Problem 10

Question

Show: (a) There is no continuous function $f: \mathbb{C}^{\bullet} \rightarrow \mathbb{C}^{\bullet}$ such that $$ (f(z))^{2}=z \text { for all } z \in \mathbb{C}^{\bullet} $$ (b) There is no continuous function $q: \mathbb{C} \rightarrow \mathbb{C}$ such that $$ (q(z))^{2}=z \text { for all } z \in \mathbb{C} $$

Step-by-Step Solution

Verified

Answer

No continuous function exists because of branch cuts and discontinuities at the origin.

1Step 1: Understanding the Problem

We need to prove two statements. The first part (a) claims that no continuous function $ f $ can be defined from the non-zero complex numbers ($ \mathbb{C}^{\bullet} $) to itself such that $(f(z))^2 = z $ for all $ z \in \mathbb{C}^{\bullet} $. The second part (b) extends this claim to all complex numbers, $ \mathbb{C} $.

2Step 2: Analyzing Part (a)

Consider the function $ f(z) = \sqrt{z} $. For complex numbers, $ \sqrt{z} $ can have two values (the principal square root and its negative), making it impossible to define a single continuous function $ f $ that works consistently over all $ \mathbb{C}^{\bullet} $. This non-single-valued behavior is due to the discontinuity when $ z $ crosses the negative real axis (a branch cut). Hence, such a continuous function cannot exist.

3Step 3: Analyzing Part (b)

To extend the argument to $ \mathbb{C} $, note that if we were to define a continuous function $ q $ over all $ \mathbb{C} $, including zero, it would need to be continuous at zeros. However, a non-zero complex square root function changes its behavior discontinuously as it includes the origin. The attempt to pass from negative to positive across zero will fail to keep continuity, hence no such continuous function $ q $ exists.

4Step 4: Conclusion

The analysis reveals that the branch cut issue (from step 2) and the behavior at the origin (from step 3) demonstrate the impossibility of defining the required continuous functions for both parts (a) and (b).

Key Concepts

Continuous FunctionNon-zero Complex NumbersSquare Root Function

Continuous Function

A continuous function is a function where small changes in the input result in small changes in the output. In simple terms, you can draw its graph without lifting your pencil from the paper. Continuity is a fundamental concept in both real and complex analysis. It ensures that extreme jumps or breaks don't occur in the values of the function.
For a function to be continuous over a set of points, it needs to have no breaks, gaps, or abrupt changes within that set.

The value at each point matches the "approaching" values from surrounding points.
Visually, this would mean the graph of the function can be sketched without any jumps.
In complex functions, this continuous "flow" of values becomes crucial when dealing with complex spaces like the set of all non-zero complex numbers, denoted as $ \mathbb{C}^{\bullet} $.

This idea is key in part (a) of the exercise where defining a continuous square root function over $ \mathbb{C}^{\bullet} $ becomes problematic, owing to discontinuity issues that occur with alternating root values.

Non-zero Complex Numbers

Non-zero complex numbers are simply complex numbers that are not zero. A complex number has the form $ a + bi $, where $ a $ and $ b $ are real numbers, and $ i $ is the imaginary unit with the property that $ i^2 = -1 $. When we talk about non-zero complex numbers, we exclude the number 0 (which would be $ 0 + 0i $).
The set $ \mathbb{C}^{\bullet} $ represents all these non-zero complex numbers.

Non-zero complex numbers exclude the origin in the complex plane.
This set becomes important when dealing with continuous square root functions as introducing zero could potentially disrupt continuity.

For part (a) of the exercise, the exclusion of zero plays a crucial role, as this allows us to specifically concentrate on the issues deriving solely from the nature of the square root function in complex numbers, particularly the multivalued nature and branch cuts associated with it.

Square Root Function

The square root function is familiar to many from real numbers, where it essentially asks, "What number squared gives me this result?" In complex analysis, however, things get more interesting. A complex number can have multiple square roots, typically two, such as $ \pm \sqrt{z} $.
This multiplicity arises because of the way angles wrap around the complex plane, leading to the need for branch cuts.

Standard practice is to define a "principal branch," which restricts the angle (or argument) so that the value chosen is continuous over an appropriate domain.
However, when you cross certain lines, such as the negative real axis (a common place for a branch cut), the output can "jump" to another value, representing discontinuity.
The non-single-valued nature of $ \sqrt{z} $ over $ \mathbb{C}^{\bullet} $ highlights why it can’t be expressed by a single continuous function.

This is central to both parts of the exercise, as creating a continuous function that encompasses all square roots of a complex number is rendered impossible due to the inherent issues of multivaluedness and branch disjointing in the complex plane.

Problem 10

Other exercises in this chapter

Problem 10

Let $P$ be a polynomial with complex coefficients: $$ P(z):=a_{n} z^{n}+a_{n-1} z^{n-1}+\cdots+a_{0} \text { with } n \in \mathbb{N}_{0}, a_{\nu} \in \mathbb{

View solution

Problem 10

Let $\left(a_{n}\right)_{n \geq 0}$ and $\left(b_{n}\right)_{n \geq 0}$ be two sequences of complex numbers such that $a_{n}=$ $b_{n}-b_{n+1}, n \geq 0$

View solution

Problem 10

Let $D \subseteq \mathbb{C}$ be open, and $D^{\prime} \subseteq \mathbb{C}$ another open subset. Let $\varphi: D \rightarrow D^{\prime}$ be analytic and e

View solution

Problem 11

(a) Let $\mathbb{H}:=\\{z \in \mathbb{C} ; \quad \operatorname{Im} z>0\\}$ be the upper half-plane. Show: \(z \in \mathbb{H} \Longleftrightarrow-1 / z \in \ma

View solution