To show that u is harmonic, we need to demonstrate that it satisfies Laplace's equation:
∇^2u = ∂^2u/∂x^2 + ∂^2u/∂y^2 = 0
Taking the partial derivatives, we have:
∂u/∂x = -e^(-x)*sin(y)
∂^2u/∂x^2 = e^(-x)*sin(y)
∂u/∂y = e^(-x)(cos(y) + cos(y))
∂^2u/∂y^2 = -e^(-x)(sin(y) + sin(y))
Adding these together, we get:
∂^2u/∂x^2 + ∂^2u/∂y^2 = e^(-x)*sin(y) - e^(-x)*sin(y) = 0
Therefore, u is harmonic.
To find the conjugate function v, we can use the Cauchy-Riemann equations:
∂u/∂x = ∂v/∂y
∂u/∂y = -∂v/∂x
Taking the partial derivatives of u with respect to x and y:
∂u/∂x = -e^(-x)sin(y)
∂u/∂y = e^(-x)(cos(y) + sin(y))
Equating these with the partial derivatives of v, we have:
∂v/∂y = -e^(-x)sin(y)
∂v/∂x = -e^(-x)(cos(y) + sin(y))
Integrating ∂v/∂y with respect to y:
v = e^(-x)*cos(y) - h(x)
Taking the partial derivative of v with respect to x:
∂v/∂x = e^(-x)*sin(y) - h'(x)
Equating this with ∂u/∂y and solving for h(x), we get:
h(x) = -e^(-x)*sin(y)
Therefore, the conjugate function of u is:
v = e^(-x)*cos(y) + e^(-x)*sin(y)