Geometric Curves, Sets, and Topology in the Complex Plane

As long as algebra is taught in school, there will be prayer in school, Cokie Roberts

Representing geometric curves in the complex plane

Complex numbers provide elegant, coordinate-free descriptions of classical geometric curves. The key insight: a complex number z= x + iy represents a point (x, y) in $\mathbb{R}^2$, and complex operations encode geometric transformations.

• Circles: The circle with center $a \in \mathbb{C}$ and radius r > 0: {$z \in \mathbb{C}$: |z -a| = r}.
  Alternative forms: $z = a + re^{i\theta}, \theta \in [0, 2\pi)$ (parametric), $(x - \mathbb{Re}(a))^2 + (y - \mathbb{Im}(a))^2 = r^2$ (Cartesian).

• Lines. Equation of the real line in $\mathbb{C}$ is {$z \in \mathbb{C}$: Im(z) = 0} = $\mathbb{R}$.
  This is the horizontal axis in the complex plane. General Line: Parametric Form.
  General line through a point $\alpha \in \mathbb{C}$ parallel to direction $\beta \in \mathbb{C} \setminus \{ 0 \}$: $z = \alpha +t\beta, t \in \mathbb{R}$
  Here, z is the variable complex number describing points on the line, $\alpha$ is a fixed complex number (point through which the line passes), $\beta$ is a fixed complex “direction” vector, and t is a real parameter moving along the line.
  As t varies over $\mathbb{R}$, z traces the infinite line through $\alpha$ in the direction $\beta$.
  From the parametric form: $\frac{z -\alpha}{\beta} = t \in \mathbb{R}$. Therefore, the equivalent form is :$Im(\frac{z -\alpha}{\beta}) = 0.$

  A complex number is real if and only if it equals its own conjugate, $w \in \mathbb{R} \Leftrightarrow w = \bar{w}$

  So: $\frac{z - \alpha}{\beta} \in \mathbb{R} \Leftrightarrow \frac{z - \alpha}{\beta} = \overline{(\frac{z - \alpha}{\beta})}$
  Recall: $\overline{(\frac{A}{B})} = \frac{\overline{(A)}}{(\overline{B})}, \overline{(z - \alpha)} = \bar{z} - \bar{\alpha}$
  $\frac{z - \alpha}{\beta} = \overline{(\frac{z - \alpha}{\beta})} ↭[\text{Cross-multiply:}] \bar{\beta}(z - \alpha) = \beta(\bar{z} - \bar{\alpha}) ↭ \bar{\beta} z - \beta \bar{z} = \bar{\beta} \alpha - \beta \bar{\alpha}$
  The right-hand side has the form $w − \bar{w}$, where w = $\bar{\beta}\alpha$. We know that for any complex number w, this identity holds $w − \bar{w} = 2iIm(w)$. Therefore,
  $\boxed{\bar{\beta} z - \beta \bar{z} = 2i\mathrm{Im}(\bar{\beta} \alpha)}$.
  When $\beta = 1$ (horizontal direction), this reduces to $\mathrm{Im}(z) = \mathrm{Im}(\alpha)$, a horizontal line.

• Ellipses. For an ellipse E in $\mathbb{C}$ whose foci are at $a, b \in \mathbb{C}$ and major axis length $c \gt |a - b|$. The sum of the distances to the foci is constant (equidistance property of the ellipse) and E may be written as: E = {$z \in \mathbb{C}$ : |z - a| + |z - b| = c}.

• Hyperbolas. For two foci $a, b \in \mathbb{C}$ and a constant 0 < c < ∣a−b∣, the hyperbola is {z : ∣∣z−a∣ − ∣z−b∣∣ = c}. The difference of distances to the foci is constant.

Describe the following sets in ℂ

• S = {z: |z| = Re(z) + 1}

Let $z \in \mathbb{C}$, and write z = x + iy, where $x, y \in \mathbb{R}$.

$z \in S$ ↭[So the condition becomes:] $\sqrt{x^2+y^2} = x + 1$

The left side is always non‑negative ($|z| = \sqrt{x^2+y^2} = x + 1$), so we must have x + 1 ≥ 0. We’ll keep this in mind.

Squaring both sides gives $x^2+y^2 = x^2 + 2x + 1 \leadsto[\text{Cancel } x^2] y^2 = 2x + 1$.

