# Exercises

## Star-Shaped Sets

### Question

Prove that every open star-shaped subset of $$\mathbb{C}$$ is simply connected.

### Answer

Let $$\Omega$$ be an open star-shaped subset of $$\mathbb{C}$$ with a center $$c$$.

For any $$z \in \mathbb{C} \setminus \Omega$$ and any $$s\geq 0$$, the point $$w = z + s (z - c)$$ belongs to $$\mathbb{C} \setminus \Omega.$$ The ray of all such points $$w$$ is unbounded and connected, thus it is included in an unbounded component of $$\mathbb{C} \setminus \Omega$$. All components of $$\mathbb{C} \setminus \Omega$$ are therefore unbounded: $$\Omega$$ is simply connected.

Alternatively, let $$\gamma$$ be a closed path of $$\Omega$$ and let $$z = c + r e^{i\alpha} \in \mathbb{C} \setminus \Omega$$. Since the ray $$\{z + s e^{i\alpha} \; | \; s \geq 0\}$$ does not intersect $$\Omega$$, for any $$t \in [0,1]$$ and any $$s\geq 0$$, $$\gamma(t) - z \neq s e^{i\alpha}$$. Thus $$e^{-i(\pi + \alpha)}(\gamma(t) - z) \in \mathbb{C} \setminus \mathbb{R}_-$$ and the function $\phi: t \in [0,1] \mapsto e^{i(\pi+\alpha)}\arg (e^{-i(\pi+\alpha)}(\gamma(t) - z))$ is defined; since it is a continuous choice of the argument $$w \mapsto \mathrm{Arg} (w - z)$$ along $$\gamma$$, $\mathrm{ind}(\gamma, z) = \frac{1}{2\pi}[\phi(1) - \phi(0)] = 0.$ Therefore, $$\Omega$$ is simply connected.

## The Argument Principle for Polynomials

### Questions

Let $$p$$ be the polynomial $p(z) = \lambda \times (z - a_1)^{n_1} \times \dots \times (z-a_m)^{n_m}$ where $$\lambda$$ is a nonzero complex number, $$a_1, \dots, a_m$$ are distinct complex numbers (the zeros or roots of the polynomial) and $$n_1, \dots, n_{m}$$ are positive natural numbers (the roots orders or multiplicities). Let $$\gamma$$ be a closed path whose image contains no root of $$p$$: $\forall \, t \in [0,1], \; p(\gamma(t)) \neq 0.$ The argument principle then states that $\mathrm{ind}(p \circ \gamma, 0) = \sum_{k=1}^m \mathrm{ind}(\gamma, a_{k}) \times n_{k}.$
1. Application: Finding the Roots of a Polynomial.

Use the figures below to determine – according to the argument principle – the number of roots $$z$$ of the polynomial $$p(z) = z^3 + z + 1$$ in the open unit disk centered on the origin. Graph of $$t \in [0,1] \mapsto \mathrm{arg} \left[ (e^{i2\pi t})^3 + (e^{i2\pi t}) + 1 \right]$$; this function has a jump of $$-2 \pi$$ at $$t=0.5$$ (where it is undefined). The dashed line represents a continuous choice of the argument of $$t \in [0,1] \mapsto (e^{i2\pi t})^3 + (e^{i2\pi t}) + 1$$. Graph of $$t \in [0,1] \mapsto |(e^{i2\pi t})^3 + (e^{i2\pi t}) + 1|$$.
2. Argument Principle Proof (Elementary). For any $$k \in \{1, \dots, m\}$$, we denote $$\theta_{k}$$ a continous choice of $$z \mapsto \mathrm{Arg}(z-a_{k})$$ on $$\gamma$$. Use the functions $$\theta_{k}$$ to build a continuous choice of $$z \mapsto \mathrm{Arg} \, z$$ on $$p \circ \gamma$$; then, prove the argument principle.

3. Argument Principle Proof (Complex Analysis). Assume that $$\gamma$$ is rectifiable; write the winding number $$\mathrm{ind}(p \circ \gamma, 0)$$ as a line integral, then find another way to prove the argument principle in this context.

### Answers

1. Let $$\gamma: t \in [0,1] \mapsto e^{i2\pi t}$$; we have $$(p \circ \gamma)(t) = (e^{i2\pi t})^3 + (e^{i2\pi t}) + 1$$. The second figure shows that the graph of $$t \mapsto |(p\circ \gamma)(t)|$$ does not vanish on $$[0,1]$$, hence the image of $$\gamma$$ contains no root of $$p$$. The second figure shows that the variation of the argument of $$z$$ on the path $$p \circ \gamma$$ is $$2\pi$$ (a variation of $$\pi$$ between $$t=0$$ and $$t=0.5$$ and also a variation of $$\pi$$ between $$t=0.5$$ and $$t=1.0$$). Accordingly, we have $\mathrm{ind}(p \circ \gamma, 0) = 1.$

On the other hand, every zero $$z$$ of $$p$$ such that $$|z|<1$$ satisfies $$\mathrm{ind}(\gamma, z) = 1$$ and every zero $$z$$ of $$p$$ such that $$|z|>1$$ satisfies $$\mathrm{ind}(\gamma, z) = 0$$. Consequently, the expression $\sum_{k=1}^m \mathrm{ind}(\gamma, a_{k}) \times n_{k}$ provides the number of roots of $$p$$ – counted with their multiplicity – within the unit circle. By the argument principle, there is a unique root of $$p$$ within the unit circle.

