 Solution

For a harmonic function, put $a=u_{xx}$ and $b=u_{xy}=u_{yx}$. Since $u_{yy}=-u_{xx}$, its Hessian matrix isA nonzero determinant is therefore negative, and the two real Hessian eigenvalues are $\pm \sqrt{a^2+b^2}$. At a stationary point the quadratic term of the Taylor expansion has both signs. Consequently every such nondegenerate stationary point is a saddle point. This is the nondegenerate stationary points of a planar harmonic function criterion. The same argument applies to $v$, whose stationary points coincide with those of $u$. A zero Hessian determinant is outside this second-order classification.