This is the equation of a parabola in the xy-plane, opening to the right (since x is expressed in terms of $y^2$), with vertex at $(-\tfrac{1}{2}, 0)$ ($y = 0 \implies x = -\tfrac{1}{2}$). The set S consists of all complex numbers whose coordinates lie on this parabola.

• S = {$z \in \mathbb{C}: z^{n-1} = \bar{z}$}, $n \in \mathbb{Z}$, n ≥ 2

Write $z=re^{i\theta}, r = |z|$. The conjugate is $\bar{z} = re^{-i\theta}$. The equation becomes, $r^{n-1}e^{i(n-1)\theta} = re^{-i\theta}$

If r = 0 (i.e., z = 0), both sides are 0, so z = 0 is a trivial solution.

Assume now r > 0 and divide both sides by r: $r^{n-2}e^{i(n-1)\theta} = e^{-i\theta}$. Next, multiply by $e^{i\theta}$ and we get: $r^{n-2}e^{in\theta} = 1$. This is a complex number equal to 1, so its modulus must be 1 and its argument must be an integer multiple of $2\pi$: $r^{n-2}=1, n\theta = 2\pi k, k \in \mathbb{Z}$.

1. n = 2, then $r^{0}=1$ holds for every r > 0. Besides, $2\theta = 2\pi k \iff \theta = \pi k$. Thus, $z = re^{i\pi k} = r\cdot (-1)^k = \pm r$, i.e. z is any non‑zero real number (positive when k even, negative when k odd). Together with z = 0, we obtain all real numbers. Thus, for n = 2, $S = \mathbb{R}$.
2. n > 2. $r^{n-2} = 1 \implies[r \gt 0] r = 1, \theta = \frac{2\pi k}{n}$.
   Distinct values of $e^{i\theta}$ occur for $k = 0, 1, \cdots, n -1$ (because $e^{i2\pi k/n}$ repeats every n).
   Thus, the non‑zero solutions are the n-th roots of unity: $z = e^{i\frac{2k\pi}{n}}, 0 \le k \le n -1$. These are the n-th roots of unity, evenly spaced around the unit circle. Therefore, S is the union of the origin and the vertices of a regular polygon in the complex plane.

Exercises

• Let $\xi = e^{\frac{2\pi·i}{n}}, n \ge 2, n \in \mathbb{Z}$. Show that $\prod_{k=0}^{n-1} (1-\xi^kz) = 1 - z^n$

Solution.

