/*! 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 1 The given function is analytic f... [FREE SOLUTION] | 91Ó°ÊÓ

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The given function is analytic for all \(z\). Show that the CauchyRiemann equations are satisfied at every point. \(f(z)=z^{3}\)

Short Answer

Expert verified
The Cauchy-Riemann equations are satisfied everywhere for \( f(z) = z^3 \).

Step by step solution

01

Write the function in terms of real and imaginary components

Express the complex function in terms of its real and imaginary components. Given the function \( f(z) = z^3 \), where \( z = x + yi \), calculate \( f(z) = (x + yi)^3 \). This expands to \( (x + yi)(x + yi)(x + yi) \).
02

Expand the function

Expand the expression \( (x + yi)^3 \) using algebraic expansion.\[(x + yi)^3 = (x+yi)(x+yi)(x+yi)\]First, compute \( (x+yi)^2 = x^2 + 2xyi - y^2 \). Then multiply \( (x^2 - y^2 + 2xyi) \) by \( (x+yi) \).
03

Simplify the expanded terms

Perform the multiplication of \( (x^2 - y^2 + 2xyi) \) by \( (x+yi) \), we get:\[(x^2 - y^2)x + 2xy^2i + 2x^2yi - (y^2yi)\]\[= x^3 - 3xy^2 + (3x^2y - y^3)i\]
04

Identify the real and imaginary parts

Identify the real part \( u(x,y) = x^3 - 3xy^2 \) and the imaginary part \( v(x,y) = 3x^2y - y^3 \) of the function \( f(x,y) \).
05

Compute the partial derivatives

Find the partial derivatives of \( u \) and \( v \):\( \frac{\partial u}{\partial x} = 3x^2 - 3y^2, \frac{\partial u}{\partial y} = -6xy \)\( \frac{\partial v}{\partial x} = 6xy, \frac{\partial v}{\partial y} = 3x^2 - 3y^2 \)
06

Verify Cauchy-Riemann equations

The Cauchy-Riemann equations state:\( \ rac{\partial u}{\partial x} = \ rac{\partial v}{\partial y} \) and \( \ rac{\partial u}{\partial y} = - \ rac{\partial v}{\partial x} \).Check:- \( 3x^2 - 3y^2 = 3x^2 - 3y^2 \) (True)- \( -6xy = -6xy \) (True)Since both equations are satisfied, the function is analytic.

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Key Concepts

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

Analytic Functions
An analytic function is a fundamental concept in complex analysis. These functions are usually defined on a complex domain. They can be represented as a power series around their points of interest. Simply put, a function is analytic if it is differentiable at every point in its domain. To check if a function is analytic, we often need to verify the Cauchy-Riemann equations. However, one key takeaway is that not every differentiable function in real analysis is analytic in complex analysis. This uniqueness arises because complex differentiability is stricter. For any function to be analytic, it must maintain its differentiability in complex terms.
Analytic functions are crucial because they preserve angles and the structure of the domain. They extend real-valued calculus into the complex plane. Students can benefit from knowing about their roles in mapping properties that are conserved. With this understanding, you can see that analytic functions provide the basis for much of the beauty in complex analysis.
Complex Analysis
Complex analysis is the study of functions that operate on complex numbers. This area of mathematics extends the principles of calculus to the complex plane. Unlike purely real-valued functions, complex functions that are analytic showcase fascinating behaviors. For instance, they are always differentiable across their domain, thanks to the stringent conditions laid by the Cauchy-Riemann equations. This differentiability leads to unique and rich properties such as infinite differentiability and the existence of power series.
  • Complex analysis simplifies solving equations that may be difficult within real analysis.
  • It helps in understanding the flow of electric circuits, fluid dynamics, and quantum mechanics.
In complex analysis, every analytic function can be expressed as a power series, much like Taylor series in real calculus. This allows for an expansion around any point within the function's domain. Understanding these concepts provides insights into why polynomial functions like \(f(z) = z^3\) are straightforwardly analytic, making it easier to ensure functions meet the Cauchy-Riemann criteria.
This extension of calculus into the complex domain significantly enhances the tools available for mathematical and engineering applications.
Partial Derivatives
Partial derivatives are a key concept when dealing with functions of multiple variables. While they may seem initially like simple calculus operations, they become a pivotal tool in complex analysis. They help us evaluate how a function changes with respect to one variable, while keeping others constant.
  • In context with analytic functions, each function is often split into its real and imaginary parts,
  • This results in the need to identify partial derivatives with respect to each independent variable,
Consider the polynomial function \(f(z) = z^3\). For verification of whether it's analytic, you would need the real part, \(u(x,y)\), and the imaginary part, \(v(x,y)\). Calculating partial derivatives for these components reveals insights into the function's behavior.Using these expressions, you can apply the Cauchy-Riemann equations, which are \(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\) and \(\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\). These equations ensure the constituents of both the real and imaginary parts are perfectly matched to fulfill conditions for complex differentiation. In summary, mastering partial derivatives is essential to verifying the analyticity of complex functions.

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Most popular questions from this chapter

Suppose the two families of curves \(u(x, y)=c_{1}\) and \(v(x, y)=c_{2}\), are orthogonal trajectories in a domain \(D .\) Discuss: Is the function \(f(z)=u(x, y)+i v(x, y)\) necessarily analytic in \(D ?\)

In this problem you are guided through the start of the proof of the proposition: If functions \(f\) and \(g\) are analytic at a point \(z_{0}\) and \(f\left(z_{0}\right)=0, g\left(z_{0}\right)=0\), but \(g^{\prime}\left(z_{0}\right) \neq 0\), then \(\lim _{z \rightarrow z_{0}} \frac{f(z)}{g(z)}=\frac{f^{\prime}\left(z_{0}\right)}{g^{\prime}\left(z_{0}\right)}\).

Show that the given function is not analytic at any point. \(f(z)=\operatorname{Re}(z)\)

The given analytic function \(f(z)=u+i v\) defines two families of level curves \(u(x, y)=c_{1}\) and \(v(x, y)=c_{2} .\) First use implicit differentiation to compute \(d y / d x\) for each family and then verify that the families are orthogonal. \(f(z)=x^{3}-3 x y^{2}+i\left(3 x^{2} y-y^{3}\right)\)

Suppose \(x=r \cos \theta, y=r \sin \theta\), and \(f(z)=u(x, y)+i v(x, y) .\) Show that $$\begin{array}{ll} \frac{\partial u}{\partial r}=\frac{\partial u}{\partial x} \cos \theta+\frac{\partial u}{\partial y} \sin \theta, & \frac{\partial u}{\partial \theta}=-\frac{\partial u}{\partial x} r \sin \theta+\frac{\partial u}{\partial y} r \cos \theta \\ \frac{\partial v}{\partial r}=\frac{\partial v}{\partial x} \cos \theta+\frac{\partial v}{\partial y} \sin \theta, & \frac{\partial v}{\partial \theta}=-\frac{\partial v}{\partial x} r \sin \theta+\frac{\partial v}{\partial y} r \cos \theta\end{array}$$ and Now use (1) in the foregoing expressions for \(v_{r}\) and \(v_{\theta} .\) By comparing your results with the expressions for \(u_{r}\) and \(u_{\theta}\), deduce the Cauchy-Riemann equations in polar coordinates given in (10).

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