Let's solve the problem step by step: ### **Step 1: Express \( w \) in terms of \( x \) and \( y \)** Given: - \( w = f(u) + g(v) \) - \( u = x + iy \) - \( v = x - iy \) - \( i^2 = -1 \) We need to compute \( w_{xx} + w_{yy} \). First, let's find the derivatives of \( w \): \[ w_x = f'(u) \frac{\partial u}{\partial x} + g'(v) \frac{\partial v}{\partial x} \] Since \( \frac{\partial u}{\partial x} = 1 \) and \( \frac{\partial v}{\partial x} = 1 \), we get: \[ w_x = f'(u) + g'(v) \] Now, compute the second derivative with respect to \( x \): \[ w_{xx} = \frac{\partial}{\partial x}[f'(u) + g'(v)] = f''(u) \frac{\partial u}{\partial x} + g''(v) \frac{\partial v}{\partial x} = f''(u) \cdot 1 + g''(v) \cdot 1 = f''(u) + g''(v) \] **Explanation:** We have used the chain rule to differentiate \( w \) with respect to \( x \), and both \( u \) and \( v \) are linear in \( x \) with coefficient 1. --- ### **Step 2: Compute \( w_{yy} \) and add to \( w_{xx} \)** Now, calculate \( w_y \): \[ w_y = f'(u) \frac{\partial u}{\partial y} + g'(v) \frac{\partial v}{\partial y} \] Since \( \frac{\partial u}{\partial y} = i \) and \( \frac{\partial v}{\partial y} = -i \): \[ w_y = f'(u) i + g'(v) (-i) = i f'(u) - i g'(v) = i[f'(u) - g'(v)] \] Now, differentiate again with respect to \( y \): \[ w_{yy} = \frac{\partial}{\partial y}\left(i[f'(u) - g'(v)]\right) \] \[ = i \left[ f''(u) \frac{\partial u}{\partial y} - g''(v) \frac{\partial v}{\partial y} \right] \] \[ = i \left[ f''(u) i - g''(v) (-i) \right] \] \[ = i^2 f''(u) + i^2 g''(v) \] Since \( i^2 = -1 \): \[ w_{yy} = -f''(u) - g''(v) \] Now add \( w_{xx} \) and \( w_{yy} \): \[ w_{xx} + w_{yy} = [f''(u) + g''(v)] + [-f''(u) - g''(v)] = 0 \] **Explanation:** The second derivatives with respect to \( x \) and \( y \) cancel each other out due to the properties of the complex unit \( i \). --- ## **Final Answer** \[ \boxed{w_{xx} + w_{yy} = 0} \]