Polar form & Euler's Formula

If you hear a voice within you say ‘you cannot paint’, then by all means paint and that voice will be silenced, Vincent van Gogh

Recall

Complex numbers are fundamental in various fields of mathematics, physics, and engineering. They provide a complete number system where every polynomial equation has a solution.

A complex number is specified by an ordered pair of real numbers (a, b) ∈ ℝ² and expressed or written in the form z = a + bi, where a and b are real numbers, and i is the imaginary unit, defined by the property $i^2 = -1 \iff[\text{or equivalently}] i = \sqrt{-1}, \boxed{\mathbb{C}= \{ a + bi ∣a, b ∈ ℝ\} }, e.g., 2 + 5i, 7\pi + i\sqrt{2}.$

Polar Form & Euler’s Formula

A complex number z = a +bi can also be expressed in polar form, which uses its modulus (magnitude) r and argument (angle) θ:

• The modulus r = |z| = $\sqrt{a^2+b^2}$. It represents the distance of the complex number from the origin in the complex plane.

  |z₁-z₂| is the distance between these two points in the complex plane. |Re(z)| ≤ |z|, |Im(z)| ≤ |z|, $z · \bar z = |z|², |z_1+z_2| \le |z_1| + |z_2|$

• The argument $θ = arg(z) = tan^{-1}(\frac{b}{a})$is the angle between the positive real axis and the line representing the complex number. The angle θ should be determined in the correct quadrant, considering the signs of a and b. Furthermore, the argument of a complex number z = a + bi (≠ 0) is the set {$\theta_0 + 2n\pi: -\pi \lt \theta_0 \le \pi, n \in \mathbb{Z}, a = rcos(\theta_0), b = rsin(\theta_0)$}, $\theta_0$ is called the principal argument of z (≠ 0), $\theta_0$ = Arg(z).

Converting Between Forms

Rectangular to Polar	Polar to Rectangular
$r = \sqrt{a^2 + b^2}$	$a = r\cos\theta$
$\theta = \arctan\left(\frac{b}{a}\right)$ (with quadrant adjustment)	$b = r\sin\theta$

The polar form of a complex number is: z = a + bi = rcos(θ) + rsin(θ)i = r(cos(θ) + isin(θ)) =[This can be written compactly using Euler’s formula], re^iθ, e.g., z = 1 + i = $\sqrt{2}e^{\frac{π}{4}i}$. It satisfies e^i(θ+β) = e^iθe^iβ.

z = 1 + i. $r = |z| = \sqrt{1^2 + 1^2} = \sqrt{2}, \theta = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}, z = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$

• $\text{arg}(z)$ refers to all possible angles that a complex number z can make with the positive real axis in the complex plane. It's a multi-valued function.
• $\text{Arg}(z)$ refers to the principal argument, which is typically restricted to a specific range, usually between $\pi$ and $\pi$. Therefore, $\text{arg}(z) = \text{Arg}(z) + 2\pi k$, where k is any integer ℤ. This accounts for the periodic nature of angles in polar coordinates, e.g., z = 1 + i = $\sqrt{2}e^{\frac{π}{4}i}, \text{arg}(z) =\frac{π}{4} + 2\pi k$.
• For a complex number z = x + iy, we can calculate Arg(z) using the following formula: $Arg(z) = ε + tan^{-1}(\frac{y}{x})$ where ε = 0, ±πdepending on the quadrant of z, it adjusts the angle based on the quadrant in which the complex number lies.

1. If x > 0, y ≥ 0, then the point is in the first quadrant: ε = 0, e.g., z = 1 + i, Arg(z) = $tan^{-1}(\frac{y}{x})$ = arctan(1) = π/4, z = $\sqrt{2} e^{\frac{\pi}{4}}$
2. If x < 0 and y ≥ 0, then the point is in the second quadrant: ε = π, e.g., z = -1 + i, Arg(z) = $tan^{-1}(\frac{y}{x}) + \pi$ = arctan(-1) + π = -π/4 + π = 3π/4, z = $\sqrt{2} e^{\frac{3\pi}{4}}$
3. If x < 0 and y < 0, then the point is in the third quadrant: ε = -π, e.g., z = -1 - i, arctan(y/x) = arctan(1) = π/4. Therefore, Arg(z) = -π + π/4 = -3π/4, z = $\sqrt{2} e^{\frac{-3\pi}{4}}$
4. If x > 0 and y < 0, then the point is in the fourth quadrant: ε = 0, e.g., z = 1 - i, arctan(y/x) = arctan(-1) = -π/4. Therefore, Arg(z) = -π/4, z = $\sqrt{2} e^{\frac{-\pi}{4}}$

