Answer to Question #220631 in Complex Analysis for Yashu

Question

harmonic conjugate of u(x,y)=e^y cosx is

Answer 1

By using one of the Cauchy Riemann equations we have $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=-e^y\cos x,$ and partially integrating with respect to $x$ gives $v=-e^y\sin x+C(y).$ Partially differentiating this expression and using the otherCauchy Riemann equation gives $\frac{\partial v}{\partial y}=-e^y\sin x+C'(x)=\frac{\partial u}{\partial x}=-e^y\sin x.$ It follows that $C'(x)=0,$ and hence $C(x)=C$ is constant. We conclude that harmonic conjugate of $u(x,y)=e^y \cos x$ is $v=-e^y\sin x+C.$