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Chapter 6: Problem 2

Show whether or not the function \(f(z)=\Re(z)=x\) is analytic.

Short Answer

Expert verified
The function is not analytic.

Step by step solution

01

Identify Cauchy-Riemann Equations

For a function to be analytic, it must satisfy the Cauchy-Riemann equations. Let's identify the real and imaginary parts of the function. Given the function is \( f(z) = \Re(z) = x \), we can express it as \( f(z) = u(x,y) + iv(x,y) = x + i(0) \), where \( u(x,y) = x \) and \( v(x,y) = 0 \). The Cauchy-Riemann equations are: \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \) and \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \).

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Key Concepts

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

Cauchy-Riemann Equations
The Cauchy-Riemann equations are foundational in complex analysis. They are used to determine if a function is analytic. When we talk about a complex function, it is typically expressed in terms of its real and imaginary components:
  • Let the complex function be expressed as \( f(z) = u(x, y) + iv(x, y) \), where \( z = x + iy \).
  • \( u(x, y) \) represents the real part, while \( v(x, y) \) is the imaginary part of the function.
To be analytic, a function must satisfy these conditions:
  • The partial derivative of \( u \) with respect to \( x \) must equal the partial derivative of \( v \) with respect to \( y \).
  • The partial derivative of \( u \) with respect to \( y \) must equal the negative of the partial derivative of \( v \) with respect to \( x \).
These are expressed as:\[\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\]\[\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\]Both conditions need to be satisfied simultaneously for the function to be considered analytic.
Analytic Functions
A function is called analytic (or holomorphic) at a point if it satisfies the Cauchy-Riemann equations at that point, and the partial derivatives involved are continuous. This means:
  • The function must be differentiable not just at a single point, but in some neighborhood around it.
  • If a function is analytic everywhere in a domain, it is termed an analytic function in that domain.
Analytic functions have powerful properties. They are infinitely differentiable and can be represented as power series within their domain of analyticity. This gives mathematicians and engineers a significant tool for analysis and synthesis.
For example, if our function \( f(z) = x \) were analytic, it would need to satisfy the Cauchy-Riemann equations at every point in its domain. However, the provided solution shows this is not the case for \( f(z) = x \), indicating it is not analytic.
Real and Imaginary Parts
When dealing with complex functions, separating them into real and imaginary parts helps us understand their behavior better:
  • The real part, \( u(x, y) \), often denotes scenarios we can relate to actual measurements or phenomena.
  • The imaginary part, \( v(x, y) \), though harder to visualize, is essential for completeness in describing the behavior of the function.
For the function in the exercise, \( f(z) = x \), we have:
  • Real part: \( u(x, y) = x \)
  • Imaginary part: \( v(x, y) = 0 \)
This decomposition helps in checking the Cauchy-Riemann equations. For instance, in our case:
  • \( \frac{\partial u}{\partial x} = 1 \)
  • \( \frac{\partial v}{\partial y} = 0 \)
  • \( \frac{\partial u}{\partial y} = 0 \)
  • \( \frac{\partial v}{\partial x} = 0 \)
These results show that \( f(z) = x \) does not fulfill the Cauchy-Riemann equations, illustrating why the function is not analytic.

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Most popular questions from this chapter

Show that the function $$ w(z)=\left(z^{2}-1\right)^{1 / 2} $$ is single-valued if we take \(-1 \leq x \leq 1, y=0\) as a cut line.

A function \(f(z)\) is analytic on and within the unit circle. Also, \(|f(z)|<1\) for \(|z| \leq 1\) and \(f(0)=0 .\) Show that \(|f(z)|<|z|\) for \(|z| \leq 1 .\) Hint. One approach is to show that \(f(z) / z\) is analytic and then express \(\left[f\left(z_{0}\right) / z_{0}\right]^{n}\) by the Cauchy integral formula. Finally, consider absolute magnitudes and take the \(n\) th root. This exercise is sometimes called Schwarz's theorem.

Using the identities $$ \cos z=\frac{e^{i z}+e^{-i z}}{2}, \quad \sin z=\frac{e^{i z}-e^{-i z}}{2 i} $$ established from comparison of power series, show that(a) \(\sin (x+i y)=\sin x \cosh y+i \cos x \sinh y\), \(\cos (x+i y)=\cos x \cosh y-i \sin x \sinh y\) (b) \(|\sin z|^{2}=\sin ^{2} x+\sinh ^{2} y, \quad|\cos z|^{2}=\cos ^{2} x+\sinh ^{2} y .\) This demonstrates that we may have \(|\sin z|,|\cos z|>\mid\) in the complex plane.

The functions \(u(x, y)\) and \(v(x, y)\) are the real and imaginary parts, respectively, of an analytic function \(w(z)\). (a) Assuming that the required derivatives exist, show that $$ \nabla^{2} u=\nabla^{2} v=0 . $$ Solutions of Laplace's equation such as \(u(x, y)\) and \(v(x, y)\) are called harmonic functions. (b) Show that $$ \frac{\partial u}{\partial x} \frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} \frac{\partial v}{\partial y}=0 $$ and give a geometric interpretation. Hint. The technique of Section \(1.6\) allows you to construct vectors normal to the curve \(u(x, y)=c_{i}\) and \(v(x, y)=c_{j}\).

A function \(f(z)\) can be expanded in a Laurent series about the origin with the coefficients \(a_{n}\) real. Show that the complex conjugate of this function of \(z\) is the same function of the complex conjugate of \(z\); that is, $$ f^{*}(z)=f\left(z^{*}\right) $$ Verify this explicitly for (a) \(f(z)=z^{n}, n\) an integer; (b) \(f(z)=\sin z\) If \(f(z)=i z,\left(a_{1}=i\right)\), show that the foregoing statement does not hold.
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