Math Problem Statement

Using Cauchy Riemann equations, check whether complex function 𝑓(𝑧) = ((π‘₯ + 1) 3 βˆ’ 3(π‘₯ + 1)𝑦 2 βˆ’ 3π‘₯) + 𝑖(𝑦 3 + 3𝑦 βˆ’ 3(π‘₯ + 1) 2𝑦) is analytic in the complex plane or not

Solution

To determine if the given complex function f(z)=u(x,y)+iv(x,y)f(z) = u(x, y) + iv(x, y) is analytic, we need to verify whether the Cauchy-Riemann equations are satisfied.

Given: f(z)=((x+1)3βˆ’3(x+1)y2βˆ’3x)+i(y3+3yβˆ’3(x+1)2y)f(z) = \left( (x+1)^3 - 3(x+1)y^2 - 3x \right) + i\left( y^3 + 3y - 3(x+1)^2 y \right)

Step 1: Express f(z)f(z) in terms of u(x,y)u(x, y) and v(x,y)v(x, y)

  • The real part u(x,y)u(x, y) is: u(x,y)=(x+1)3βˆ’3(x+1)y2βˆ’3xu(x, y) = (x+1)^3 - 3(x+1)y^2 - 3x

  • The imaginary part v(x,y)v(x, y) is: v(x,y)=y3+3yβˆ’3(x+1)2yv(x, y) = y^3 + 3y - 3(x+1)^2 y

Step 2: Cauchy-Riemann Equations

The Cauchy-Riemann equations are: βˆ‚uβˆ‚x=βˆ‚vβˆ‚yandβˆ‚uβˆ‚y=βˆ’βˆ‚vβˆ‚x\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}

We need to compute the partial derivatives of u(x,y)u(x, y) and v(x,y)v(x, y).

Step 3: Compute partial derivatives of u(x,y)u(x, y)

  • βˆ‚uβˆ‚x\frac{\partial u}{\partial x}: βˆ‚uβˆ‚x=3(x+1)2βˆ’3y2βˆ’3\frac{\partial u}{\partial x} = 3(x+1)^2 - 3y^2 - 3

  • βˆ‚uβˆ‚y\frac{\partial u}{\partial y}: βˆ‚uβˆ‚y=βˆ’6(x+1)y\frac{\partial u}{\partial y} = -6(x+1)y

Step 4: Compute partial derivatives of v(x,y)v(x, y)

  • βˆ‚vβˆ‚x\frac{\partial v}{\partial x}: βˆ‚vβˆ‚x=βˆ’6(x+1)y\frac{\partial v}{\partial x} = -6(x+1)y

  • βˆ‚vβˆ‚y\frac{\partial v}{\partial y}: βˆ‚vβˆ‚y=3y2+3βˆ’3(x+1)2\frac{\partial v}{\partial y} = 3y^2 + 3 - 3(x+1)^2

Step 5: Check the Cauchy-Riemann equations

  • First Cauchy-Riemann equation: βˆ‚uβˆ‚x=3(x+1)2βˆ’3y2βˆ’3andβˆ‚vβˆ‚y=3y2+3βˆ’3(x+1)2\frac{\partial u}{\partial x} = 3(x+1)^2 - 3y^2 - 3 \quad \text{and} \quad \frac{\partial v}{\partial y} = 3y^2 + 3 - 3(x+1)^2 Clearly, βˆ‚uβˆ‚xβ‰ βˆ‚vβˆ‚y\frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y}.

Since the first Cauchy-Riemann equation is not satisfied, we conclude that the function is not analytic.

Conclusion:

The given function is not analytic in the complex plane because the Cauchy-Riemann equations are not satisfied.


Would you like any further details or clarifications?

Here are some related questions to expand on this topic:

  1. What are the conditions for a function to be analytic in a region?
  2. How are the Cauchy-Riemann equations derived from the definition of differentiability in the complex plane?
  3. What are some common examples of analytic functions?
  4. How do harmonic functions relate to the Cauchy-Riemann equations?
  5. Can you provide an example of a function that is not analytic and explain why?

Tip: A function is analytic at a point if it is differentiable at that point and in some neighborhood around it.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Cauchy-Riemann Equations
Analytic Functions

Formulas

-

Theorems

Cauchy-Riemann Equations

Suitable Grade Level

Advanced undergraduate