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

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 $$A$$?

• yes, it is the same – Masacroso Apr 15 at 10:41

## 1 Answer

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).