Math Problem Statement

Prove that (+) Re(f(z))² = 2lf'(z)|² where f(z) is analytic.

Solution

To prove the given identity:

Re(f(z))2=2f(z)2\text{Re}(f(z))^2 = 2 |f'(z)|^2 where f(z)f(z) is analytic, we'll work step by step and show whether this statement holds under the assumption that f(z)f(z) is an analytic function.

Step 1: Express f(z)f(z) in terms of real and imaginary parts

Let f(z)=u(x,y)+iv(x,y)f(z) = u(x, y) + i v(x, y), where u(x,y)u(x, y) is the real part and v(x,y)v(x, y) is the imaginary part of f(z)f(z), with z=x+iyz = x + iy. Since f(z)f(z) is analytic, it satisfies the Cauchy-Riemann equations: ux=vy,uy=vx\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}

Step 2: Compute f(z)2|f'(z)|^2

To compute f(z)2|f'(z)|^2, we first find the derivative f(z)f'(z). Since f(z)f(z) is analytic, the derivative is:

f(z)=dfdz=ux+ivxf'(z) = \frac{df}{dz} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}

Now, the modulus of f(z)f'(z) is:

f(z)2=(ux)2+(vx)2|f'(z)|^2 = \left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial v}{\partial x} \right)^2

Using the Cauchy-Riemann equations, we know that:

vx=uy\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}

Thus,

f(z)2=(ux)2+(uy)2|f'(z)|^2 = \left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y} \right)^2

Step 3: Compute Re(f(z))2\text{Re}(f(z))^2

The real part of f(z)f(z) is u(x,y)u(x, y), so:

Re(f(z))2=u(x,y)2\text{Re}(f(z))^2 = u(x, y)^2

Step 4: Relating u(x,y)2u(x, y)^2 to f(z)2|f'(z)|^2

For the given identity Re(f(z))2=2f(z)2\text{Re}(f(z))^2 = 2 |f'(z)|^2 to hold, the square of the real part u(x,y)2u(x, y)^2 would need to be related to the derivatives of u(x,y)u(x, y). However, there is no general reason for the square of the real part of an analytic function to be twice the modulus squared of its derivative in the general case.

Conclusion

The given identity:

Re(f(z))2=2f(z)2\text{Re}(f(z))^2 = 2 |f'(z)|^2 does not hold in general for an arbitrary analytic function f(z)f(z). This can be seen by considering specific counterexamples or by the fact that there is no inherent relationship between the square of the real part of a function and the modulus of its derivative that would lead to the factor of 2.

Would you like me to clarify any step in this process?


Here are some relative questions to further expand your understanding:

  1. What is the significance of the Cauchy-Riemann equations in complex analysis?
  2. Can you provide an example of an analytic function where Re(f(z))22f(z)2\text{Re}(f(z))^2 \neq 2 |f'(z)|^2?
  3. How is the modulus of a complex derivative calculated in general?
  4. What conditions must an analytic function satisfy in terms of its real and imaginary components?
  5. How do the Cauchy-Riemann equations ensure a function is analytic?

Tip: In complex analysis, the modulus of a derivative provides insight into the behavior of a function's growth, and the Cauchy-Riemann equations ensure the function's smoothness and differentiability in the complex plane.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Analytic Functions
Cauchy-Riemann Equations

Formulas

Re(f(z))^2
|f'(z)|^2
f'(z) = df/dz = ∂u/∂x + i ∂v/∂x

Theorems

Cauchy-Riemann Equations
Modulus of a Derivative

Suitable Grade Level

University Level (Advanced Complex Analysis)