Answer to Question #142504 in Complex Analysis for Doll

Question

Find the original function without finding the corresponding conjugate
u(x,y)=e^xcosy

Answer 1

$\displaystyle u(x, y) = e^x \cos(y)\\ \frac{\partial u}{\partial x} = e^x \cos(y) \\ \textsf{A harmonic function satisfies the}\\\textsf{Cauchy Riemann's equation}\\ \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = e^x \cos(y)\\ \therefore v = \int e^x \cos(y)\, \mathrm{d}y = e^x\sin(y) + C\\ \begin{aligned} f(z) &= u(x, y) + iv(x, y) \\&= e^x(\cos(y) + i\sin(y)) + C \\&= e^x\cdot e^{iy} + C \\&= e^{x + iy} + C = e^z + C \end{aligned} \therefore \textsf{The original function is}\, f(z) = e^{z} + C\\ \textsf{where}\, C\, \textsf{is an arbitrary constant.}$