Complex Analysis Prob

Suppose u is a twice continuously differentiable real-valued harmonic function on a disk D(z_0;r) centered at z_0 = x_0 +iy_0. For (x_1,y_1) contained in D(z_0;r), show that the equation,

v(x_1,y_1) = c + ∫[y_1,y_0] ∂u∂x(x_1,y) dy − ∫[x_1,x_0] ∂u∂y(x,y_0) dy

defines a harmonic conjugate for u on D(z_0;r) with v(x_0,y_0) = c.

** ∫[y_1,y_0] means the integration from y_1 (top) , y_0 (bottom)

Thanks.

The formula for v should be

v(x_1,y_1) = c + ∫[y_1,y_0] ∂u/∂(x_1) (x_1,y) dy - ∫[x_1,x_0] ∂u/∂(y_1)(x,y_1) dx.

Setting x_1 = x_0 and y_1 = y_0 causes the integrals to be zero, leaving v(x_0, y_0) = c. By the fundamental theorem of calculus,

∂/∂(x_1) ∫[x_1,x_0] ∂u/∂(y_1)(x,y_1) dx = ∂u/∂(y_1)(x_1,y_1),

∂/∂(y_1) ∫[y_1,y_0] ∂u/∂(x_1)(x_1,y) dy = ∂u/∂(x_1)(x_1,y_1).

Thus

∂v/∂(x_1) = 0 + 0 - ∂u/∂(x_1) = -∂u/∂(x_1),

∂v/∂(y_1) = 0 + ∂u/∂(y_1) - 0 = ∂u/∂(y_1),

which implies u, v satisfy the Cauchy-Riemann equations on D(z_0; r). Hence v is a harmonic conjugate for u on D(z_0; r)

Edit: Thanks Bob for checking. Still, I think there is an error in the book's formula. The derivatives ∂/∂x and ∂/∂y in the integrals do not contain the (x_1)- or (y_1)-variables. With the way it's shown above, ∂v/∂(x_1)(x_1,y_1) ≠ -∂u/∂(y_1)(x_1,y_1), so he Cauchy-Riemann equations are satisfied by u,v.