Question

The real part of an analytic function \( f(z) = u + iv \), where \( v(x, y) = \left( \frac{e^{-y} - e^y}{2} \right) \sin x \), is:

• A. \( \cos x \sinh y + C \)
• B. \( \cos x \cosh y + C \)
• C. \( -\cos x \sinh y + C \)
• D. \( -\cos x \cosh y + C \)

Hint:

When given the imaginary part of an analytic function, use Cauchy-Riemann equations to find the real part by integration.

Answer 1

The Correct Option is B

Given that \( f(z) = u + iv \) is analytic, the Cauchy-Riemann equations must hold: \[ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \] We are given: \[ v(x, y) = \left( \frac{e^{-y} - e^y}{2} \right) \sin x = -\sinh y \cdot \sin x \] Compute: \[ \frac{\partial v}{\partial y} = -\cosh y \cdot \sin x \Rightarrow \frac{\partial u}{\partial x} = -\cosh y \cdot \sin x \] \[ \frac{\partial v}{\partial x} = -\sinh y \cdot \cos x \Rightarrow \frac{\partial u}{\partial y} = \sinh y \cdot \cos x \] Now integrate: \[ \frac{\partial u}{\partial x} = -\cosh y \cdot \sin x \Rightarrow u = \cosh y \cdot \cos x + C \]