2. If $$\theta_0$$ is an argument of $$\lambda$$, the sum $\theta: t \in [0,1] \mapsto \theta_0 + n_1 \theta_1(t) \times \dots + n_m \theta_{m}(t)$ is continuous and $\begin{split} e^{i\theta(t)} = &\, e^{i\theta_0} \times e^{i n_1\theta_1(t)} \times \dots \times e^{i n_m \theta_m(t)} \\ = &\, \frac{\lambda}{|\lambda|} \times \frac{(\gamma(t) - a_1)^{n_1}}{|\gamma(t) - a_1|^{n_1}} \times \dots \times \frac{(\gamma(t) - a_m)^{n_m}}{|\gamma(t) - a_m|^{n_m}} \\ = &\, \frac{(p \circ \gamma)(t)}{|(p \circ \gamma)(t)|}, \end{split}$ therefore $$\theta$$ is a choice of the argument of $$z \mapsto z$$ on $$p \circ \gamma$$. Consequently, $\begin{split} [z \mapsto \mathrm{Arg} \, z]_{p \circ \gamma} &= \, \theta(1) - \theta(0) \\ &= \, \theta_0 - \theta_0 + \sum_{k=1}^m n_{k}(\theta_{k}(1) - \theta_{k}(0)) \\ &= \, \sum_{k=1}^{m} n_{k} \times [z \mapsto \mathrm{Arg} (z-a_{k})]_{\gamma}. \end{split}$ A division of both sides of this equation by $$2\pi$$ concludes the proof.

3. The integral expression of the winding number is $\mathrm{ind}(p \circ \gamma, 0) = \frac{1}{i2\pi} \int_{p \circ \gamma} \frac{dz}{z}.$ The polynomial $$p$$ is holomorphic on $$\mathbb{C}$$, hence we can perform the change of variable $$z = p(w)$$, which yields $\mathrm{ind}(p \circ \gamma, 0) = \frac{1}{i2\pi} \int_{\gamma} \frac{p'(w)}{p(w)} dw.$ If we factor $$p(w)$$ as $$(w-a_{k})^{n_{k}} q(w)$$, we see that $\frac{p'(w)}{p(w)} = \frac{n_{k}}{w-a_{k}} + \frac{q'(w)}{q(w)};$ applying this process repeatedly for every $$k \in \{1,\dots,m\}$$, until $$q$$ is a constant, provides $\frac{p'(w)}{p(w)} = \sum_{k=1}^m \frac{n_{k}}{w-a_{k}}$ and consequently $\begin{split} \mathrm{ind}(p \circ \gamma, 0) = &\, \frac{1}{i2\pi}\int_{\gamma} \left[ \sum_{k=1}^m \frac{n_{k}}{w-a_{k}}\right] dw \\ = &\, \sum_{k=1}^m \left[\frac{1}{i2\pi}\int_{\gamma} \frac{dw}{w-a_{k}}\right] \times n_{k} \\ = &\, \sum_{k=1}^m \mathrm{ind}(\gamma, a_{k}) \times n_{k}. \end{split}$

## Set Operations & Simply Connected Sets

### Questions

Suppose that $$A$$, $$B$$ and $$\mathbb{C} \setminus C$$ are open subsets of $$\mathbb{C}.$$ For each of the three statements below,
• determine whether or not the statement is true (either prove it or provide a counter-example);

• if the statement is false, find a sensible assumption that makes the new statement true (and provide a proof).

The statements are:

1. Intersection. The intersection $$A \cap B$$ of two simply connected sets $$A$$ and $$B$$ is simply connected.

2. Complement. The relative complement $$A \setminus C$$ of a connected set $$C$$ in a simply connected set $$A$$ is simply connected.

3. Union. The union $$A \cup B$$ of two connected and simply connected sets $$A$$ and $$B$$ is simply connected.

### Answers

1. Intersection. The statement holds true. Indeed, let $$\gamma$$ be a closed path of $$A \cap B$$; it is a path of $$A$$ and a path of $$B$$. As both sets are simply connected, the interior of $$\gamma$$ is included in $$A$$ and in $$B$$, that is in $$A \cap B$$: this intersection is simply connected.

Alternatively, let $$C$$ be a component of $\mathbb{C} \setminus (A \cap B) = (\mathbb{C} \setminus A) \cup (\mathbb{C} \setminus B),$ and let $$z \in C;$$ we have $$z \in \mathbb{C} \setminus A$$ or $$z \in \mathbb{C} \setminus B$$. If $$z \in \mathbb{C} \setminus A$$, the component of $$\mathbb{C} \setminus A$$ that contains $$z$$ is unbounded; it is a connected set that contains $$z$$ and is included in $$\mathbb{C} \setminus (A \cap B)$$, hence, it is also included in $$C$$. Consequently, $$C$$ is unbounded. If instead $$z \in \mathbb{C} \setminus B$$, a similar argument provides the same result. Consequently, all components of $$\mathbb{C} \setminus (A \cap B)$$ are unbounded: $$A \cap B$$ is simply connected.