Quadrant	Condition	$\theta$
I	$a > 0, b > 0$	$\arctan(b/a)$
II	$a < 0, b > 0$	$\pi + \arctan(b/a)$
III	$a < 0, b < 0$	$-\pi + \arctan(b/a)$
IV	$a > 0, b < 0$	$\arctan(b/a)$

Euler’s Formula

Euler’s formula is a fundamental formula in complex analysis. It establishes the mathematical relationship between trigonometric functions and the complex exponential function. It states that $\boxed{e^{i\theta} = \cos\theta + i\sin\theta}$. This allows us to represent complex numbers compactly in exponential form. It simplifies operations like multiplication, division, and exponentiation.

Proof.

The exponential function $e^x$ for any complex number z is defined by its Taylor series: $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$.

Substituting $z = i\theta$, we get: $e^{i\theta} = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} = \sum_{n=0}^{\infty} \frac{i^n\theta^n}{n!}$

The imaginary unit i has a cyclic pattern: $i^0=1, i^1 = 1, i^2 = -1, i^3 = -i, i^4 = 1$, and so one. This periodicity allows us to split the series into even (n = 2k) and odd (n = 2k+1) terms:

1. Even terms (n = 2k): $\sum_{k=0}^{\infty} \frac{i^{2k}\theta^{2k}}{(2k)!} = \frac{(-1)^k\theta^{2k}}{(2k)!}$ (since $i^{2k} = (i^2)^k = (-1)^k$). This is the Taylor series for $\cos(\theta)$.
2. Odd terms (n = 2k +1): $\sum_{k=0}^{\infty} \frac{i^{2k+1}\theta^{2k+1}}{(2k+1)!} = i\sum_{k=0}^{\infty} \frac{(-1)^k\theta^{2k+1}}{(2k+1)!}$ (since $i^{2k+1} = i \cdot i^{2k} = i(-1)^k$). This is the Taylor series for $\sin(\theta)$.

Adding the even and odd series gives: $e^{i\theta} = \cos(\theta) + i\sin(\theta)$.

Euler’s Identity

Setting $\theta = \pi$ in Euler’s formula: $\boxed{e^{i\pi} + 1 = 0}$. This is often called “the most beautiful equation in mathematics” as it connects five fundamental constants: $e$, $i$, $\pi$, $1$, and $0$.

Deriving Trigonometric Identities

$\boxed{\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}}$

The identities for cosine and sine in terms of complex exponentials follow directly from Euler’s formula, which states: $e^{i\theta} = cos(\theta) + i\sin(\theta)$.

Taking the complex conjugate (or replacing θ with −θ) gives: $e^{-i\theta} = cos(\theta) - i\sin(\theta)$.

Adding the two equations eliminates the sine term: $e^{i\theta} + e^{-i\theta} = 2cos(\theta) \implies \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$.

Subtracting them eliminates the cosine term: $e^{i\theta} - e^{-i\theta} = 2isin(\theta) \implies \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2}$

Exponential Form of Complex Numbers

The exponential form of complex numbers $\boxed{z = re^{i\theta}}$ is a powerful representation derived directly from Euler’s formula. It reduces complex arithmetic to simple algebraic rules (e.g., adding exponents for multiplication).

Different ways of representing complex numbers

Form	Expression	Comment
Rectangular	z = a + bi	Conversion: $r = \sqrt{a^2 + b^2}, \theta= arctan(b/a)$ (adjusted for quadrant)
Polar (trigonometric)	$z = r(\cos\theta + i\sin\theta)$	Directly equivalent to $z = re^{i\theta}$ by Euler’s formula
Polar (exponential)	$z = re^{i\theta}$	Simplifies calculus operations

Multiplication and Division of Complex Numbers in Polar Form

The polar form $z=re^{i\theta}$ is not just an alternative notation; it provides a deep geometric interpretation of complex arithmetic, making multiplication and division intuitive and computationally simple.

