By using one of the Cauchy Riemann equations we have ∂x∂v=−∂y∂u=−eycosx, and partially integrating with respect to x gives v=−eysinx+C(y). Partially differentiating this expression and using the otherCauchy Riemann equation gives ∂y∂v=−eysinx+C′(x)=∂x∂u=−eysinx. It follows that C′(x)=0, and hence C(x)=C is constant. We conclude that harmonic conjugate of u(x,y)=eycosx is v=−eysinx+C.
Comments