2. Complement. The statement does not hold: consider $$A = D(0,3)$$ and $$C = \overline{D(0,1)}$$. The set $$A$$ is open and simply connected and the set $$C$$ is closed and connected. The set $$C$$ is actually a component of $$A \setminus C$$: it is included in $$A \setminus C$$, connected and maximal.

However, the statement holds if additionally the set $$C \setminus A$$ is not empty. Let $$\gamma$$ be a closed path of $$A \setminus C$$ and let $$z \in \mathbb{C} \setminus (A \setminus C)$$. If $$z \in \mathbb{C} \setminus A$$, as $$A$$ is simply connected, $$z$$ belongs to the exterior of $$\gamma$$. Otherwise, $$z \in A \cap C$$; as $$C$$ is a connected subset that does not intersect the image of $$\gamma$$, the function $$w \in C \mapsto \mathrm{ind}(\gamma, w)$$ is constant. There is a $$w \in C \setminus A$$ and $$\mathrm{ind} (\gamma, z)=\mathrm{ind} (\gamma, w) = 0$$. Therefore $$z$$ also belongs to the exterior of $$\gamma$$: $$A \setminus C$$ is simply connected.

Alternatively, let $$D$$ be a component of $\mathbb{C} \setminus (A \setminus C) = (\mathbb{C} \setminus A) \cup C.$ Some of its elements are in $$\mathbb{C} \setminus A$$: otherwise, $$C$$ would be a connected superset of $$D$$ that is included in $$\mathbb{C} \setminus (A \setminus C)$$; we would have $$C=D$$ and therefore $$C \setminus A$$ would be empty. Now, as $$D$$ contains at least a point $$z$$ of $$\mathbb{C} \setminus A$$, it contains the component of $$\mathbb{C} \setminus A$$ that contains $$z$$; therefore $$D$$ is unbounded. Consequently, $$A \setminus C$$ is simply connected.

3. Union. The statement doesn’t hold: consider $A_s = \{e^{i 2 \pi t} \; | \; t \in [0, 1/2]\}, \; B_s = \{e^{i 2 \pi t} \; | \; t \in [1/2, 1]\}.$ and the associated dilations $A = \{z \in \mathbb{C} \; | \; d(z, A_s) < 1 \}, \; B = \{z\in \mathbb{C} \; | \; d(z, B_s) < 1 \}.$ They are both open, connected and simply connected (their complement in the plane has a single path-connected component and it is unbounded) but their union $$A \cup B$$ is the annulus $$D(0,3) \setminus D(0,1)$$. We already considered this set in question 2: it is not simply connected.

However, the statement holds if additionally, the intersection $$A \cap B$$ is connected. Let $$\gamma$$ be a closed path of $$A \cup B$$ and let $$z \in \mathbb{C} \setminus (A \cap B)$$. We have to prove that $$\mathrm{ind}(\gamma, z) = 0$$.

There exist1 a sequence $$(\gamma_1, \dots, \gamma_n)$$ of consecutive paths of $$A \cup B$$ whose concatenation is $$\gamma$$ and such that for any $$k \in \{1,\dots, n\}$$, $$\gamma_{k}([0,1]) \subset A$$ or $$\gamma_{k}([0,1]) \subset B$$.

Let $$a_{k}$$ be the initial point of $$\gamma_{k}$$ and let $$w \in A \cap B$$. As $$A$$, $$B$$ and $$A \cap B$$ are connected, for any $$k \in \{1,\dots, n\}$$, there is a path $$\beta_{k}$$ from $$w$$ to $$a_{k}$$ such that $$\beta_{k}([0,1]) \subset A$$ if $$a_{k} \in A$$ and $$\beta_{k} ([0,1]) \subset B$$ if $$a_{k} \in B$$. We denote $$\beta_{n+1}=\beta_1$$ for convenience; define the paths $$\alpha_{k}$$ as the concatenations $\alpha_{k} = \beta_{k} \, | \, \gamma_{k} \, | \, \beta_{k+1}.$ By construction $[x \mapsto \mathrm{Arg} (x-z)]_{\gamma} = \sum_{k=1}^{n} [x \mapsto \mathrm{Arg} (x-z)]_{\alpha_{k}}.$ Every path $$\alpha_{k}$$ is closed, hence this is equivalent to $\mathrm{ind}(\gamma,z) = \sum_{k=1}^{n} \mathrm{ind}(\alpha_{k}, z),$ but every $$\alpha_{k}$$ belongs either to $$A$$ or $$B$$, which are simply connected, hence the right-hand-side is equal to zero. (This proof was adapted from Ronnie Brown’s argument on Math Stack Exchange)