Circle Equation in the Complex Plane

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Circle Equation in the Complex Plane

The equation |z - a| = r defines a circle in the complex plane(the distance between any point z on the circle and the center a is equal to the radius r) where:

• z = x + iy represents any point on the circle
• a = h + ki is the center of the circle
• r is the radius of the circle, a positive real number.

The absolute value |z - a| represents the distance between points z and a in the complex plane. This distance formula is:

|z - a| = |x + yi - (h + ki)| = |(x - h) + (y - k)i| = $\sqrt{(x - h)² + (y - k)²}$

When this distance equals r, we get the familiar circle equation: (x - h)² + (y - k)² = r².

Special Cases

• Circle centered at origin (a = 0): |z| = r. This represents all complex numbers with magnitude r. It forms a circle of radius r centered at the origin.
• Circle centered on real axis (a is real): |z - a| = r where a = h ∈ ℝ. Center is at (h, 0) in the coordinate system.
• Circle centered on imaginary axis (a is purely imaginary): |z - ki| = r where a = ki, k ∈ ℝ. Center is at (0, k) in the coordinate system.

Interior/exterior of circles & Annular regions:

• |z - a| < r (interior of the circle)
• |z - a| > r (exterior of the circle)
• |z - a| ≤ r (closed disk)
• r₁ < |z - a| < r₂ (ring-shaped region between two circles)

Product of Two Complex Numbers in Polar Form

Let $z_1 = r_1(cos(\theta_1) + isin(\theta_1)), z_2 = r_2(cos(\theta_2) + isin(\theta_2))$.

Then, $z_1 · z_2 = r_1(cos(\theta_1) + isin(\theta_1))·r_2(cos(\theta_2) + isin(\theta_2)) = r_1r_2[cos(\theta_1)cos(\theta_2)-sin(\theta_1)sin(\theta_2) + i(sin(\theta_1)cos(\theta_2)+cos(\theta_1)sin(\theta_2))]$ =[The sum formulas for sine and cosine are: sin(α + β) = sin α cos β + cos α sin β and cos(α + β) = cos α cos β - sin α sin β] = $r_1r_2(cos(\theta_1 + \theta_2) + isin(\theta_1+\theta_2))$

• Modulus: $|z_1||z_2| = r_1r_2$
• Argument: $arg(z_1·z_2) = \theta_1 + \theta_2 = arg(z_1)+arg(z_2)$.

In words, the modulus of the product is the product of the moduli and the argument of the product is the sum of the arguments (Figure 1).

De Moivre’s Theorem

In particular, $z = r(cos(\theta) + isin(\theta)), z² = r²(cos(2\theta)+isin(2\theta)).$ Let’s prove that $z^n = r^n(cos(n\theta)+isin(n\theta))$

Base cases: It is obviously a true statement for n = 0 ($z^0 = 1 = r^0(cos(0) + isin(0))$) and 1.

Inductive step: Assume that it holds for n = k, $z^k = r^k(cos(k\theta)+isin(k\theta)) \leadsto z^{k+1} = z^k·z = r^k(cos(k\theta)+isin(k\theta))·r(cos(\theta)+isin(\theta)) = r^{k+1}[cos(k\theta)cos(\theta)-sin(k\theta)sin(\theta) + i(cos(k\theta)sin(\theta) + sin(k\theta)cos(\theta))]$ =[The sum formulas for sine and cosine are: sin(α + β) = sin α cos β + cos α sin β and cos(α + β) = cos α cos β - sin α sin β] = $r^{k+1}(cos((k+1)\theta) + i(sin(k+1))\theta).$ ∎

Negative exponents, n = -1, $z^{-1}=\frac{1}{z} = \frac{1}{r(cos(\theta)+isin(\theta))}= \frac{1}{r(cos(\theta)+isin(\theta))}\frac{cos(\theta)-isin(\theta)}{cos(\theta)-isin(\theta)} = \frac{cos(\theta)-isin(\theta)}{r} = r^{-1}(cos(\theta)-isin(\theta)) = r^{-1}(cos(-\theta)+isin(-\theta))$

