring $z^6 - 1$ introduces two real roots ($\pm 1$). Grouping the remaining conjugate pairs gives:
$$\begin{aligned}
z^6 - 1 &= (z-1)(z+1)\left(z^2 - 2z\cos\dfrac{2\pi}{6} + 1\right)\left(z^2 - 2z\cos\dfrac{4\pi}{6} + 1\right) \\
&= (z-1)(z+1)\left(z^2 - 2z\cos\dfrac{\pi}{3} + 1\right)\left(z^2 - 2z\cos\dfrac{2\pi}{3} + 1\right) \\
&= (z-1)(z+1)(z^2 - z + 1)(z^2 + z + 1)
\end{aligned}$$
Can you find a similar factorization for $z^8 - 1$?
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:
$$\begin{cases}
r^n = \rho \implies r = \sqrt[n]{\rho} \\
n\theta = \phi + 2k\pi \implies \theta = \dfrac{\phi + 2k\pi}{n}
\end{cases}$$
For integer values $k = 0, 1, 2, \dots, n-1$, the $n$ distinct roots are defined formally as:
$$z_k = \sqrt[n]{\rho} \text{cis}\left(\dfrac{\phi + 2k\pi}{n}\right)$$
EXAMPLE 4
Solve the specific equation $z^3 = 8i$.
Solution: The constant $8i$ is translated to the polar form $8\text{cis}\dfrac{\pi}{2}$. Equating $r^3\text{cis}(3\theta) = 8\text{cis}\dfrac{\pi}{2}$ yields the radius $r = 2$ and the angle equation
$$3\theta = \dfrac{\pi}{2} + 2k\pi$$
Solving for the argument $\theta$ produces $\theta = \dfrac{\pi}{6} + \dfrac{2k\pi}{3} = \dfrac{\pi + 4k\pi}{6}$. Computing for $k=0, 1, 2$ establishes the three distinct roots:
$$\begin{aligned}
z_0 &= \mathbf{2\text{cis}\dfrac{\pi}{6}} \\
z_1 &= \mathbf{2\text{cis}\dfrac{5\pi}{6}} \\
z_2 &= 2\text{cis}\dfrac{9\pi}{6} = \mathbf{2\text{cis}\dfrac{3\pi}{2}}
\end{aligned}$$
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 $\dfrac{\pi}{6}$ ($30^\circ$) rather than at $0$.