Problems and Solutions of Complex Calculus

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

Representing geometric curves in the complex plane

• Circles: {z: |z -a| = r} where $a \in \mathbb{C}$ and r > 0 is the radius.
• Lines.

Equation of the real line in ℂ is {z: Im(z) = 0}.

General line through α parallel to direction β ≠ 0: z = α + βt, t ∈ ℝ. Here, z is the variable complex number describing points on the line, α is a fixed complex number (point through which the line passes), β is a fixed complex “direction” vector, and t is a real parameter moving along the line

$\frac{z -\alpha}{\beta} = t \in \mathbb{R}$. Therefore, the equivalent form is :$Im(\frac{z -\alpha}{\beta}) = 0.$

Alternative representation using complex conjugates:$\\{z: \bar{\beta}z - \beta\bar{z} = 2i\mathrm{Im}(\bar{\beta} \alpha)}$ for line through α with direction β.

$w \in \mathbb{R} \ \Leftrightarrow\ w = \bar{w}$

So: $\frac{z - \alpha}{\beta} \in \mathbb{R} ↭ \frac{z - \alpha}{\beta} = \overline{(\frac{z - \alpha}{\beta})}$

Recall: $\overline{(\frac{A}{B})} = \frac{\overline{(A)}}{(\bar{B})}, \overline{(z - \alpha)} = \bar{z} - \bar{\alpha}$

$\frac{z - \alpha}{\beta} = \overline{(\frac{z - \alpha}{\beta})} ↭ \bar{\beta}(z - \alpha) = \beta(\bar{z} - \bar{\alpha}) ↭ \bar{\beta} z - \beta \bar{z} = \bar{\beta} \alpha - \beta \bar{\alpha}$

Note: The right-hand side is of the form the right-hand side is of the form $w − \bar{w}$, where w = $\bar{\beta}\alpha$. We know that for any complex number w, the identity is $w − \bar{w} = 2iIm(w)$. Therefore:

Therefore: $\boxed{\bar{\beta} z - \beta \bar{z} = 2i\mathrm{Im}(\bar{\beta} \alpha)}$.

• Ellipses. For an ellipse E in ℂ whose mayor axis is c ∈ ℝ and whose foci are at a, b ∈ ℂ. The sum of the distances to both fixed foci is constant (equidistance property of the ellipse) and E may be written as: {z : |z - a| + |z - b| = c} where c > ∣a − b∣ (ensuring the ellipse exists)

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

Describe the following sets in ℂ

• S = {z: |z| = Re(z) + 1}

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

$z \in S ↭[\text{ So the condition becomes: }] \sqrt{x^2+y^2} = x + 1 ↭[\text{ Squaring both sides: }] x^2+y^2 = x^2 + 2x + 1 \leadsto y^2 = 2x + 1$. This is the equation of a parabola in the xy-plane, opening to the right, with vertex at $(-\tfrac{1}{2}, 0)$. Since every complex number z = x + iy corresponds to a point (x, y), the set S consists of all complex numbers whose coordinates lie on this parabola.

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

z = 0 is a trivial solution, it satisfies the equation since both sides are zero.

$z \in S ↭[\text{ So the condition becomes: }] z^n = z\bar{z} ↭ z^n = |z|^2 ↭[\text{ Take moduli: }] |z|^n = |z|^2 ↭[\text{ Assuming } z \ne 0] z^{n-2} = 1 ↭ |z| = 1 ↭[\text{ So nonzero solutions lie on the unit circle in the complex plane }] z = e^{i\theta}$

Therefore, $z^{n-1} = \bar{z} \leadsto e^{i(\theta(n-1))} = e^{-i\theta} \leadsto \theta(n-1) - (-\theta) = 2k\pi, k \in \mathbb{Z}$.

$n\theta = 2k\pi \leadsto \theta = \frac{2k\pi}{n}$. The nonzero solutions are: $z = e^{i\frac{2k\pi}{n}}, 0 \le k \le n -1$. 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.

Exercise

• 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.

The equation zⁿ - 1 = 0 is the defining equation for the n-th roots of unity. The solutions are the complex numbers z which, when raised to the n-th power, equal 1.

The principal n-th root of unity, denoted by ξ, is the root with the smallest positive argument (angle). It’s defined as: $\xi = e^{i\frac{2\pi}{n}}= cos(\frac{2\pi}{n}+isin(\frac{2\pi}{n}))$. The full set of n solutions is given by taking integer powers of this principal root: zᵏ = ξᵏ for k = 0, 1, 2, …, n − 1.

The n roots of unity are the vertices of a regular n-sided polygon inscribed in the unit circle on the complex plane.

