Let's solve the problem step by step: Given: - \( w = f(u) + g(v) \) - \( u = x + iy \) - \( v = x - iy \) - \( i = \sqrt{-1} \) We are to show that \( w_{xx} + w_{yy} = 0 \). --- ### Step 1: Compute the derivatives #### First derivatives Let’s compute \( w_x \) and \( w_y \): - \( w_x = f'(u) u_x + g'(v) v_x \) - \( w_y = f'(u) u_y + g'(v) v_y \) Now, - \( u_x = 1 \), \( u_y = i \) - \( v_x = 1 \), \( v_y = -i \) So, - \( w_x = f'(u) \cdot 1 + g'(v) \cdot 1 = f'(u) + g'(v) \) - \( w_y = f'(u) \cdot i + g'(v) \cdot (-i) = i f'(u) - i g'(v) = i(f'(u) - g'(v)) \) --- #### Second derivatives Now, calculate \( w_{xx} \): \[ w_{xx} = \frac{\partial}{\partial x} (w_x) = \frac{\partial}{\partial x} \left( f'(u) + g'(v) \right) \] \[ = f''(u) \cdot u_x + g''(v) \cdot v_x \] \[ = f''(u) \cdot 1 + g''(v) \cdot 1 = f''(u) + g''(v) \] Now, \( w_{yy} \): \[ w_{yy} = \frac{\partial}{\partial y} (w_y) = \frac{\partial}{\partial y} \left( i (f'(u) - g'(v)) \right) \] \[ = i \left( \frac{\partial}{\partial y} f'(u) - \frac{\partial}{\partial y} g'(v) \right) \] \[ = i \left( f''(u) u_y - g''(v) v_y \right) \] \[ = i \left( f''(u) \cdot i - g''(v) \cdot (-i) \right) \] \[ = i \left( i f''(u) + i g''(v) \right) \] \[ = i^2 (f''(u) + g''(v)) \] \[ = -1 \cdot (f''(u) + g''(v)) \] \[ = - (f''(u) + g''(v)) \] --- ### Step 2: Add the derivatives \[ w_{xx} + w_{yy} = (f''(u) + g''(v)) + [ - (f''(u) + g''(v)) ] = 0 \] --- ### **Conclusion** \[ \boxed{w_{xx} + w_{yy} = 0} \] This verifies the statement given in the problem.