Question

Let \(D\) be the closed unit disc in the complex plane. (Thus \(z \in D\) if and only if \(|z| \leq 1 .)\) Let \(g\) be a continuous mapping of \(D\) into the unit circle \(T\). (Thus, \(|g(z)|=1\) for every \(z \in D .)\) Prove that \(g(z)=-z\) for at least one \(z \in T\). Hint: For \(0 \leq r \leq 1,0 \leq t \leq 2 \pi\), put $$ \gamma_{,}(t)=g\left(r e^{t}\right), $$ and put \(\psi(t)=e^{-u} \gamma_{1}(t) .\) If \(g(z) \neq-z\) for every \(z \in T\), then \(\psi(t) \neq-1\) for every \(t \in[0,2 \pi]\). Hence Ind \((\psi)=0\), by Exercises 24 and 26. It follows that Ind \(\left(\gamma_{1}\right)=1\). But Ind \(\left(\gamma_{0}\right)=0 .\) Derive a contradiction, as in Exercise \(27 .\)

Short answer

Define the function \(\gamma_r(t)\) and \(\psi(t)\) as mentioned in the solution. Explain why assuming there is no such \(z\) implies \(\psi(t)\neq -1\), and use the winding number to find a contradiction. Finally, conclude that \(g(z) = -z\) for some \(z \in T\).

Step-by-step solution

Define \(\gamma_r\) and \(\psi\) functions

Define the function \(\gamma_r(t) = g(re^{it})\) and \(\psi(t) = e^{-it}\gamma_1(t)\).

Show that assuming there is no such \(z\) implies \(\psi(t)\neq -1\)

Now, assume that there exists no \(z \in T\) such that \(g(z) = -z\). This means that for every \(z \in T\), \(g(z) \neq -z\). Equivalently, \(\gamma_1(t) \neq -e^{it}\) for every \(t \in [0, 2\pi]\). Thus, we have that \(\psi(t) = e^{-it}\gamma_1(t) \neq -1\) for every \(t \in [0, 2\pi]\).

Use winding number to find a contradiction

Using the properties of the winding number, we can conclude that Ind\((\psi) = 0\). It follows that \(\text{Ind}(\gamma_1) = 1\). On the other hand, consider the case where \(r = 0\). In this case, \(\gamma_0(t) = g(0)\) for all \(t\). Since \(g(0)\) is a constant function independent of \(t\), \(\text{Ind}(\gamma_0) = 0\). Now, consider a continuous family of curves that connects \(\gamma_0(t)\) to \(\gamma_1(t)\). As we gradually move from \(r = 0\) to \(r = 1\), the winding number (which is always an integer) cannot change discontinuously. Therefore, the winding number of \(\gamma_1\) must be equal to the winding number of \(\gamma_0\). However, we have found that \(\text{Ind}(\gamma_1) = 1\) and \(\text{Ind}(\gamma_0) = 0\). This is a contradiction, as we cannot move continuously from \(\gamma_0\) to \(\gamma_1\) and change the winding number at the same time.

Conclude that \(g(z) = -z\) for some \(z \in T\)

The contradiction implies that our original assumption (that there exists no \(z \in T\) such that \(g(z) = -z\)) must be false. Therefore, we can conclude that \(g(z) = -z\) for at least one \(z \in T\).

Key concepts

Winding Number

The winding number is a fundamental concept in mathematical analysis, particularly within the realm of complex analysis. It is a way to determine how many times a continuous curve winds around a point in the complex plane without crossing itself. Imagine it as counting how many complete circles the curve makes around a certain point.

For a continuous curve \[\begin{equation}\text{Ind}(\text{winding number}) = \frac{1}{2\text{π}i} \text{integral of }\frac{df}{f}=\text{number of times } f(\text{curve}) \text{ encloses a point.}\text{\end{equation}\]}In our exercise, by assuming that \[\begin{equation}\text{Ind}(\psi)=0\text{\end{equation}\]}it implies that the curve \[\begin{equation}\text{Ind}(\gamma_1(ti))\text{\end{equation}\]}does not wind around the point -1 even once, when it is expected to do so once, thus creating a contradiction and leading us to improve our understanding of this concept.

Continuous Mapping

Continuous mapping is another cornerstone of mathematical analysis, central to both real and complex analysis. A function is continuous if small changes to the input result in small changes to the output, thereby ensuring no sudden jumps or breaks in the curve it defines. This is like drawing a line without lifting the pencil from the paper.

In the context of our exercise, a continuous mapping \[\begin{equation}g: D \rightarrow T\text{\end{equation}\]}is given, ensuring that the mapping from the closed unit disc \[\begin{equation}D\text{\end{equation}\]}to the unit circle\[\begin{equation}T\text{\end{equation}\]}is smooth without any ruptures or gaps. One important aspect of continuous mappings in complex analysis is that they naturally preserve the winding number during deformation, as expected when we gradually move from \[\begin{equation}\gamma_0\text{\end{equation}\]}to \[\begin{equation}\gamma_1\text{\end{equation}\]}along a continuous path.

Complex Plane

The complex plane is a two-dimensional plane used to visualize and manipulate complex numbers. It is akin to a Cartesian coordinate system but with the horizontal axis representing the real part of complex numbers and the vertical axis representing the imaginary part. Each point in this plane corresponds to one unique complex number.

For the problem at hand, the exercise revolves around the behavior of a function within the closed unit disc \[\begin{equation}D\text{\end{equation}\]}on the complex plane, where every point \[\begin{equation}z\text{\end{equation}\]}within \[\begin{equation}D\text{\end{equation}\]}satisfies \[\begin{equation}\vert z \vert \leq 1\text{\end{equation}\]}. The unit circle \[\begin{equation}T\text{\end{equation}\]}is another important concept in the complex plane representing all points with a distance of one from the origin. This circle is crucial in understanding the behavior of the function \[\begin{equation}g\text{\end{equation}\]}and its winding number around the unit circle.