answer:To solve this boundary value problem using complex analysis, we first recognize that the given partial differential equation is the Laplace's equation in two dimensions: frac{partial^2 u(x,y)}{partial x^2} + frac{partial^2 u(x,y)}{partial y^2} = 0. We will use the method of conformal mapping to transform the Laplace's equation into a simpler problem. Let's consider the following conformal mapping: z = x + iy = a cosh(w), where w = u + iv and u(x,y) is the real part of w. The mapping z = a cosh(w) maps the strip 0 < u < a and -infty < v < infty onto the upper half-plane Im(z) > 0. The boundary conditions become: u(0,y) = 0 Rightarrow w = iv on the imaginary axis, frac{partial u}{partial x}(a,y) + u(a,y) = 0 Rightarrow w = a + iv on the line Re(w) = a. Now, we consider the function f(w) = frac{1}{pi} log(z - a) - frac{1}{pi} log(z). The function f(w) is harmonic in the upper half-plane since it is the real part of an analytic function. Moreover, it satisfies the boundary conditions: f(0) = 0 and f(a) = 0. Now, we need to find the inverse of the conformal mapping to obtain the solution u(x,y) in terms of x and y. From the mapping z = a cosh(w), we have: w = cosh^{-1} left(frac{z}{a}right). Substituting z = x + iy, we get: w = cosh^{-1} left(frac{x + iy}{a}right) = u(x,y) + iv(x,y). Now, we can find the real part of w which is the solution u(x,y): u(x,y) = Releft(cosh^{-1} left(frac{x + iy}{a}right)right). Finally, we substitute the harmonic function f(w) into the expression for u(x,y): u(x,y) = Releft(frac{1}{pi} logleft(frac{x + iy - a}{a}right) - frac{1}{pi} log(x + iy)right). This is the solution to the given boundary value problem using complex analysis.

answer:Let f(z) = z^2 + 1, where z = x + iy (with x and y being real numbers). Then f(z) = (x + iy)^2 + 1 = (x^2 - y^2 + 1) + i(2xy). Let u(x, y) be the harmonic conjugate of f(z). Then the function g(z) = f(z) + iu(z) is analytic in the upper half-plane. Now, let v(x, y) = 2xy be the imaginary part of f(z). We want to find u(x, y) such that g(z) = u(x, y) + iv(x, y) is analytic. To do this, we use the Cauchy-Riemann equations, which are necessary and sufficient conditions for differentiability: ∂u/∂x = ∂v/∂y and ∂u/∂y = -∂v/∂x. Let's compute the partial derivatives of v(x, y): ∂v/∂x = 2y and ∂v/∂y = 2x. Now we can use the Cauchy-Riemann equations to find the partial derivatives of u(x, y): ∂u/∂x = 2y and ∂u/∂y = -2x. Now we need to find u(x, y) by integrating these partial derivatives: 1) Integrate ∂u/∂x = 2y with respect to x: u(x, y) = 2xy + h(y), where h(y) is an arbitrary function of y. 2) Differentiate u(x, y) = 2xy + h(y) with respect to y: ∂u/∂y = 2x + h'(y). Now we compare this with the second Cauchy-Riemann equation, ∂u/∂y = -2x: 2x + h'(y) = -2x. This implies that h'(y) = -4x. However, this is a contradiction since h'(y) should be a function of y only, not x. The correct approach is to notice that the second Cauchy-Riemann equation is already satisfied by the given u(x, y) = 2xy + h(y): ∂u/∂y = 2x + h'(y) = -2x. This implies that h'(y) = -4x, which is a function of x only. Therefore, h(y) must be a constant, say C. So, the harmonic conjugate is: u(x, y) = 2xy + C. Now let's verify that g(z) = f(z) + iu(z) satisfies the Laplace equation in the upper half-plane. The Laplace equation states that: ∇²g = ∂²g/∂x² + ∂²g/∂y² = 0. We have g(z) = (x^2 - y^2 + 1) + i(2xy + C). Let's compute the second partial derivatives: ∂²g/∂x² = 2 and ∂²g/∂y² = -2. Thus, ∇²g = 2 - 2 = 0, which means that g(z) satisfies the Laplace equation in the upper half-plane.

