/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Show that the function \(f\) is ... [FREE SOLUTION] | 91Ó°ÊÓ

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Show that the function \(f\) is discontinuous at the given point.\(f(z)=\operatorname{Arg}(i z) ; z_{0}=i\)

Short Answer

Expert verified
The function \(f(z) = \operatorname{Arg}(i z)\) is discontinuous at \(z_0 = i\).

Step by step solution

01

Review the Function Definition

The function given is \(f(z) = \operatorname{Arg}(i z)\). The function \(\operatorname{Arg}\) returns the argument or angle of a complex number, which is the angle from the positive real axis to the line representing the complex number in the complex plane.
02

Evaluate the Function at the Given Point

To find \(f(i)\), substitute \(z = i\) into \(f(z)\). We have \(i \times i = -1\), so \(f(i) = \operatorname{Arg}(-1)\). The principal argument of \(-1\) is \(\pi\). Therefore, \(f(i) = \pi\).
03

Analyze Limits Around the Point

To show discontinuity, we need to look at the limits as \(z\) approaches \(i\) from different paths.* Path 1: Let \(z = i + t\), then \(i z = -t + i\). So \(f(z) = \operatorname{Arg}(-t + i)\). As \(t \to 0\), \(\operatorname{Arg}(-t + i)\) approaches \(\pi/2\).* Path 2: Let \(z = i(t + 1)\), then \(i z = -1 - it\). So \(f(z) = \operatorname{Arg}(-1 - it)\). As \(t \to 0\), \(\operatorname{Arg}(-1 - it)\) approaches \(-\pi\).
04

Determine the Continuity

Determine whether \(\lim_{z \to i} f(z)\) exists and is equal to \(f(i)\). In this case, \(\lim_{z \to i} f(z)\) does not exist because the limit depends on the path taken, as shown by the differing values of \(\pi/2\) and \(-\pi\). Since these limits are not equal and do not equal \(f(i) = \pi\), the function \(f\) is discontinuous at \(z_0 = i\).

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Key Concepts

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

Discontinuity
In complex analysis, discontinuity indicates points where a function does not behave smoothly or predictably. Functions are typically continuous when you can draw them without lifting your pencil. A discontinuous function might have a jump, hole, or asymptote, which implies different limit values when approached from various paths.
For the function \(f(z) = \operatorname{Arg}(i z)\) analyzed at \(z_0 = i\), discontinuity arises because \(f\) yields different limit values when approached from distinct directions in the complex plane. This happens because the limit of the arguments from two paths aren't the same. Such behavior inhibits the establishment of a single limit value at \(z_0\), indicating discontinuity.
To identify discontinuity:
  • Compute the function's value at the point, if possible.
  • Analyze the function as it approaches the point from various paths.
  • Look for inconsistent limits from different directions.
Understanding this can help identify how changes in a path affect the limit values, proving whether a function is discontinuous.
Complex Numbers
Complex numbers are an extension of real numbers, incorporating a real part and an imaginary part. The imaginary unit \(i\) is defined as \(\sqrt{-1}\). A complex number is expressed as \(z = x + yi\) where \(x\) is the real part and \(y\) is the imaginary component.
In the context of the given exercise, the function involves multiplying the complex number \(z\) by \(i\), switching it in the complex plane. The operation \(i z\) involves rotation by \(\frac{\pi}{2}\) (90 degrees counterclockwise). This affects how \(f(z)\) behaves.
The Arg function, which outputs the argument or angle, indicates the direction from the positive real axis to a complex number's position in the plane. For \(f(z) = \operatorname{Arg}(i z)\):
  • Calculate \(i z\) as a product, and determine its angle.
  • Evaluate how the calculations differ when approaching \(z_0\) from various paths.
Mastery of complex numbers is essential to navigate these operations as they convey the behavior of complex functions and their geometrical representations.
Limit Analysis
Limit analysis in complex analysis is crucial for understanding continuity and discontinuity of functions. This involves examining how a function behaves as \(z\) approaches a specific point, often from multiple paths.
In the exercise, we examined limit behavior as complex \(z\) approached \(z_0 = i\) along different paths:
  • For path \(z = i + t\): It moved vertically, resulting in \( \operatorname{Arg}(-t + i) \), nearing \( rac{\pi}{2}\).
  • For path \(z = i(t + 1)\): It moved horizontally, yielding \( \operatorname{Arg}(-1 - it) \), tending to \(-\pi\).
These differences showed the nonexistence of a unified limit as \(z o i\), illustrating the principle that varying paths can result in distinctive limits.
When performing limit analysis:
  • Identify potential paths approaching the target point.
  • Evaluate the function on these paths to observe limit behavior.
  • Compare the outcomes to ascertain continuity or discontinuity.
Such exploration of limits unveils how intricacies in complex functions manifest about specific points.

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Most popular questions from this chapter

Find the image of the half-plane \(\operatorname{Im}(z) \geq 0\) under each of the following principal \(n\) th root functions. (a) \(f(z)=z^{1 / 2}\) (b) \(f(z)=z^{1 / 3}\) (c) \(f(z)=z^{1 / 4}\)

Let \(f\) be a complex function. Explain the relationship between the vector field associated with \(f(z)\) and the vector field associated with \(g(z)=i f(z) .\) Illustrate with sketches using a simple function for \(f\).

Consider the complex function \(f(z)=\frac{1+\imath}{z}+2\) defined on the annulus \(1 \leq|z| \leq 2\) (a) Use mappings to determine upper and lower bounds on the modulus of \(f(z)=\frac{1+i}{z}+2 .\) That is, find real values \(L\) and \(M\) such that \(L \leq\left|\frac{1+i}{z}+2\right| \leq M\) (b) Find values of \(z\) that attain your bounds in (a). In other words, find \(z_{0}\) and \(z_{1}\) such that \(z_{0}\) and \(z_{1}\) are in the annulus \(1 \leq|z| \leq 2\) and \(\left|f\left(z_{0}\right)\right|=L\) and \(\left|f\left(z_{1}\right)\right|=M\).

Suppose that the set \(S\) is mapped onto the set \(S^{\prime}\) by the complex mapping \(w=f(z)\). If \(S=S^{\prime}\) as subsets of a single copy of the complex plane, then \(S\) is said to be invariant under \(f\). Notice that it is not necessary that \(f(z)=z\) for all \(z\) in \(S\) in order for \(S\) to be invariant under \(f\). (a) Explain why the closed disk \(|z| \leq 2\) is invariant under the rotation \(R(z)=a z,|a|=1\) (b) What are the invariant sets under a translation \(T(z)=z+b, b \neq 0 ?\) (c) What are the invariant sets under a magnification \(M(z)=a z, a>0 ?\)

Find the image of the region \(|z| \leq 8, \pi / 2 \leq \arg (z) \leq 3 \pi / 4\), under each of the following principal \(n\) th root functions. (a) \(f(z)=z^{1 / 2}\) (b) \(f(z)=z^{1 / 3}\) (c) \(f(z)=z^{1 / 4}\)

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