0
$\begingroup$

Show that $h(z)=x^3+3xy^2-3x+i(y^3+3x^2y-3y)$ is differentiable on the coordinate axes but is nowhere analytic.

I know that if the Cauchy-Riemann equations do not hold then we can say that the complex function is nowhere analytic, however I don't know how to show that it is differentiable on the coordinate axes. The reason is because I am not sure I understand what it means to be differentiable on just the coordinate axes.

In this case we have
$\frac{\partial u}{\partial x}=3x^2+3y^2-3$
$\frac{\partial v}{\partial y}=3y^2+3x^2-3$
$\frac{\partial u}{\partial y}=6xy$
$\frac{\partial v}{\partial x}=6xy$
Since $\frac{\partial u}{\partial y}=\frac{\partial v}{\partial x}$ we can say that the function is nowhere analytic, but what how do I show that the function is differentiable on the coordinate axes?

  • 1
    One way is to use the limit definition. – 2017-02-26
  • 1
    Notice that $\partial u / \partial y = - \partial v / \partial x$ does hold on the coordinate axes! – 2017-02-26

1 Answers 1

0

"Differentiable on the coordinate axes" simply means differentiable at each $(x,y)$ satisfying $xy=0$. Since the C-R equations hold on the coordinate axes and the partials of $u,v$ are continuous, we are done.

  • 0
    I understand now, however if I am asked if a complex function is differentiable at let's say $(3,7)$, then all I have to do is get the partial derivatives for the C-R equations and substitute $x=3$ and $y=7$ and if the equations still hold then it is analytic at that point, if not it isn't analytic at that point? – 2017-02-26
  • 1
    Yes, provided the partial derivatives of $u$ and $v$ are continuous (as they are here). – 2017-02-26