Furthermore, the multiplicative inverse $\frac{1}{\xi^k}$ of any root of unity is also a root of unity: $(\frac{1}{\xi^k})^n - 1 = (\frac{1}{\xi^n})^k - 1 = (\frac{1}{1})^k - 1 = 0.$ There’s a deeper and more elegant reason for this. For any complex number z on the unit circle (meaning ∣z∣ = 1), its inverse is equal to its complex conjugate: $\frac{1}{z} = \bar{z}$. Since all roots of unity ξᵏ lie on the unit circle, we have: $\frac{1}{ξᵏ} = \bar{ξᵏ}$, meaning that the complex conjugate of any root of unity is also a root of unity.

The Factor Theorem states that if r is a root of a polynomial P(z) (meaning P(r) = 0, e.g., r = $\frac{1}{\xi^k}$, P(z) = zⁿ - 1), then the linear polynomial (z − r, e.g., $z - \frac{1}{\xi^k}$) is a factor of P(z).

Therefore, z - $\frac{1}{ξᵏ}$ is a factor of zⁿ - 1, k = 0, 1, 2, …, n − 1. Since this is true for all the roots, we can write the complete factorization of the polynomial as the product of all its linear factors: zⁿ - 1 = $\prod_{k=0}^{n-1} (z - \frac{1}{ξᵏ})$

$\prod_{k=0}^{n-1} (z - \frac{1}{ξᵏ}) = \prod_{k=0}^{n-1} \frac{(ξᵏz - 1)}{ξᵏ}$

Notice that $\prod_{k=0}^{n-1} \frac{1}{ξᵏ} = \prod_{k=0}^{n-1} e^{\frac{-2k\pi i}{n}} = e^{\frac{-2\pi i}{n}(0 + 1 + \cdots + n-1)} = e^{\frac{-2\pi i}{n}\frac{n(n-1)}{2}} = e^{-\pi i (n-1)}$

Case 1. If n is odd, $\prod_{k=0}^{n-1} \frac{1}{ξᵏ} = e^{-\pi i (n-1)} =[-\pi i · \text{ an even number}] = 1$

zⁿ - 1 = $\prod_{k=0}^{n-1} (z - \frac{1}{ξᵏ}) = \prod_{k=0}^{n-1} \frac{(ξᵏz - 1)}{ξᵏ} = \prod_{k=0}^{n-1} (ξᵏz - 1) \leadsto [\text{ since n-1 is even (number of factors) because n is odd}] 1 - zⁿ = \prod_{k=0}^{n-1} (1-\xi^kz)$

Case 2. If n is even, $\prod_{k=0}^{n-1} \frac{1}{ξᵏ} = e^{-\pi i (n-1)} =[-\pi i · \text{ an odd number}] = -1$

zⁿ - 1 = $\prod_{k=0}^{n-1} (z - \frac{1}{ξᵏ}) = \prod_{k=0}^{n-1} \frac{(ξᵏz - 1)}{ξᵏ} = (-1)\prod_{k=0}^{n-1} (ξᵏz - 1) \leadsto [\text{ since n-1 is odd (number of factors) because n is even}] 1 - zⁿ = \prod_{k=0}^{n-1} (1-\xi^kz)$

• 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 ∈ S, i ≠ j}.

This set D is finite and contains only positive real numbers. Let $m = \min D \gt 0$.

Theorem. Every non‑empty finite subset of 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 m > 0.

Take any point $z_k \in S$ and choose ε such that 0 < ε < m. The open ball B(z_k, ε) contains no other points of S except z_k itself, because every other point is at least m units away from z_k. Therefore: $B(z_k, \varepsilon) \cap S = \\{z_k\\}$.

Recall. A subset S ⊆ ℂ is open if every point of S is an interior point. Formally, ∀z₀ ∈ S, ∃R > 0 : (such that) D_R(z₀) ⊆ S -the ball D_R(z₀) sits entirely inside S. 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 $\mathbb{R}^n$).

(ii) S is closed.

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

Since S is finite and y ∉ S, every element of D_y is positive, and D_y has a minimum $m=\min D_y \gt 0$.

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

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

Let z = x + iy, $\frac{\bar{z}}{z} = \frac{x -iy}{x + iy} = \frac{x -iy}{x + iy}\frac{x -iy}{x - iy} = \frac{x^2-y^2-2ixy}{x^2 + y^2} = \frac{x^2-y^2}{x^2 + y^2} - \frac{2ixy}{x^2 + y^2}$

If we approach z → 0 along y = x, $\frac{\bar{z}}{z} = 0 -\frac{2ix^2}{2x^2} = -i$

If we approach z → 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.