1. Defining the Roots of Unity. The equation $w^n - 1 = 0$ is the defining equation for the n-th roots of unity. The solutions are the complex numbers $w$ which, when raised to the n-th power, equal 1.
   The principal n-th root of unity is: $\xi = e^{\frac{2\pi i}{n}} = \cos\left(\frac{2\pi}{n}\right) + i\sin\left(\frac{2\pi}{n}\right)$.
   The full set of $n$ solutions is given by taking integer powers of this principal root: $w_k = \xi^k$ for $k = 0, 1, 2, \dots, n - 1$.
   These roots form the vertices of a regular n-sided polygon inscribed in the unit circle on the complex plane.
2. The Symmetry of the Roots. The multiplicative inverse $\frac{1}{\xi^k}$ (or $\xi^{-k}$) of any root of unity is also a root of unity.
   There is an elegant geometric reason for this: for any complex number on the unit circle ($|w| = 1$), its inverse is exactly equal to its complex conjugate ($\frac{1}{w} = \bar{w}$).
   Because the roots of unity are symmetrically distributed across the real axis, the conjugate of any root is simply another root in the set.
   Therefore, the set of inverse roots $\{\xi^{-0}, \xi^{-1}, \dots, \xi^{-(n-1)}\}$ is exactly the same as the original set of roots $\{\xi^0, \xi^1, \dots, \xi^{n-1}\}$, just listed in a different order.
3. Applying the Factor Theorem. The Factor Theorem states that if $r$ is a root of a polynomial $P(w)$, then $(w - r)$ is a factor.
   Because $\xi^{-k}$ represents all the roots of $w^n - 1 = 0$, we can factor the polynomial completely, so we get:
   $$w^n - 1 = \prod_{k=0}^{n-1} (w - \xi^{-k})$$
   Now, we replace the dummy variable $w$ with our target variable $z$:
   $$z^n - 1 = \prod_{k=0}^{n-1} (z - \xi^{-k})$$
4. Algebraic Manipulation.
   Let’s manipulate the product to reach our target identity.
   $$z^n - 1 = \prod_{k=0}^{n-1} \frac{\xi^k z - 1}{\xi^k} = \left( \prod_{k=0}^{n-1} \xi^{-k} \right) \left( \prod_{k=0}^{n-1} (\xi^k z - 1) \right)$$
   Let’s evaluate the two parts separately:
   The product of the denominators. $\prod_{k=0}^{n-1} \frac{1}{ξᵏ} = \prod_{k=0}^{n-1} e^{\frac{-2k\pi i}{n}} = e^{\frac{-2\pi i}{n}(0 + 1 + \cdots + n-1)} = e^{\frac{-2\pi i}{n}\frac{n(n-1)}{2}} = e^{-\pi i (n-1)}$.
   The sum of the exponents forms an arithmetic progression, resulting in $e^{-\pi i (n-1)}$. By Euler’s formula, this simplifies perfectly to $(-1)^{n-1}$.
   The product of the numerators: We want terms that look like $(1 - \xi^k z)$, not $(\xi^k z - 1)$. We can factor out a $-1$ from every single one of the $n$ factors. This pulls out exactly $n$ negative signs, giving us $(-1)^n \prod_{k=0}^{n-1} (1 - \xi^k z)$.
   Now, substitute both parts back into the equation and we get:
   $$z^n - 1 = (-1)^{n-1} \cdot (-1)^n \prod_{k=0}^{n-1} (1 - \xi^k z)$$
   Multiply the constants: $(-1)^{n-1} \cdot (-1)^n = (-1)^{2n-1} = (-1)$.
   We just need to multiply both sides by $-1$, and we arrive at the final identity:
   $$\boxed{1 - z^n = \prod_{k=0}^{n-1} (1 - \xi^k z)} \blacksquare$$

• Let S be a finite set of $\mathbb{C}$ (or $\mathbb{R}^n$). (i) Then, S is open if and only if $S = \varnothing.$ (ii) S is closed.

Solution.

(i) Let S be a finite set.

If S = $\varnothing$, it is trivially open: by definition, every point of S (there are none) satisfies the condition for openness.

Suppose $S \neq \varnothing$. Since S is finite, we can consider the set of pairwise distances between distinct points of S: D = $\{ |z_i - z_j|: z_i, z_j \in S, i \ne j \}$.

This set D is finite (there are ${n \choose 2}$ pairs, so D has at most ${n \choose 2}$ elements) and contains only positive real numbers. Define $m = \min D \gt 0$.

Theorem. Every non‑empty finite subset of a totally ordered set ($\mathbb{R}$ is a totally ordered set) has both a minimum and a maximum element. D is bounded below by 0 (and none of those distances are actually 0, because we only compare distinct points). Being finite, it must have a smallest element — that’s the minimum distance or separation m > 0 - no two distinct points of S are closer than m.

Show no point is interior. Take any point $z_k \in S$ and choose any ε such that 0 < ε < m.

The open ball $B(z_k, \varepsilon)$ contains no other points of S except $z_k$ itself, because every other point is at least m units away from $z_k$. Therefore: $B(z_k, \varepsilon) \cap S = \{z_k \}$ (only the center point).

Recall. A subset $S \subset \mathbb{C}$ is open if every point of S is an interior point. In other words, an open set contains all their interior points, i.e., every point has a neighborhood entirely within the set. However, $B(z_k, \varepsilon)$ contains infinitely many points (in $\mathbb{C}$ or $\mathbb{R}^n$).

Besides, for any $\varepsilon >0$: $B(z_k, \varepsilon)$ has infinitely many points. Since S is finite, there must be points in $B(z_k, \varepsilon)$ that are not in S. Therefore, $B(z_k, \varepsilon) \nsubseteq S$ for every $\varepsilon > 0$.

(ii) S is closed.

Let $y \in \mathbb{C} \setminus S$. Consider the finite set of distances $D_y = \{d(y,s): s \in S \}$.

Since S is finite and $y \notin S$, every element of D_y is positive, and D_y has a minimum $m=\min D_y \gt 0$. This is the distance from y to the nearest point of S.

