Problem 7

Question

$f(z)=\tan z$ is discontinuous at $z=\pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}, \ldots$ but is analytic within and on the circle $|z|=1$

Step-by-Step Solution

Verified

Answer

$\tan z$ is analytic within and on $|z|=1$ because no discontinuities are on or inside the unit circle.

1Step 1: Understanding the Problem

We are given the function $f(z) = \tan z$ and need to analyze its continuity and analyticity within and on the circle $|z|=1$. We know $\tan z$ is discontinuous at points where $\cos z = 0$, specifically at $z = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \ldots$. Our task is to determine its behavior within and on $|z|=1$.

2Step 2: Locate Discontinuous Points

The function $\tan z$ is defined as $\frac{\sin z}{\cos z}$. Discontinuities occur at points where $\cos z = 0$, leading to vertical asymptotes. These points are $z = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \ldots$. We need to find which of these points, if any, lie on $|z|=1$.

3Step 3: Evaluate Points on the Circle

The circle $|z|=1$ is defined as all points $z = e^{i\theta}$ for $\theta$ in $[0,2\pi)$. We check these points: none of them coincide with $\pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \ldots$ since these points do not map to points where $|z|=1$ directly using simple transformations.

4Step 4: Conclusion of Analyticity

Since none of the discontinuities of $\tan z$ are on $|z|=1$ and it is analytic wherever it is continuous, $\tan z$ is analytic within and on the circle $|z|=1$.

Key Concepts

Analytic FunctionsDiscontinuities in Complex FunctionsUnit Circle in Complex Plane

Analytic Functions

An analytic function, in complex analysis, is a function that is expressible as a power series around any point within its domain. This characteristic allows analytic functions to be infinitely differentiable. Furthermore, such functions are known for maintaining a perfect balance between their real and imaginary parts through unique derivatives at every point within their radius of convergence.

Analytic functions have the distinctive property of being locally represented by a convergent power series.
When a complex function is analytic, its behavior is smooth and predictable.
They are often considered in contexts where continuity and differentiability are vital.

In the case of our function, $f(z) = \tan z$, even though it has discontinuities at specific points, it is analytic at all other points within its domain, including and inside the unit circle $|z|=1$, as long as none of those discontinuous points fall within or on the boundary of the domain.

Discontinuities in Complex Functions

Discontinuities in complex functions often occur where the function is undefined or tends towards infinity. They are crucial in determining where a function may not be analytic. For the given function, $ \tan z = \frac{\sin z}{\cos z} $, discontinuities arise wherever $\cos z = 0$ because this would make the denominator zero, resulting in undefined values.

Common sources of discontinuities are poles, where functions go to infinity.
Functions can be continuous in specific regions while having discontinuities elsewhere, which means analyticity can be selectively determined.
Analyzing these interruptions helps in identifying regions of functionality.

In $f(z) = \tan z$, the discontinuities occur at points such as $z = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \ldots$, which are vertical asymptotes. Understanding these breaks is vital to establishing where the function retains analytic characteristics.

Unit Circle in Complex Plane

The unit circle in the complex plane, defined by $|z|=1$, is a circle with radius 1 centered at the origin of the complex plane. It represents a critical boundary in many areas of mathematics and physics and is particularly significant in complex analysis for studying function behaviors.

The parametrization of any point on the unit circle is given by $z = e^{i\theta}$ for $\theta$ ranging from 0 to $2\pi$.
Within the context of complex functions, the unit circle serves as an essential check-point to determine continuity and analyticity.
It allows us to map and explore behavior patterns of complex functions within and around the circle.

For the function $f(z) = \tan z$, it is crucial to note its continuities and discontinuities relative to the unit circle. Since none of the discontinuous points $\pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \ldots$ lie exactly on $|z|=1$, $\tan z$ remains analytic on and inside this circle.

Problem 7

Other exercises in this chapter

Problem 7

(a) By Theorem 18.9 with $f(z)=\frac{z^{2}}{z+2 i}$ \\[ \oint_{C} \frac{\frac{z^{2}}{z+2 i}}{\frac{z+2 i}{z-2 i}} d z=2 \pi i\left(-\frac{4}{4 i}\right)=-2 \p

View solution

Problem 7

$$\int_{1-i}^{1+i} z^{3} d z=\left.\frac{1}{4} z^{4}\right|_{1-i} ^{1+i}=0$$

View solution

Problem 7

Using $z=e^{i t}=\cos t+i \sin t, d z=(-\sin t+i \cos t) d t$ and $x=\cos t$, \(\begin{aligned} \oint_{C} \operatorname{Re}(z) d z &=\int_{0}^{2 \pi} \cos t

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Problem 8

$f(z)=\frac{z^{2}-9}{\cosh z}$ is discontinuous at $\frac{\pi}{2} i, \pm \frac{3 \pi}{2} i, \ldots$ but is analytic within and on the circle $|z|=1$

View solution