Math Problem Statement

Намерете аналитична функция w(z) = u + iv, w(0) = i, ако u(x,y) = y^3 - 3x^2y + x^2 - y^2.

Solution

To solve this problem, we need to find the analytic function w(z)=u(x,y)+iv(x,y)w(z) = u(x, y) + iv(x, y), where the real part u(x,y)u(x, y) is given, and w(0)=iw(0) = i. This problem can be approached using the Cauchy-Riemann equations to find the imaginary part v(x,y)v(x, y) and then determine the full analytic function.

Given:

  • u(x,y)=y33x2y+x2y2u(x, y) = y^3 - 3x^2 y + x^2 - y^2
  • w(0)=iw(0) = i implies that at z=0z = 0, w(0)=u(0,0)+iv(0,0)=iw(0) = u(0, 0) + iv(0, 0) = i.

Step 1: Use the Cauchy-Riemann equations.

For w(z)=u(x,y)+iv(x,y)w(z) = u(x, y) + iv(x, y) to be analytic, the Cauchy-Riemann equations must hold:

ux=vyanduy=vx\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}

First, we compute the partial derivatives of u(x,y)u(x, y).

ux=6xy+2x\frac{\partial u}{\partial x} = -6xy + 2x uy=3y23x22y\frac{\partial u}{\partial y} = 3y^2 - 3x^2 - 2y

Step 2: Use Cauchy-Riemann equations to find v(x,y)v(x, y).

From ux=vy\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}:

vy=6xy+2x\frac{\partial v}{\partial y} = -6xy + 2x

Now integrate with respect to yy:

v(x,y)=3xy2+2xy+f(x)v(x, y) = -3xy^2 + 2xy + f(x)

Next, use uy=vx\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}:

vx=3y2+2y+f(x)\frac{\partial v}{\partial x} = -3y^2 + 2y + f'(x)

Equating this to uy-\frac{\partial u}{\partial y}, we have:

(3y23x22y)=3y2+2y+f(x)- (3y^2 - 3x^2 - 2y) = -3y^2 + 2y + f'(x)

Simplifying:

3x2=f(x)3x^2 = f'(x)

Integrating with respect to xx:

f(x)=x3+Cf(x) = x^3 + C

Thus, v(x,y)=3xy2+2xy+x3+Cv(x, y) = -3xy^2 + 2xy + x^3 + C.

Step 3: Determine the constant CC.

We know that w(0)=iw(0) = i. At z=0z = 0 (i.e., x=0x = 0 and y=0y = 0):

w(0)=u(0,0)+iv(0,0)=iw(0) = u(0, 0) + iv(0, 0) = i

Since u(0,0)=0u(0, 0) = 0 and v(0,0)=Cv(0, 0) = C, we have:

C=1C = 1

Thus, v(x,y)=3xy2+2xy+x3+1v(x, y) = -3xy^2 + 2xy + x^3 + 1.

Step 4: Write the analytic function.

The analytic function w(z)w(z) is:

w(z)=u(x,y)+iv(x,y)=(y33x2y+x2y2)+i(3xy2+2xy+x3+1)w(z) = u(x, y) + iv(x, y) = \left( y^3 - 3x^2 y + x^2 - y^2 \right) + i \left( -3xy^2 + 2xy + x^3 + 1 \right)

Or in terms of z=x+iyz = x + iy, this is the final analytic function.


Do you have any questions or need further details?

Here are 5 related questions:

  1. What are the Cauchy-Riemann equations and how are they used to determine analyticity?
  2. How do you integrate functions involving multiple variables like in this problem?
  3. What is the significance of the condition w(0)=iw(0) = i?
  4. How can the real and imaginary parts of an analytic function be interpreted geometrically?
  5. Can the form of an analytic function always be uniquely determined from its real part?

Tip: When solving for analytic functions, always start by verifying the Cauchy-Riemann equations before attempting to integrate for the imaginary part.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Cauchy-Riemann Equations
Analytic Functions
Partial Derivatives

Formulas

Cauchy-Riemann equations: ∂u/∂x = ∂v/∂y and ∂u/∂y = -∂v/∂x
v(x, y) = ∫ (partial derivative of u w.r.t x or y)
w(z) = u(x, y) + iv(x, y)

Theorems

Cauchy-Riemann Theorem
Analytic Function Theorem

Suitable Grade Level

Undergraduate