1. **State the problem:** Prove that the function $u = e^{-x}(x \sin y - y \cos y)$ is harmonic and find $v$ such that $f(z) = u + iv$ is analytic. 2. **Check if $u$ is harmonic:** A function $u(x,y)$ is harmonic if it satisfies Laplace's equation: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0.$$ 3. **Calculate $u_x$ and $u_y$:** - Using the product rule, $$u_x = \frac{\partial}{\partial x} \left(e^{-x}(x \sin y - y \cos y)\right) = -e^{-x}(x \sin y - y \cos y) + e^{-x}(\sin y) = e^{-x}(-x \sin y + y \cos y + \sin y).$$ $$u_y = \frac{\partial}{\partial y} \left(e^{-x}(x \sin y - y \cos y)\right) = e^{-x}(x \cos y - \cos y + y \sin y) = e^{-x}((x - 1) \cos y + y \sin y).$$ 4. **Calculate $u_{xx}$:** $$u_{xx} = \frac{\partial}{\partial x} \left(e^{-x}(-x \sin y + y \cos y + \sin y)\right)$$$$= -e^{-x}(-x \sin y + y \cos y + \sin y) + e^{-x}(-\sin y) = e^{-x}(x \sin y - y \cos y - \sin y - \sin y) = e^{-x}(x \sin y - y \cos y - 2 \sin y).$$ 5. **Calculate $u_{yy}$:** $$u_{yy} = \frac{\partial}{\partial y} \left(e^{-x}((x - 1) \cos y + y \sin y)\right) = e^{-x}(-(x - 1) \sin y + \sin y + y \cos y) = e^{-x}(- (x - 1) \sin y + \sin y + y \cos y).$$ Simplify the terms inside: $$- (x - 1) \sin y + \sin y = -x \sin y + \sin y + \sin y = -x \sin y + 2 \sin y.$$ So, $$u_{yy} = e^{-x}(-x \sin y + 2 \sin y + y \cos y).$$ 6. **Sum $u_{xx} + u_{yy}$:** $$u_{xx} + u_{yy} = e^{-x}(x \sin y - y \cos y - 2 \sin y) + e^{-x}(-x \sin y + 2 \sin y + y \cos y) = e^{-x}(0) = 0.$$ Hence, $u$ satisfies Laplace's equation and is harmonic. 7. **Find $v$ such that $f(z) = u + iv$ is analytic:** Since $f$ is analytic, $u$ and $v$ satisfy the Cauchy-Riemann equations: $$u_x = v_y, \quad u_y = -v_x.$$ From step 3, $$u_x = e^{-x}(-x \sin y + y \cos y + \sin y),$$ $$u_y = e^{-x}((x - 1) \cos y + y \sin y).$$ 8. **Integrate $v_y = u_x$ to find $v$ up to a function of $x$:** $$v = \int v_y dy = \int u_x dy = \int e^{-x}(-x \sin y + y \cos y + \sin y) dy.$$ Separate integrals: $$v = e^{-x} \left(-x \int \sin y dy + \int y \cos y dy + \int \sin y dy \right) + h(x).$$ Compute each integral: - $\int \sin y dy = -\cos y + C$ - $\int y \cos y dy$: Use integration by parts: Let $u = y$, $dv = \cos y dy$, so $du = dy$, $v = \sin y$. Then, $$\int y \cos y dy = y \sin y - \int \sin y dy = y \sin y + \cos y + C.$$ So, $$v = e^{-x} \left(-x (-\cos y) + y \sin y + \cos y - \cos y \right) + h(x) = e^{-x}(x \cos y + y \sin y) + h(x).$$ 9. **Find $v_x$ from the above expression:** $$v_x = \frac{\partial}{\partial x} \left(e^{-x}(x \cos y + y \sin y) + h(x)\right) = -e^{-x}(x \cos y + y \sin y) + e^{-x}(\cos y) + h'(x) = e^{-x}(-x \cos y - y \sin y + \cos y) + h'(x).$$ 10. **Use Cauchy-Riemann equation $u_y = -v_x$:** $$u_y = e^{-x}((x - 1) \cos y + y \sin y) = -v_x = -e^{-x}(-x \cos y - y \sin y + \cos y) - h'(x) = e^{-x}(x \cos y + y \sin y - \cos y) - h'(x).$$ Set equal: $$e^{-x}((x - 1) \cos y + y \sin y) = e^{-x}(x \cos y + y \sin y - \cos y) - h'(x).$$ Simplify left and right: Left: $e^{-x}((x - 1) \cos y + y \sin y)$ Right: $e^{-x}(x \cos y + y \sin y - \cos y) - h'(x)$ Subtract right side terms from left side: $$e^{-x}((x - 1) \cos y + y \sin y - x \cos y - y \sin y + \cos y) = -h'(x).$$ Simplify inside parentheses: $$(x - 1) \cos y + y \sin y - x \cos y - y \sin y + \cos y = \cos y(-1 + 1) + \sin y (y - y) = 0.$$ So, $$0 = -h'(x) \implies h'(x) = 0 \implies h(x) = C,$$ constant. 11. **Write final $v$:** $$v = e^{-x}(x \cos y + y \sin y) + C.$$ **Answer:** - $u = e^{-x}(x \sin y - y \cos y)$ is harmonic. - $v = e^{-x}(x \cos y + y \sin y) + C$ makes $f(z) = u + iv$ analytic.