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This proof is a demonstration of the fact that if a complex function \( f(z) \) is analytic on a domain \( D \) and its derivative \( f'(z) = 0 \) throughout \( D \), then \( f(z) \) must be a constant function on \( D \). Here’s a detailed explanation of the steps:

### Problem Statement
The problem states:
- Let \( f(z) \) be analytic on a domain \( D \).
- If \( f'(z) = 0 \) on \( D \), show that \( f(z) \) is constant on \( D \).

### Solution Explanation

1. **Analyticity and Cauchy-Riemann (C-R) Equations**:
   - Since \( f(z) \) is analytic, it must satisfy the **Cauchy-Riemann equations** in the domain \( D \). This implies that \( f(z) = u(x, y) + iv(x, y) \), where \( u \) and \( v \) are the real and imaginary parts of \( f(z) \), respectively.
   - The Cauchy-Riemann equations state that:
     \[
     \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.
     \]

2. **Condition \( f'(z) = 0 \)**:
   - The derivative \( f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \).
   - Given \( f'(z) = 0 \), we have:
     \[
     \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = 0.
     \]
   - This implies that both the real and imaginary parts must be zero:
     \[
     \frac{\partial u}{\partial x} = 0 \quad \text{and} \quad \frac{\partial v}{\partial x} = 0.
     \]

3. **Using the Cauchy-Riemann Equations**:
   - From the first part, we already have \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = 0 \) and \( \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} = 0 \).
   - This implies that all partial derivatives \( \frac{\partial u}{\partial x} \), \( \frac{\partial u}{\partial y} \), \( \frac{\partial v}{\partial x} \), and \( \frac{\partial v}{\partial y} \) are zero.

4. **Implication of Zero Partial Derivatives**:
   - Since all partial derivatives of \( u \) and \( v \) are zero, both \( u \) and \( v \) must be constant functions (as they do not vary with \( x \) or \( y \)).
   - Consequently, \( f(z) = u + iv \) is also constant because both \( u \) and \( v \) are constants.

### Conclusion
Thus, if \( f(z) \) is analytic on \( D \) and \( f'(z) = 0 \) on \( D \), then \( f(z) \) must be constant on \( D \).

Would you like more details on any specific part, or do you have any questions?

Here are some related questions to explore:
1. What are the Cauchy-Riemann equations, and why are they significant in complex analysis?
2. How does analyticity of a function in a domain differ from differentiability?
3. Can you give an example of a non-constant analytic function that does not satisfy \( f'(z) = 0 \)?
4. Why do zero partial derivatives imply that a function is constant?
5. What are the implications of the Cauchy-Riemann equations in the context of conformal mappings?

**Tip:** For any complex function, if you know it’s analytic and its derivative is zero, it’s immediately a constant function due to the properties of analytic functions.

If f(z) is analytic on a domain D, and if f'(z) = 0 on D, then f(z) is constant.

Understand the proof that if a complex function f(z) is analytic on a domain D and its derivative f'(z) is zero throughout D, then f(z) must be constant. This solution uses complex analysis concepts such as the Cauchy-Riemann equations and the Constant Function Theorem.

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