For $\varepsilon=m/2$, the open ball $B(y,\varepsilon)$ satisfies $B(y,\varepsilon)\cap S=\varnothing$. Hence every $y \in \mathbb{C} \setminus S$ is an interior point of $ℂ\setminus S$. Since $\mathbb{C} \setminus S$ is open, S is closed.

• Prove that $\lim_{z \to 0} \frac{\bar{z}}{z}$ does not exist.

Let z = x + iy, $\frac{\bar{z}}{z} = \frac{x -iy}{x + iy} = \frac{x -iy}{x + iy}\frac{x -iy}{x - iy} = \frac{x^2-y^2-2ixy}{x^2 + y^2} =[\text{Separating real and imaginary parts:}] \frac{x^2-y^2}{x^2 + y^2} - \frac{2ixy}{x^2 + y^2}$

1. Path 1. If we approach $z \to 0$ along y = x, $\frac{\bar{z}}{z} = 0 -\frac{2ix^2}{2x^2} = -i$
2. Path 2. If we approach $z \to 0$ along y = 2x, $\frac{\bar{z}}{z} = \frac{x^2-4x^2}{5x^2} -\frac{2i2x^2}{5x^2} = \frac{-3}{5}-\frac{4i}{5}$. Since different paths yield different limits, the limit at 0 does not exist. The limit depends on the path of approach.

• Let f be a complex function $f: \mathbb{C} \to \mathbb{C}$ defined by $f(z) = \frac{z}{1 + |z|}$. Show that f is continuous, one-to-one and onto on B(0; 1).

Part 1. Continuity

|z| is the modulus of z, defined as |z| = $\sqrt{x^2 + y^2}$. The modulus is continuous because squaring, addition, and square root are all continuous on $\mathbb{R}$.

Denominator is Never Zero (For all $z \in \mathbb{C}, |z| \geq 0$, so 1 + |z| > 0). This means the denominator of f(z) is never zero and also continuous (it is the sum of the constant function 1 and the continuous function |z|), so f(z) is defined everywhere on $\mathbb{C}$.

The quotient of two continuous functions is continuous as long as the denominator is non-zero, which we’ve already confirmed.

Part 2. $f(z) = \frac{z}{1 + |z|}$ is onto the open unit ball.

We need to show $f(\mathbb{C}) \subseteq B(0;1)$ — the image of f lies inside the open unit ball; and $B(0;1) \subseteq f(\mathbb{C})$ — every point in the ball is hit by some input $z \in \mathbb{C}$.

2.a. The image of f lies inside the open unit ball. Take any z $\in \mathbb{C}$. We want to show: $\left| \frac{z}{1 + |z|} \right| < 1$. Recall $B(0;1) = \{ w \in \mathbb{C} : |w| < 1 \}$

Let’s compute the modulus: $\left| \frac{z}{1 + |z|} \right| = \frac{|z|}{1 + |z|}$

Since $|z| \geq 0$, we have: $\left| \frac{z}{1 + |z|} \right| = \frac{|z|}{1 + |z|} < 1$ Why? Because: $|z| < 1 + |z| \Rightarrow \frac{|z|}{1 + |z|} < 1$. Thus, $|f(z)| \lt 1, \forall z \in \mathbb{C} \implies f(\mathbb{C}) \subseteq B(0; 1)$.

2.b. f is onto B(0; 1). Let’s show that for every $w \in B(0;1)$, there exists a $z \in \mathbb{C}$ such that $f(z) = \frac{z}{1 + |z|} = w$

Case 1. If w = 0, choose z = 0, f(0) = 0.

Case 2. If $w \ne 0$, write w in polar form: $w = re^{i\theta}, 0 \lt r \lt 1, \theta \in arg(w)$

Look for z of the same argument $z = R e^{i\theta}$, where R > 0 is to be determined.

We want $f(z) = \frac{R e^{i\theta}}{1 + R} = r e^{i\theta} ↭ r = \frac{R}{1 + R} ↭ r + rR = R ↭ R(r -1) + r = 0$ ↭[Since 0 < r < 1, we can solve:] $R = \frac{r}{1-r}$. Hence, $\boxed{z = \frac{r}{1-r}e^{i\theta}}$ satisfies f(z) = w. Therefore, every point of the open unit ball is attained, so f is onto B(0;1).

Part 3. $f(z) = \frac{z}{1 + |z|}$ is one-to-one.

