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Chapter 1: Problem 6

Let \(f(z)\) be a continuous function for all \(z\). Show that if \(f\left(z_{0}\right) \neq 0\), then there must be a neighborhood of \(z_{0}\) in which \(f(z) \neq 0\).

Short Answer

Expert verified
If \( f(z_0) eq 0 \), by continuity there is a \( \delta \)-neighborhood around \( z_0 \) where \( f(z) eq 0 \).

Step by step solution

01

Understand the Problem

The problem asks to show that if a continuous function is nonzero at a point, it remains nonzero in a neighborhood around that point.
02

Consider the Definition of Continuity

Since the function is continuous at the point \( z_0 \), for any point \( z \) close enough to \( z_0 \), the function values \( f(z) \) will be close to \( f(z_0) \).
03

Apply the Definition

Use the epsilon-delta definition of continuity, which states that for every \( \varepsilon > 0 \) there exists a \( \delta > 0 \) such that if \( |z - z_0| < \delta \), then \( |f(z) - f(z_0)| < \varepsilon \).
04

Choose an Appropriate Epsilon

Since \( f(z_0) eq 0 \), choose \( \varepsilon = \dfrac{|f(z_0)|}{2} \). This ensures that the function values are sufficiently close to \( f(z_0) \).
05

Establish the Delta Neighborhood

Find a \( \delta \) corresponding to the chosen \( \varepsilon \) from the epsilon-delta definition of continuity. This gives us the neighborhood \( |z - z_0| < \delta \) where \( |f(z) - f(z_0)| < \dfrac{|f(z_0)|}{2} \).
06

Conclusion from the Neighborhood

Within this \( \delta \)-neighborhood, we have \( |f(z) - f(z_0)| < \dfrac{|f(z_0)|}{2} \). This implies \( |f(z)| > \dfrac{|f(z_0)|}{2} \), ensuring that \( f(z) eq 0 \) within this neighborhood.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

continuous functions
In mathematics and particularly in complex analysis, a function is considered continuous if, at a given point and any nearby points, the function values converge closely to the function value at that point. For example, if \(f(z)\) is continuous at \(z_0\), then as \(z\) gets closer to \(z_0\), the value of \(f(z)\) will get closer to \(f(z_0)\). This concept is critical as it helps us understand behavior and properties of functions over specific regions. In simpler terms, a continuous function does not have any sudden jumps or breaks.
epsilon-delta definition
The epsilon-delta definition is a formal way to express the continuity of a function. It states that for any positive number \(\varepsilon\) (no matter how small), there is a positive number \(\delta\) such that whenever the distance between \(z\) and \(z_0\) is smaller than \(\delta\) (i.e., \(|z - z_0| < \delta\)), the distance between \(f(z)\) and \(f(z_0)\) will be smaller than \(\varepsilon\) (i.e., \(|f(z) - f(z_0)| < \varepsilon\)). This definition is fundamental for proving various properties of continuous functions. It gives us a precise way to confirm that small changes in the input result in small changes in the output.
neighborhoods in complex plane
In the context of complex analysis, a neighborhood of a point \(z_0\) refers to a region around \(z_0\) where every point within a certain distance \(\delta\) from \(z_0\) is included. Mathematically, this is written as \(|z - z_0| < \delta\). This concept ensures that we are considering all points sufficiently close to \(z_0\) when analyzing functions. When we discuss a function being nonzero in a neighborhood, it means that around the point \(z_0\), within a specific radius \(\delta\), the function does not take the value zero. This is essential for understanding the behavior of functions near specific points.
nonzero functions
A nonzero function at a point \(z_0\) is a function where \(f(z_0) eq 0\). To show that a function is nonzero in a neighborhood around \(z_0\), we use the properties of continuity. If \(f(z_0) eq 0\), we can select an appropriate \(\varepsilon\), such as \(\dfrac{|f(z_0)|}{2}\). Then, for the corresponding \(\delta\), the function \(f(z)\) remains nonzero, as \(|f(z) - f(z_0)| < \dfrac{|f(z_0)|}{2}\) guarantees that \(|f(z)| > \dfrac{|f(z_0)|}{2}\). Thus, within this neighborhood, \(f(z) eq 0\), demonstrating the robustness of nonzero values around \(z_0\) in continuous functions.

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Most popular questions from this chapter

Solve for the roots of the following equations: (a) \(z^{3}=4\) (b) \(z^{4}=-1\) (c) \((a z+b)^{3}=c\), where \(a, b, c>0\) (d) \(z^{4}+2 z^{2}+2=0\)

(a) Use the real Taylor series formulae $$ \begin{gathered} e^{x}=1+x+O\left(x^{2}\right), \quad \cos x=1+O\left(x^{2}\right) \\ \sin x=x\left(1+O\left(x^{2}\right)\right) \end{gathered} $$ where \(O\left(x^{2}\right)\) means we are omitting terms proportional to power \(x^{2}\) (i.e, \(\lim _{x \rightarrow 0}\left(O\left(x^{2}\right)\right) /\left(x^{2}\right)=C\), where \(C\) is a constant), to establish the following: $$ \lim _{z \rightarrow 0}\left(e^{z}-(1+z)\right)=\lim _{r \rightarrow 0}\left(e^{r \cos \theta} e^{i r \sin \theta}-(1+r(\cos \theta+i \sin \theta))\right)=0 $$ (b) Use the above Taylor expansions to show that (c.f. Eq. (1.3.18)) $$ \begin{aligned} \lim _{\Delta z \rightarrow 0}\left(\frac{e^{\Delta z}-1}{\Delta z}\right) &=\lim _{r \rightarrow 0}\left\\{\frac{\left(e^{r \cos \theta} \cos (r \sin \theta)-1\right)+i e^{r \cos \theta} \sin (r \sin \theta)}{r(\cos \theta+i \sin \theta)}\right\\} \\ &=1 \end{aligned} $$

Suppose we are given the following differentialequations: (a) \(\frac{d^{3} w}{d t^{3}}-k^{3} w=0\) (b) \(\frac{d^{6} w}{d t^{6}}-k^{6} w=0\) where \(t\) is real and \(k\) is a real constant. Find the general real solution of the above equations. Write the solution in terms of real functions.

Use Euler's formula for the exponential and the well-known series expansions of the real functions \(e^{x}, \sin y\), and \(\cos y\) to show that $$ e^{z}=\sum_{j=0}^{\infty} \frac{z^{j}}{j !} $$ Hint: Use $$ (x+i y)^{k}=\sum_{j=0}^{k} \frac{k !}{j !(k-j) !} x^{j}(i y)^{k-j} $$

Let \(z_{1}=x_{1}\) and \(z_{2}=x_{2}\), with \(x_{1}, x_{2}\) real, and the relationship $$ e^{i\left(x_{1}+x_{2}\right)}=e^{i x_{1}} e^{i x_{2}} $$ to deduce the known trigonometric formulae $$ \begin{aligned} &\sin \left(x_{1}+x_{2}\right)=\sin x_{1} \cos x_{2}+\cos x_{1} \sin x_{2} \\ &\cos \left(x_{1}+x_{2}\right)=\cos x_{1} \cos x_{2}-\sin x_{1} \sin x_{2} \end{aligned} $$ and therefore show $$ \begin{aligned} &\sin 2 x=2 \sin x \cos x \\ &\cos 2 x=\cos ^{2} x-\sin ^{2} x \end{aligned} $$
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