1
$\begingroup$

Let $f(z):=u(x,y)+iv(x,y)$ be a complex-valued function. It is known that the associated Cauchy-Riemann equations are

$$u_x=2x, u_v = 1$$ $$v_x = -1, v_y = 2$$

So $f$ is differentiable at $z=1+iy$, $y\in\mathbb{R}$. Does this mean that $f$ is holomorphic at the same points? The definition of holomorphicity states that a holomorphic function is complex-differentiable in the neighbourhoods of all the points in its domain, but here we have a line, a closed set.

Would appreciate a clarification.

  • 0
    Hint: Can you find an open neighborhood of any point to make it holomorphic in it? – 2017-02-10
  • 0
    In complex analysis, a neighborhood of $a$ is $\{z \in \mathbb{C}, |z-a|< \epsilon\}$ for some $\epsilon > 0$, and the open sets containing it – 2017-02-10
  • 0
    http://math.stackexchange.com/questions/464217/clarification-of-use-of-cauchy-riemann-equations?rq=1 – 2017-02-10
  • 0
    @ZTransformer There is no open neighbourhood of any point on the line in which the function could be holomorphic, but can a function be holomorphic at just one point, and not at some neighbourhood of that point? – 2017-02-10

0 Answers 0