$z^{-n} = (z^n)^{-1} = \frac{1}{z^n} =[\text{By previous demonstration, the equality holds } \forall n \in \mathbb{N}] \frac{1}{r^n(cos(n\theta)+i(sin(n\theta)))} = r^{-n}\frac{1}{cos(n\theta)+i(sin(n\theta))}\frac{cos(n\theta)-i(sin(n\theta))}{cos(n\theta)-i(sin(n\theta))} = r^{-n}\big(cos(n\theta)-i(sin(n\theta))\big) = r^{-n}\big(cos(-n\theta)+i(sin(-n\theta))\big)$

De Moivre's Theorem. $z^n = r^n(cos(n\theta)+isin(n\theta))$ and the equality holds $\forall n \in \mathbb{Z}.$

Alternatively, given any arbitrary real number x and a integer n, (cos(x)+isin(x))ⁿ = cos(nx) + isin(nx). One can derive de Moivre’s formula using Euler’s formula and the exponential law for integer powers, (e^ix)ⁿ = e^inx.

nth Roots of a Complex Number

To solve $w^n = z = re^{i\theta}$, write w = $r^{\frac{1}{n}}e^{i\frac{\theta + 2\pi·k}{n}}$, k = 0, 1, ···, n - 1.

Example: Find the fourth roots of z = -3

z = -3 =[Write −3 in polar form] 3(cos(π)+isin(π)) = 3(cos(2kπ + π)+isin(2kπ + π)) = 3e^iπ.

We are interested on calculating $w=z^{\frac{1}{4}} \leadsto w = 3^{\frac{1}{4}}e^{\frac{1}{4}} =[\text{The 4th roots are:}] \sqrt[4]{3} \big(cos(\frac{2k\pi +\pi}{4})+isin(\frac{2k\pi + \pi}{4})\big),$ k = 0, 1, 2, 3.

For k = 0, 1, 2, and 3, we get all the possible values of w, that is, $w = 3^{\frac{1}{4}}(\plusmn \frac{1}{\sqrt{2}} \plusmn \frac{i}{\sqrt{2}})$.

Stereographic projection

• Embedding ℂ as the Plane z = 0 in ℝ³ We view each complex number z = x +iy ↭ (x, y) as the point (x, y, 0) in ℝ³. In Calculus I–II, this is like embedding ℝ on the x-axis; here we embed ℂ on the xy-plane.

• The Unit Sphere & the North Pole.

Consider the unit sphere in ℝ³, centered at the origin. It consists of all points (x, y, z) satisfying x² + y² + z² = 1, S = {(x, y, z): x² + y² + z² = 1}. The north pole of the sphere is the point N = (0, 0, 1). This set intersects the x-y plane (z = 0) in the unit circle: z = 0, x² + y² = 1 (S).

Defining the Stereographic Projection

The stereographic projection maps a complex point z identified with the plane z = 0 in ℝ³, (x, y, 0) to a point Z = (x₁, y₁, z₁) on the unit sphere S by drawing a straight line (a ray) from the north pole N through z until it intersects S. (Figure 2). That ray strikes S at exactly one other point Z = (x₁, y₁, z₁). The map π: ℂ→S∖{N}, z = (x, y) ⟼Z = (x₁, y₁, z₁) is the stereographic projection.

Parametrizing the Ray

The line connecting N(0, 0, 1) and z(x, y, 0) can be parameterized as: Z(t) = t(N) + (1 - t)·(x, y, 0) = t(0, 0, 1) + (1-t)(x, y, 0), t ∈ ℝ ↭ $\big((1-t)x, (1-t)y, t\big)$. To find the intersection point, we substitute the coordinates of Z(t) into the equation of the sphere: (1-t)²x² + (1-t)y² + t² = 1 ↭ (1-t)²(x²+y²) + t² = 1 ↭ (1-t)²|z|² + t² = 1.

Rearrange to a quadratic in t. ∣z∣²(1 -2t + t²) + t² - 1 = 0 ↭ t²(|z|²+1) -2t|z|² + |z|² - 1 = 0. Using the quadratic equation, we get $t = \frac{2|z|² \plusmn \sqrt{4|z|⁴-4(|z|²+1)(|z|² - 1)}}{2(|z|²+1)} = \frac{2|z|² \plusmn \sqrt{4z⁴-4z⁴+ 4}}{2(|z|²+1)} = \frac{2|z|² \plusmn 2}{2(|z|²+1)}$

We have two solutions for t.

