$f$ is $\mathbb{R}$ - differentiable iff $Re(f)$ and $Im(f)$ are $\mathbb{R}$ - differentiable

Question

I just read that a sufficient condition for a function $f:A \rightarrow \mathbb{C},f(z) = u(z)+ i v(z)$ to be holomorphic is:

• $A$ open.
• f is $\mathbb{R}$ - differentiable in $A$.
• The Cauchy Riemann equations hold in $A$.

In the book i' m reading $\mathbb{R}$ - differentiability is defined as:

$f: D \rightarrow \mathbb{C}$ is $\mathbb{R}$ - differentiable in $z_0$ if there exists an $\mathbb{R}$ - linear map $T$ such that

$$\lim_{z \rightarrow z_0}\frac{f(z)-f(z_0)-T(z-z_0)}{| z-z_0 |}=0$$

Is this equivalent to saying that $Re(z)$ and $Im(z)$ (which are real valued) are differentiable in $A$?

Answer 1

You can look at a complex-valued function as being represented by two real-valued functions, say $f(z) = f(x,y) = u(x,y) + iv(x,y)$ for $z = x + iy$. Now we want $f$ to be real-differentiable before even considering the Cauchy-Riemann equations (in fact, we need both to be complex-differentiable). What does it mean for a function to be real-differentiable? This is simple in one variable, but you might want to look up what it means for a function to be differentiable in multiple variables and think about what the differential actually is (spoiler: it is basically the best linear approximation of our function at a point).