Question

A complex function \( f(z) = u + iv \) with \( u = \operatorname{Re}(z) + \operatorname{Im}(z) \) is analytic if \( v = \) ________

• A. \( v = \operatorname{Re}(z) + \operatorname{Im}(z) + c \), where \( c \) is an arbitrary constant
• B. \( v = -\operatorname{Re}(z) + \operatorname{Im}(z) + c \), where \( c \) is an arbitrary constant
• C. \( v = \operatorname{Re}(z) - \operatorname{Im}(z) + c \), where \( c \) is an arbitrary constant
• D. \( v = -\operatorname{Re}(z) - \operatorname{Im}(z) + c \), where \( c \) is an arbitrary constant

Hint:

Use Cauchy-Riemann equations to test whether a complex function is analytic. Express \( f(z) \) as \( u(x,y) + iv(x,y) \), then solve accordingly.

Answer 1

The Correct Option is B

To ensure analyticity, \( f(z) = u(x, y) + i v(x, y) \) must satisfy the Cauchy-Riemann equations: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \] Given \( u(x, y) = x + y \), we have: \[ \frac{\partial u}{\partial x} = 1, \quad \frac{\partial u}{\partial y} = 1 \] So we require: \[ \frac{\partial v}{\partial y} = 1, \quad \frac{\partial v}{\partial x} = -1 \] Integrating, \[ \frac{\partial v}{\partial x} = -1 \Rightarrow v = -x + g(y) \] \[ \frac{\partial v}{\partial y} = g'(y) = 1 \Rightarrow g(y) = y + c \Rightarrow v = -x + y + c \] So \( v = -\operatorname{Re}(z) + \operatorname{Im}(z) + c \)