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

Continuity:

1. |z| is the modulus of z, defined as |z| = $\sqrt{x^2 + y^2}$ if z = x + iy. The modulus is continuous because square, addition, and square root are all continuous on ℝ.
2. 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, so f(z) is defined everywhere on ℂ.
3. The quotient of two continuous functions (The numerator z is continuous, the denominator 1 + |z| is continuous) is continuous as long as the denominator is non-zero, which we’ve already confirmed.

$f(z) = \frac{z}{1 + |z|}$ is onto the open unit ball $B(0;1) \subset \mathbb{C}$. We need to show both:

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

Let 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: $|z| < 1 + |z| \Rightarrow \frac{|z|}{1 + |z|} < 1$.

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

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

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

Let’s suppose $z = r_1 e^{i\theta}$, where $r_1$ > 0 is to be determined.

We want $f(z) = \frac{r_1 e^{i\theta}}{1 + r_1} = r e^{i\theta} = w ↭ r = \frac{r_1}{1+r_1} ↭ r + rr_1 = r_1 ↭ r_1(r -1) + r = 0 ↭ r_1 = \frac{r}{1-r}$. Hence, z = $\frac{r}{1-r}e^{i\theta}$, f(z) = w

$f(z) = \frac{z}{1 + |z|}$ 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: z₁ = z₂∎

• Suppose that G is a non-empty region in ℂ (open and connected set in the complex plane) and that f : G → $\tilde{G}$ is a continuous and surjective (onto) map, where $\tilde{G}$ is an open set. Probe that $\tilde{G}$ is a region.

Solution:

Definition. Let D ⊆ ℂ be a subset of the set of complex numbers. D is an open region of ℂ if and only if D is an open set and connected.

A key property we will use is that for open sets in ℂ (or any Euclidean space ℝⁿ), 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 ∈ G s.t. 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 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: γ : [0, 1] → G such that γ(0) = x_a and γ(1) = x_b.

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

Since G is connected and f is continuous, f(G) is connected. Thus, every pair a, b ∈ $\tilde{G}$ is joined by a continuous path in $\tilde{G}$. Hence $\tilde{G}$ is path-connected, and therefore connected. Since $\tilde{G}$ is also open (given), it is a region.

Note: Since G is connected and f is continuous, f(G) is connected. Assume f(G) is disconnected, f(G) = U ∪ V with non-empty, disjoint, open sets U, V in the subspace topology of f(G).

Then, $G = f^{-1}(U)\cup f^{-1}(V) \text{ ,with } f^{-1}(U), f^{-1}(V)$ open (by continuity), disjoint, and non-empty (since U, V are non-empty and f is surjective onto f(G)). This separates G, contradicting connectedness. □

• Derive the Cauchy Riemann equations in the polar form.

Solution:

Let f be a complex function, $f(z)=u(x,y)+iv(x,y), z=x+iy=r·e^{i\theta}, x=r\cos(\theta), y=r\sin(\theta)$. f is complex differentiable in a neighborhood with r>0 (care is needed at r=0).

f(re^iθ) = u(r, θ) + iv(r, θ).

Using the Chain Rule in Partial derivatives, $\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}·\frac{\partial x}{\partial r} + \frac{\partial u}{\partial y}·\frac{\partial y}{\partial r} \leadsto [x = rcos(\theta), y = rsin(\theta)] u_r = u_x·cos(\theta) + u_y·sin(\theta)$

Similarly, $u_\theta = u_x·x_\theta + u_y·y_\theta = [x = rcos(\theta), y = rsin(\theta)] u_x·r(-sin(\theta)) + u_y·rcos(\theta) = -r(u_xsin(\theta) -u_ycos(\theta))$

$v_r = v_x·x_r + v_y·y_r = v_x·cos(\theta) + v_y·sin(\theta)$

$v_\theta = v_x·x_\theta + v_y·y_\theta = v_x·r(-sin(\theta)) + v_y·rcos(\theta)$

Use the Cartesian Cauchy–Riemann equations, $u_x = v_y, u_y = -v_x$

Substitute into the chain-rule formulas:

$v_r = v_x·cos(\theta) + v_y·sin(\theta) = -u_y·cos(\theta) + u_x·sin(\theta) =[u_\theta = -r(u_xsin(\theta) -u_ycos(\theta))] \frac{-1}{r}u_\theta$

$v_\theta = v_x·r(-sin(\theta)) + v_y·rcos(\theta) = -u_y·r(-sin(\theta)) + u_x·rcos(\theta) = r(u_ysin(\theta)+u_xcos(\theta)) =[u_r = u_x·cos(\theta) + u_y·sin(\theta)] ru_r$

Finally, the Polar Cauchy–Riemann equations are: $\boxed{-rv_r = u_\theta, v_\theta = ru_r}$