Answer to Question #242986 in Complex Analysis for Msa

Question

Show that u(x,y)=e^xcosy and v(x,y)=e^xsiny are harmonic for all values of (x,y).

Answer 1

Let us show that $u(x,y)=e^x\cos y$ and $v(x,y)=e^x\sin y$ are harmonic for all values of $(x,y).$ Taking into account that $\frac{\partial u(x,y)}{\partial x}=e^x\cos y, \frac{\partial u^2(x,y)}{\partial x^2}=e^x\cos y, \frac{\partial u(x,y)}{\partial y}=-e^x\sin y, \frac{\partial u^2(x,y)}{\partial y^2}=-e^x\cos y,$

we conclude that $\frac{\partial u^2(x,y)}{\partial x^2}+\frac{\partial u^2(x,y)}{\partial y^2}=e^x\cos y-e^x\cos y=0,$ and hence the function $u(x,y)=e^x\cos y$ is harmonic.

Since $\frac{\partial v(x,y)}{\partial x}=e^x\sin y, \frac{\partial v^2(x,y)}{\partial x^2}=e^x\sin y, \frac{\partial v(x,y)}{\partial y}=e^x\cos y, \frac{\partial v^2(x,y)}{\partial y^2}=-e^x\sin y,$

we conclude that $\frac{\partial v^2(x,y)}{\partial x^2}+\frac{\partial v^2(x,y)}{\partial y^2}=e^x\sin y-e^x\sin y=0,$ and hence the function $v(x,y)=e^x\sin y$ is harmonic.