Partial derivatives of a complex valued real variable function

Let $\Delta_R$ be a disc centered at the origin on the complex plane.

Let $f:\Bbb R\to \Delta_R$ defined as,

$f(x)=u(x) + i v(x)$

The following are my questions:

$Q1)$ What can we say about $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y }$ for the points on disc?Will they be $0$?

$Q2)$ I am a little confused because, as $\Delta_R$ is not even the domain of $f$ then how can we talk about the partial derivatives of $f$ on $\Delta_R$?

Solution 1:

Main ideas: The actual question is in your comment. What you wrote in the question doesn't make any sense, as you yourself noticed in Q2).

You are given a real valued function $u$ in an open disc $D$ with the property that $f=u^{-1} + iu$ is analytic on $D.$ Notice that $u$ is never $0$ in $D.$ Because $f$ is analytic in $D,$ the CR equations tell us

$$ (u^{-1})_x = u_y,\,\, (u^{-1})_y = -u_x.$$

Now $(u^{-1})_x = -u^{-2}u_x,$ etc. Play around with this. Your strategy is to show $u_x=u_y \equiv 0.$

Another approach, if you know the open mapping theorem: The range of $f$ is contained in the curve $\{(1/x,x): x\in \mathbb R, x\ne 0\}.$ That is not an open set.