Given two complex numbers in polar form, $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$, their product is: $z_1 z_2 = r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} =[e^a \cdot e^b = e^{a+b}] r_1 r_2 e^{i(\theta_1 + \theta_2)}$, hence $|z_1 z_2| = |z_1| |z_2|, \arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$, e.g., $z_1 = 2e^{\frac{i\pi}{3}}$ and $z_2=3e^{\frac{i\pi}{2}}, z_1z_2 = 6e^{\frac{5\pi i}{6}}$

Product Property of Exponents. The product of two exponentials with the same base is equal to the base raised to the sum of their exponents, $a^m·a^n = a^{m+n}$

The multiplication of two complex numbers in polar form results in a new complex number whose modulus is the product of the moduli ($|z_1z_2| = r_1r_2 = |z_1| |z_2|$) and whose argument is the sum of the arguments ($\arg(z_1 z_2) = \theta_1 + \theta_2 = \arg(z_1) + \arg(z_2)$).

Geometrically, multiplication by a complex number performs a rotation and a scaling.Multiplication by $e^{i\phi}$ rotates the complex number by angle $\phi$ counterclockwise. Multiplication by a real scalar $r$ stretches or shrinks the modulus by factor r.

For division, with $z_2 \ne 0$: $\frac{z_1}{z_2} = \frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}$, e.g., $z_1 = 2e^{\frac{i\pi}{3}}$ and $z_2=3e^{\frac{i\pi}{2}}, \frac{z_1}{z_2} = \frac{2}{3}e^{\frac{-\pi i}{6}}$

To divide two complex numbers:

1. Divide their moduli: $|\frac{z_1}{z_2}| = \frac{r_1}{r_2} = \frac{|z_1|}{|z_2|}$
2. Subtract their arguments: $arg(\frac{z_1}{z_2}) = \theta_1 - \theta_2 = arg(z_1) - arg(z_2)$.

The Ambiguity of the Argument

A fundamental subtlety in working with polar form is that the argument arg(z) is not unique.

Given a complex number z, it can be expressed as z = |z|e^iθ₁ = |z|e^iθ₂ ⇒[From this equality, we can infer:] e^iθ₁ = e^iθ₂ ⇒[Using Euler’s formula, we can express the complex exponentials in terms of sine and cosine:] cos(θ₁) + isin(θ₁) = cos(θ₂) + isin(θ₂) ⇒[From this equality, we can separate the real and imaginary parts:] cos(θ₁) = cos(θ₂) and sin(θ₁) = sin(θ₂). This is true if and only if: $\theta_2 = \theta_1 + 2\pi k \iff \theta_2 - \theta_1 = 2\pi k, \forall k \in \mathbb{Z}$. Thus, the argument is a set of angles: arg(z) = {$\theta + 2\pi k, \forall k \in \mathbb{Z}$}

In complex analysis, the argument (arg) of a complex number z = a+bi is the angle θ formed with the positive real axis. However, because angles can differ by multiples of 2π radians, the argument isn’t unique.

The beautiful property is that the formulas hold perfectly for these sets:

1. $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) = \{ \theta_1 + \theta_2 + 2\pi k | k \in \mathbb{Z} \}$
2. $\arg(\frac{z_1}{z_2}) = \arg(z_1) - \arg(z_2) = \{ \theta_1 + \theta_2 + 2\pi k | k \in \mathbb{Z} \}$

When you add or subtract the infinite sets of possible angles, you get precisely the infinite set of possible angles for the result. For any complex numbers z₁ and z₂, this property ensures that the sum of their arguments corresponds correctly to the argument of their product, accounting for the periodic nature of angles.

The same does not apply to $\text{Arg}(z_1)+\text{Arg}(z_2)$ which is typically restricted to $-\pi < \text{Arg}(z) \leq \pi$. Since $\text{Arg}(z)$ is confined to a specific range, the sum $\text{Arg}(z_1) + \text{Arg}(z_2)$ can fall outside this range ( $\text{Arg}(z_1) + \text{Arg}(z_2)$ may not equal $\text{Arg}(z_1z_2)$). The sum may fall outside the principal range (−π, π], requiring an adjustment by ±2π to bring it back into range.

Theorem. For $z \ne 0$, the multiplicative inverse $z^{-1}$ is given by $z^{-1} = \frac{\bar z}{|z|^2}$

Fundamental property. For any complex number z, $z\bar z$ = |z|². Since $z \ne 0, |z|^2 \ne 0$. Dividing both sides by |z|², we get $z(\frac{\bar z}{|z|²}) = 1$ ⇒ Hence, the multiplicative inverse of z is $z^{-1} =\frac{\bar z}{|z|²}$, e.g., $(1+i)^{-1} =\frac{1-i}{|\sqrt{2}|²} = \frac{1-i}{2}$.