• The “+” choice gives $\frac{2|z|² + 2}{2(|z|²+1)}$ = 1 which is N itself —discarded. Since we are projecting from the north pole, we are interested in the other solution.
• The “−” choice gives $t = \frac{|z|²-1}{|z|² +1}, 1 - t = \frac{2}{|z|² +1}$

Coordinates on the Sphere

Substitute into Z(t) = ((1−t)x, (1−t)y, t). We obtain the projection:

π(z) = (x₁, y₁, z₁) = $(\frac{2x}{|z|²+1}, \frac{2y}{|z|²+1}, \frac{|z|²-1}{|z|²+1})$ =[Equivalently, writing z = x +iy and $\bar z$ = x −iy:] $(\frac{z + \bar z}{|z|²+1}, \frac{-i(z -\bar z)}{|z|²+1}, \frac{|z²|-1}{|z|²+1})$

Inverse Projection

Given a point (x₁, y₁, z₁) on the sphere S (other than the north pole with z₁ ≠ 1), we can find the corresponding complex number z in the plane: If we are given a point ($Z \ne (0, 0, 1) \leadsto z_1 \ne 1$) on S, then z on ℂ can be found $z = \frac{x_1+iy_1}{1-z_1} \in ℂ$. Notice that $\frac{x_1 + iy_1}{1-z_1} = \frac{\frac{z + \bar z}{|z|²+1}+\frac{z -\bar z}{|z|²+1}}{1-\frac{|z²|-1}{|z|²+1}} = (\frac{z + \bar z}{|z|²+1} +\frac{z -\bar z}{|z|²+1})·\frac{1}{\frac{2}{|z|²+1}} = \frac{2z}{|z|²+1}·\frac{|z|²+1}{2} = z$. This check shows the formula is exactly the inverse of the forward stereographic projection.

The stereographic projection establishes a one-to-one correspondence between the complex plane ℂ and the unit sphere S excluding the north pole N. To complete the correspondence, we add a point at infinity (∞) to ℂ and map it to the north pole N. This extended complex plane ℂ ∪ {∞} is called the Riemann sphere.

In the extended complex plane (the Riemann sphere), we declare π⁻¹(N) = ∞. The resulting bijection ℂ ∪ {∞} ≅ S is the Riemann sphere.

The Chordal (Spherical) Metric

The chordal metric (or spherical metric) defines a distance between two points on the Riemann sphere (and therefore between two complex numbers, including ∞).

If z and z’ are two finite complex numbers, the chordal distance is the Euclidean distance between their projections on the sphere:

$d(z, z’) = \begin{cases} d(Z, Z’), &z, z’ \ne \infty\\ d(Z, ∞) = \frac{2}{\sqrt{1+|z|²}}, &z’ = \infty \end{cases}$

$d(z, z’) = \begin{cases} \sqrt{(x_1-x_1’)²+(x_2-x_2’)²+(x_3-x_3’)²}, &z, z’ \ne \infty\\ \frac{2}{\sqrt{1+|z|²}}, &z’ = \infty \end{cases}$

Notice: The point at infinity ∞ on the Riemann sphere corresponds exactly to the north pole N = (0, 0, 1) of the unit sphere S. d(z, ∞) = $\sqrt{(x_1-0)^2+(y_1-0)^2+(z_1-1)^2}$

Plugging in the coordinates:

1. $(x_1^2)+(x_2^2) = \frac{4(x^2+y^2)}{(1+|z|^2)^2} = \frac{4|z|^2}{(1+|z|^2)^2}$
2. $z_1-1 = \frac{|z²|-1}{|z|²+1}-1 = \frac{|z²|-1-(|z|²+1)}{|z|²+1} = \frac{-2}{|z|²+1}$

Therefore, $d(z, ∞)^2 = (x_1-0)^2+(y_1-0)^2+(z_1-1)^2 = \frac{4|z|^2}{(1+|z|^2)^2} + \frac{4}{(1+|z|^2)^2} = \frac{4·(1+|z|^2)}{(1+|z|^2)^2} = \frac{4}{1+|z|^2}$

Taking the square root gives the previous formula: $d(Z, ∞) = \frac{2}{\sqrt{1+|z|²}}$

This correspondence along with the distance is called the stereographic projection. We define the operations: a + ∞ = ∞, a·∞ = ∞ (a ≠ 0), a/∞ = 0, a/0 = ∞ (a ≠ 0), a ∈ ℂ