Prove: If f is holomorphic in the region G ⊆ C and always real valued, then f is constant in
G. (Hint: Use the Cauchy-Riemann equations to show that f ′ = 0.)

To prove that if \( f \) is holomorphic in a region \( G \subset \mathbb{C} \) and always real-valued, then \( f \) is constant in \( G \), we can use the **Cauchy-Riemann equations** to show that the derivative \( f' = 0 \). Here's the step-by-step proof:

### Given Information:
- \( f: G \to \mathbb{C} \) is **holomorphic** in the region \( G \subset \mathbb{C} \).
- \( f(z) \) is always **real-valued** for all \( z \in G \).

### Goal:
We want to prove that \( f \) is constant in \( G \).

### Step 1: Write \( f(z) \) in terms of real and imaginary parts
Since \( f(z) \) is holomorphic, we can express it as:
\[
f(z) = u(x, y) + iv(x, y)
\]
where \( z = x + iy \) (with \( x, y \in \mathbb{R} \)), and \( u(x, y) \) and \( v(x, y) \) are the real and imaginary parts of \( f(z) \), respectively.

### Step 2: Analyze the real-valued condition
Given that \( f(z) \) is real-valued, this implies that for all \( z = x + iy \in G \):
\[
f(z) = u(x, y) + iv(x, y) \quad \text{is real}.
\]
For this to hold, the imaginary part must vanish, i.e., \( v(x, y) = 0 \) for all \( x, y \in G \). Hence, the function reduces to:
\[
f(z) = u(x, y),
\]
where \( u(x, y) \) is a real-valued function, and \( v(x, y) = 0 \) for all \( z \in G \).

### Step 3: Use the Cauchy-Riemann equations
Since \( f(z) = u(x, y) \) is holomorphic, it satisfies the **Cauchy-Riemann equations**:
\[
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.
\]
However, we know that \( v(x, y) = 0 \) everywhere in \( G \). Thus, substituting \( v = 0 \) into the Cauchy-Riemann equations gives:
\[
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = 0 \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} = 0.
\]

### Step 4: Conclusion from the partial derivatives
From the Cauchy-Riemann equations, we now have:
\[
\frac{\partial u}{\partial x} = 0 \quad \text{and} \quad \frac{\partial u}{\partial y} = 0.
\]
This means that \( u(x, y) \) is constant with respect to both \( x \) and \( y \). Therefore, \( u(x, y) \) must be a constant function, say \( u(x, y) = C \) for some constant \( C \).

Since \( f(z) = u(x, y) \), we conclude that \( f(z) = C \) for all \( z \in G \), which shows that \( f \) is constant in \( G \).

### Final Conclusion:
If \( f \) is holomorphic in a region \( G \subset \mathbb{C} \) and always real-valued, then \( f \) must be constant in \( G \).

---

Would you like more details on any part of the proof? Here are five related questions for deeper understanding:

1. What does it mean for a function to be holomorphic in a region?
2. How do the Cauchy-Riemann equations relate to complex differentiability?
3. Why does the condition that \( f \) is real-valued imply that its imaginary part vanishes everywhere?
4. Can this proof be extended to show that any harmonic function that is constant on a boundary must be constant inside a region?
5. How does Liouville's theorem generalize this result to bounded holomorphic functions?

**Tip**: When using the Cauchy-Riemann equations, always check if real or imaginary parts of the function vanish, as this simplifies many proofs.

Prove: If f is holomorphic in the region G ⊆ C and always real valued, then f is constant in G. (Hint: Use the Cauchy-Riemann equations to show that f ′ = 0.)

Explore the proof that a holomorphic function which is always real-valued must be constant in a region. This solution applies the Cauchy-Riemann equations to show that the derivative is zero, making it suitable for undergraduate-level studies in complex analysis.

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