Let $z = \sqrt{2}e^{\frac{i\pi}{4}}$

1. $\bar{z} = \sqrt{2}(cos(\frac{π}{4}) -isin(\frac{π}{4})) = \sqrt{2}e^{\frac{-i\pi}{4}}, |z|^2 = (\sqrt{2})^2=2, z^{-1} = \frac{\sqrt{2}e^{\frac{-i\pi}{4}}}{2} = \frac{e^{-\frac{π}{4}i}}{\sqrt{2}}$
2. $z^{-1} =\frac{1}{z} = \frac{1}{\sqrt{2}e^{i\frac{π}{4}}} =[(e^{iθ})^{-1} = e^{-iθ}] \frac{e^{-\frac{π}{4}i}}{\sqrt{2}}$

Special Case: Unit Circle (∣z∣ = 1)

Let $z = e^{iθ} = cos(θ) + isin(θ) ≠ 0, z^{-1}=\frac{\bar z}{|z|²} = \frac{cos(θ)-sin(θ)i}{cos(θ)²+sin(θ)²}$ =[Since cosine is even (cos(−θ) = cos(θ)) and sine is odd (sin(−θ) = −sin(θ))] $\frac{cos(-\theta)+isin(-\theta)}{1} = e^{-iθ}$. Therefore, $\boxed{z^{-1} = e^{-iθ}}$. This aligns with the property of exponents: $(e^{iθ})^{-1} = e^{-iθ}$

De Moivre’s Theorem

For any integer n, De Moivre’s Theorem asserts: $\boxed{(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)}$. Equivalently, using Euler’s formula: $\boxed{(e^{i\theta})^n = e^{in\theta}}$.

This theorem bridges complex numbers, trigonometry, and exponentiation, with profound implications in mathematics and engineering.

Proof.

Start with a complex number in polar form: $z=r(\cos(\theta) +i\sin(\theta))=re^{i\theta }.$

Raise it to the integer power n: $z^n=(re^{i\theta })^n.$

Use the laws of exponents: $(re^{i\theta })^n=r^n(e^{i\theta })^n=r^ne^{in\theta}$.

Apply Euler’s identity again: $e^{in\theta}=\cos (n\theta)+i\sin(n\theta).$ Thus, $\boxed{z^n=r^n(\cos(n\theta) +i\sin(n\theta))}.\quad \blacksquare$

When r = 1, $z=e^{i\theta }=\cos(\theta) +i\sin(\theta)$, and the general formula becomes $(\cos(\theta) +i\sin(\theta))^n=\cos(n\theta) +i\sin (n\theta)$.

Applications of De Moivre’s Theorem

1. Powers of Complex Numbers. Compute $(1 + i)^8$. Convert to polar form: $1 + i = \sqrt{2}e^{i\pi/4}$ and apply De Moivre: $(1 + i)^8 = (\sqrt{2})^8 e^{i \cdot 8 \cdot \pi/4} = 16e^{i \cdot 2\pi} = 16 \cdot 1 = 16$.
   Compute $(1 - i)^{10}$. Convert to polar form: $1 - i = \sqrt{2}e^{-i\pi/4}$ and apply De Moivre: $(1 - i)^{10} = (\sqrt{2})^{10} e^{-i \cdot 10 \cdot \pi/4} = 32e^{-i \cdot 5\pi/2} = 32e^{-i(2\pi + \frac{\pi}{2})}=32e^{-i\pi/2} = 32(-i) = -32i$. z = -1 -i, z¹⁰⁰. z = -1 -i = $\sqrt{2} e^{\frac{-\pi}{4}}$ = $\sqrt{2}e^{\frac{5π}{4}i}$.

   $z^{100} =[ \text{De Moivre's Theorem } ] \sqrt{2}^{100}e^{\frac{5π∙100}{4}i} = 2^{50}e^{125πi} = 2^{50}(e^{πi})^{125} = [e^{πi} = -1] -2^{50}$

2. Deriving Multiple Angle Formulas Double-Angle Formulas: Square $\cos(\theta)+i\sin(\theta): (\cos(\theta) +i\sin(\theta))^2=\cos^2(\theta) +2i\cos(\theta)\sin(\theta) -\sin ^2(\theta) =[\text{Group real and imaginary parts}] \cos^2(\theta) -\sin ^2(\theta) +2i\cos(\theta)\sin(\theta).$

   By De Moivre for n = 2: $(\cos(\theta) +i\sin(\theta))^2 = \cos(2\theta ) + i\sin(2\theta)$.