Let’s assume f(z₁) = f(z₂) for two complex numbers z₁ and z₂.

$\frac{z_1}{1 + |z_1|} = \frac{z_2}{1 + |z_2|}$

First, we take the modulus (absolute value) of both sides of the equation.

$|\frac{z_1}{1 + |z_1|}| = |\frac{z_2}{1 + |z_2|}| ↭ \frac{|z_1|}{1 + |z_1|} = \frac{|z_1|}{1 + |z_2|}$

Let’s consider the real-valued function g(x) = $\frac{x}{1+x}$ for x ≥ 0. Its derivative is g′(x) = $\frac{1}{(1+x)^2}$, which is always positive. This means g(x) is a strictly increasing function. A strictly increasing function is always one-to-one.

Since g(∣z₁∣) = g(∣z₂∣), it must be true that: ∣z₁∣ = ∣z₂∣

Now we can substitute this result back into our original equation. Letting ∣z₁∣ = ∣z₂∣ = r: $\frac{z_1}{1 + r} = \frac{z_2}{1 + r}$

Since r ≥ 0, the denominator (1 + r) is a non-zero real number, so we can multiply both sides by it (1 + r) to obtain $z_1 = z_2 \blacksquare$. Thus, f is injective.

• Let $G \subseteq \mathbb{C}$ be a region (non-empty, open, and connected set in the complex plane). Let $f: G \to \tilde{G}$ be continuous and surjective (onto) map, where $\tilde{G}$ is an open subset of $\mathbb{C}$. Then, $\tilde{G}$ is a region (open and connected).

Solution:

Method 1: Path-Connectedness (Topological).

We must prove $\tilde{G}$ is connected (it is already open by hypothesis).

A key property we will use is that for open sets in $\mathbb{C}$ (or any Euclidean space $\mathbb{R}^n$), connectedness is equivalent to path-connectedness. Therefore, we can prove $\tilde{G}$ is connected by showing it is path-connected.

Let a and b any two arbitrary distinct points in $\tilde{G}$.

Because the function f is surjective, every point in $\tilde{G}$ has at least one pre-image in G. Therefore, there must exist points $x_a, x_b \in G: f(x_a) = a, f(x_b) = b$.

We are given that G is a region, so it is connected. As an open set in $\mathbb{C}$, it must also be path-connected. This means we can find a continuous path, let’s call it γ, that lies entirely within G and connects $x_a$ and $x_b$. We can represent this path as a continuous function: $\gamma : [0, 1] \to G$ such that $\gamma(0) = x_a$ and $\gamma(1) = x_b$.

Consider the composition, $f \circ \gamma : [0, 1] \to \tilde{G}$. $f \circ \gamma$ is continuous and satisfies $(f \circ \gamma)(0) = a$ and $(f \circ \gamma)(1) = b$. Furthermore, its image lies in f(G) = $\tilde{G}$.

Conclusion: Every pair of points in $\tilde{G}$ is joined by a continuous path. Thus, $\tilde{G}$ is path-connected, hence connected.

Method 2: Direct Proof via Connectedness (Set-Theoretic)

Suppose for the sake of contradiction that $\tilde{G}$ is disconnected, $\tilde{G} = U \cup V$ with non-empty, disjoint, open sets U, V in the subspace topology of $\tilde{G} = f(G)$.

1. $f^{-1}(U) \subseteq G$ is open in G (f is continuous, U is open in $\tilde{G}$).
2. $f^{-1}(V) \subseteq G$ is open in G (same reasoning).
3. $f^{-1}(U)\cap f^{-1}(V)=f^{-1}(U\cap V)=f^{-1}(\emptyset )=\emptyset$, they are disjoint since $U \cap V = \emptyset$.
4. $G=f^{-1}(\tilde {G})=f^{-1}(U\cup V)=f^{-1}(U)\cup f^{-1}(V)$ since f is surjective and every point maps to U or V.
5. Because U and V are non-empty and f is surjective onto $\tilde{G}$, there exist points in G mapping into each: $\exists z_1\in G$ with $f(z_1) \in U$, so $f^{-1}(U)\neq \emptyset$, $\exists z_2\in G$ with $f(z_2)\in V$, so $f^{-1}(V)\neq \emptyset$.

This separates G into two disjoint non-empty open sets, contradicting connectedness. □