Math Problem Statement

) If 𝑓(𝑧) and 𝑓(𝑧) Μ…Μ…Μ…Μ…Μ…Μ… are both analytic.prove that 𝑓(𝑧) is constant.

Solution

To prove that if both f(z)f(z) and f(z)β€Ύ\overline{f(z)} are analytic, then f(z)f(z) must be constant, we can use the fact that an analytic function satisfies the Cauchy-Riemann equations and apply some properties related to complex differentiability.

Let f(z)=u(x,y)+iv(x,y)f(z) = u(x, y) + iv(x, y), where uu and vv are the real and imaginary parts of f(z)f(z), respectively. Since f(z)f(z) is analytic, uu and vv satisfy the Cauchy-Riemann equations: 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}.

Since f(z)β€Ύ\overline{f(z)} is also analytic, we can express f(z)β€Ύ\overline{f(z)} as f(z)β€Ύ=u(x,y)βˆ’iv(x,y)\overline{f(z)} = u(x, y) - iv(x, y), and for this to be analytic, uu and βˆ’v-v must also satisfy the Cauchy-Riemann equations: βˆ‚uβˆ‚x=βˆ‚(βˆ’v)βˆ‚y=βˆ’βˆ‚vβˆ‚yandβˆ‚uβˆ‚y=βˆ’βˆ‚(βˆ’v)βˆ‚x=βˆ‚vβˆ‚x.\frac{\partial u}{\partial x} = \frac{\partial (-v)}{\partial y} = -\frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial (-v)}{\partial x} = \frac{\partial v}{\partial x}.

Now we have two sets of conditions:

  1. From the analyticity of f(z)f(z): 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}.
  2. From the analyticity of f(z)β€Ύ\overline{f(z)}: βˆ‚uβˆ‚x=βˆ’βˆ‚vβˆ‚yandβˆ‚uβˆ‚y=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}.

Comparing these two sets of equations, we observe that they imply: βˆ‚vβˆ‚y=βˆ’βˆ‚vβˆ‚yβ‡’βˆ‚vβˆ‚y=0\frac{\partial v}{\partial y} = -\frac{\partial v}{\partial y} \quad \Rightarrow \quad \frac{\partial v}{\partial y} = 0 and βˆ‚vβˆ‚x=βˆ’βˆ‚vβˆ‚xβ‡’βˆ‚vβˆ‚x=0.\frac{\partial v}{\partial x} = -\frac{\partial v}{\partial x} \quad \Rightarrow \quad \frac{\partial v}{\partial x} = 0.

Thus, vv is constant. With vv being constant, the Cauchy-Riemann equations imply that uu must also be constant, because the partial derivatives of uu would then be zero as well.

Therefore, f(z)=u+ivf(z) = u + iv is constant.

This completes the proof that if both f(z)f(z) and f(z)β€Ύ\overline{f(z)} are analytic, then f(z)f(z) must be constant.

Would you like more details or have any questions?

Here are some related questions you might find interesting:

  1. How do the Cauchy-Riemann equations characterize the analyticity of a function?
  2. What is the significance of a function being constant in complex analysis?
  3. Can a function that is real-valued and analytic be non-constant?
  4. How does the concept of harmonic functions relate to analytic functions?
  5. What is the maximum modulus principle in complex analysis?

Tip: In complex analysis, analyticity is a powerful property, and the Cauchy-Riemann equations serve as a fundamental tool to determine if a function is analytic.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Analytic Functions
Complex Differentiability

Formulas

Cauchy-Riemann Equations: βˆ‚u/βˆ‚x = βˆ‚v/βˆ‚y and βˆ‚u/βˆ‚y = -βˆ‚v/βˆ‚x

Theorems

Cauchy-Riemann Equations
Constant Function Theorem in Complex Analysis

Suitable Grade Level

Advanced Undergraduate - Graduate Level