   Equating real and imaginary parts: $\boxed{\cos(2\theta)= \cos^2(\theta)−\sin^2(\theta), \sin(2\theta) = 2\sin(\theta)\cos(\theta)}$

   Triple-Angle Formulas: Cube $\cos(\theta)+i\sin(\theta): (\cos(\theta)+i\sin(\theta))^3 ​ =\cos^3(\theta) +3i\cos^2(\theta)\sin(\theta) −3\cos(\theta)\sin^2(\theta) −i\sin^3(\theta) = ​(\cos^3(\theta)−3\cos(\theta)\sin^2(\theta))+i(3\cos^2(\theta)\sin(\theta)-\sin^3(\theta))$

   By De Moivre for n = 3: $(\cos(\theta) +i\sin(\theta))^3 = \cos(3\theta ) + i\sin(3\theta)$.

   Equating parts: $\boxed{\cos(3\theta) = \cos^3(\theta)−3\cos(\theta)\sin^2(\theta), \sin(3\theta) = 3\cos^2(\theta)\sin(\theta)-\sin^3(\theta)}$

3. Roots of Complex Numbers. To find the n-th roots of $z = re^{i\theta}$, solve $w^n = z$. Let $w = \rho e^{i\phi}$:

   $\rho^n e^{in\phi} = re^{i\theta} \implies \rho = r^{1/n}, n\phi = \theta + 2\pi k$ (k = 0, 1, …,n − 1). Thus, the n-th roots are: $w_k = r^{1/n}e^{\frac{i(\theta + 2\pi k)}{n}}$ (k = 0, 1, ...,n − 1), e.g., cube roots of 8 (i.e., $8e^{i0}$): $w_k = 2e^{\frac{i(0 + 2\pi k)}{3}}$ (k = 0, 1, 2) $\implies 2, -1 + i\sqrt{3}, -1 -i\sqrt{3}$

   There are exactly $n$ distinct $n$th roots. The $n$-th roots of $z$ are equally spaced on a circle of radius $\sqrt[n]{|z|}$, separated by angles of $\dfrac{2\pi}{n}$.

4. Roots of Unity. The n-th roots of $1$ are: $\omega_k = e^{2\pi i k/n}, k = 0, 1, \ldots, n-1$. Let $\omega = e^{2\pi i/n}$ (primitive n-th root of unity), then the n-th roots of unity are: $\{1, \omega, \omega^2, \ldots, \omega^{n-1}\}$ ​

5. Solving Equations. Solve $z^4 = −16$. The number −16 lies on the negative real axis. Its modulus is ∣−16∣ = 16, and its principal argument is π. Thus, $−16=16e^{i\pi}$.

   Fourth roots: $z_k​=16^{1/4}e^{i(\pi+2\pi k)/4} = 2e^{i(\pi/4+\pi k/2)}$ (k = 0, 1, 2, 3).

   Roots: $2e^{i\pi/4}, 2e^{i3\pi/4}, 2e^{i5\pi/4}, 2e^{i7\pi/4}.$

   $w_0 = 2e^{i\pi/4} = 2(cos(\pi/4) + isin(\pi/4)) = 2(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}) = \sqrt{2}(1+i)$.
   $w_1 = 2e^{i3\pi/4} = 2(cos(3\pi/4) + isin(3\pi/4)) = 2(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}) = \sqrt{2}(-1+i)$.
   $w_2 = 2e^{i5\pi/4} = 2(cos(5\pi/4) + isin(5\pi/4)) = 2(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}) = \sqrt{2}(-1-i)$.
   $w_3 = 2e^{i7\pi/4} = 2(cos(7\pi/4) + isin(7\pi/4)) = 2(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}) = \sqrt{2}(1-i)$.
   The four roots are equally spaced on a circle of radius 2 in the complex plane. Their arguments differ by π/2 (90°), forming the vertices of a square centered at the origin. This illustrates a general property: the nth roots of any nonzero complex number lie on a circle and are the vertices of a regular n-gon.