Question

If $f'(z) = 0$ everywhere in a connected open set $G \subset \mathbb{C}$, then $f(z)$ is

• A. $|z|$
• B. constant
• C. $z^2 + z + 1$
• D. $z + 10$

Hint:

Connectedness of $G$ is essential! If $G$ were disconnected (e.g., two disjoint open disks), $f(z)$ could take different constant values on each component.

Answer 1

The Correct Option is B

Step 1: Concept
In complex analysis, a domain $G \subset \mathbb{C}$ is a connected open set. A fundamental theorem states that if an analytic function $f(z)$ has derivative zero everywhere on a domain $G$, then $f(z)$ must be constant on $G$.

Step 2: Key Formulas and Approach
Let $f(z) = u(x, y) + i v(x, y)$ be an analytic function defined on $G$, where $u(x, y)$ and $v(x, y)$ are real-valued functions. The complex derivative $f'(z)$ is related to partial derivatives via Cauchy-Riemann equations: \[ f'(z) = u_x + i v_x = v_y - i u_y \]

Step 3: Step-by-step Explanation

• Given $f'(z) = 0$ for all $z \in G$: \[ u_x + i v_x = 0 \implies u_x = 0 \quad \text{and} \quad v_x = 0 \]
• By Cauchy-Riemann equations ($u_x = v_y$ and $u_y = -v_x$): \[ v_y = u_x = 0 \quad \text{and} \quad u_y = -v_x = 0 \]
• Therefore, all first-order partial derivatives of $u$ and $v$ vanish identically on $G$: \[ \frac{\partial u}{\partial x} = 0, \quad \frac{\partial u}{\partial y} = 0, \quad \frac{\partial v}{\partial x} = 0, \quad \frac{\partial v}{\partial y} = 0 \]
• Since $G$ is a connected open set, any two points in $G$ can be joined by a polygonal path in $G$. Integrating along the path shows that $u(x, y) = c_1$ and $v(x, y) = c_2$ are constant functions.
• Thus, $f(z) = c_1 + i c_2 = C$ (a complex constant) on $G$.

Step 4: Final Answer
An analytic function with derivative identically zero on a domain is constant. Thus, Option (B) is correct.