My question is about Taylors theorem but in the context of proving the following theorem (C-R equations).
Let $f = u + iv$ be defined on domain $D$ in $\mathbb{C}$, where $u$ and $v$are real-valued. Then $f(z)$ is analytic on $D \iff$ $u(x,y)$ and $v(x,y)$ have continuous first-order partial derivatives that satisfy $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$ and, $$\frac{\partial u}{\partial y} = \frac{-\partial v}{\partial x}.$$
In the part where, if the partial derivatives of $u$ and $v$ exist, are continuous, and satisfy the C-R equations, then $f = u + iv$ is analytic, you can use Taylor's theorem.
Fix $z \in D$. Taylors theorem with remainder provides an approximation $$u(x+\Delta x, y+\Delta y) = u(x,y) + \frac{\partial u}{\partial x}(x,y)\Delta x + \frac{\partial v}{\partial y}(x,y)\Delta y + R(\Delta x, \Delta y)$$ where $\frac{R(\Delta x, \Delta y)}{|\Delta z|} \to 0$ as $\Delta z \to 0$. You do similarly for $v(x+\Delta x, y+\Delta y)$ with the remainder $S(\Delta x, \Delta y)$.
Using the C-R equations and $\Delta z = \Delta x + i\Delta y$ we get $$f(z+\Delta z) = f(z) + (\frac{\partial u}{\partial x}(x,y) + i\frac{\partial v}{\partial x})\Delta z + R(\Delta z) + iS(\Delta z) \iff \frac{f(z+\Delta z) - f(z)}{\Delta z} = \frac{\partial u}{\partial x}(x,y) + i\frac{\partial v}{\partial x}(x,y) + \frac{R(\Delta z) + iS(\Delta z)}{\Delta z},$$ and $\frac{f(z+\Delta z) - f(z)}{\Delta z} \to \frac{\partial u}{\partial x}(x,y) + i\frac{\partial v}{\partial x}(x,y)$ as $\Delta z \to 0$.
My question is not so much about the C-R equations (they just provide context for my actual question) but my question is actually about Taylor's theorem. I can't recognize the form of the theorem at all, is this Taylor's theorem for several variables or a complex version?
Thanks.