answer:Since f(z) is analytic in the region x > 0 and y > 0, it satisfies the Cauchy-Riemann equations: (1) ∂u/∂x = ∂v/∂y (2) ∂u/∂y = -∂v/∂x We are given the boundary conditions: f(z) |x=0 = y^2 => u(0, y) = y^2, v(0, y) = 0 f(z) |y=0 = x^3 => u(x, 0) = x^3, v(x, 0) = 0 Using the boundary conditions, we can apply the Cauchy-Riemann equations on the boundaries: For x = 0: (1) ∂u/∂x = ∂v/∂y => ∂(y^2)/∂x = ∂(0)/∂y => 0 = 0 (2) ∂u/∂y = -∂v/∂x => ∂(y^2)/∂y = -∂(0)/∂x => 2y = 0 For y = 0: (1) ∂u/∂x = ∂v/∂y => ∂(x^3)/∂x = ∂(0)/∂y => 3x^2 = 0 (2) ∂u/∂y = -∂v/∂x => ∂(x^3)/∂y = -∂(0)/∂x => 0 = 0 From the Cauchy-Riemann equations on the boundaries, we have: 2y = 0 => y = 0 3x^2 = 0 => x = 0 This implies that the function f(z) is constant in the region x > 0 and y > 0. Since f(z) is continuous on its boundary, we can determine the value of f(z) in the region x > 0 and y > 0 by evaluating f(z) at any point on the boundary. Let's evaluate f(z) at z = 2i: f(2i) = f(0 + 2i) = u(0, 2) + iv(0, 2) = (2^2) + i(0) = 4 Therefore, f(2i) = 4.

answer:To solve the Dirichlet problem, we will use the Poisson integral formula. The Poisson integral formula for a harmonic function u(x, y) in the unit disc D with boundary values given by a continuous function f(theta) on the boundary partial D is given by: u(x, y) = frac{1}{2pi} int_{0}^{2pi} P(x, y, theta) f(theta) dtheta where P(x, y, theta) = frac{1 - r^2}{1 - 2rcos(theta - phi) + r^2} is the Poisson kernel, with (x, y) = (rcosphi, rsinphi). In our problem, the boundary values are given by u(x, y) = x^2 - y^2 on partial D. We can rewrite this in polar coordinates as f(theta) = cos^2theta - sin^2theta. Now, we can plug this into the Poisson integral formula: u(x, y) = frac{1}{2pi} int_{0}^{2pi} P(x, y, theta) (cos^2theta - sin^2theta) dtheta We can split the integral into two parts: u(x, y) = frac{1}{2pi} int_{0}^{2pi} P(x, y, theta) cos^2theta dtheta - frac{1}{2pi} int_{0}^{2pi} P(x, y, theta) sin^2theta dtheta Now, we can use the double-angle formulas for cos^2theta and sin^2theta: cos^2theta = frac{1 + cos(2theta)}{2}, quad sin^2theta = frac{1 - cos(2theta)}{2} Plugging these into the integral, we get: u(x, y) = frac{1}{2pi} int_{0}^{2pi} P(x, y, theta) frac{1 + cos(2theta)}{2} dtheta - frac{1}{2pi} int_{0}^{2pi} P(x, y, theta) frac{1 - cos(2theta)}{2} dtheta Simplifying and combining the integrals, we get: u(x, y) = frac{1}{2pi} int_{0}^{2pi} P(x, y, theta) cos(2theta) dtheta Now, we can plug in the expression for the Poisson kernel: u(x, y) = frac{1}{2pi} int_{0}^{2pi} frac{1 - r^2}{1 - 2rcos(theta - phi) + r^2} cos(2theta) dtheta This integral can be solved using contour integration techniques from complex analysis. The final result is: u(x, y) = x^2 - y^2 Thus, the solution to the Dirichlet problem is u(x, y) = x^2 - y^2, which is the same as the given boundary values.