Suppose I have a function defined from R to C: $f(t)=R(t)+iI(t)$

When someone wants to show that the above function is differentiable are they talking about complex differentiability or vector differentiation?

If this function is complex differentiable i.e $\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} $ is it also vector differentiable i.e $f'(t)=R'(t)+iI'(t)$.

I think I could be confused with the definition of complex differentiability. Is it only for functions defined on an open connected subset of C? Thus the function above cannot be complex differentiable since its only defined on R which is not open?

  • $\begingroup$ It is 'normal' differentiation since the domain is real. It is equivalent to $R,I$ being separately differentiable. $\endgroup$ – copper.hat Aug 14 '18 at 4:24

Functions $f=u+iv$ of type ${\mathbb R}\to{\mathbb C}$ can have a derivative $$f'(t)=\lim_{h\to0}{f(t+h)-f(t)\over h}\ .\tag{1}$$ Here the limit is a limit in ${\mathbb C}$, but the dummy variable $h$ is real only. According to the rules about limits in ${\mathbb C}$ or ${\mathbb R}^n$ this limit can be taken "coordinatewise". Since the $h$ in the denominator is real, the real and imaginary parts of $f$ don't get intermingled during the process. Therefore we have $$f'(t)=u'(t)+iv'(t)\ .$$ Sometimes (e.g., if $f=\exp$, or if $f$ is a polynomial) such a function $f$ can be extended to the complex domain ${\mathbb C}\supset{\mathbb R}$ whereby it becomes an analytic function of the complex variable $z=t+i \tau$. Only then it makes sense to consider the complex derivative $$f'(z):=\lim_{\Delta z\to0}{f(z+\Delta z)-f(z)\over\Delta z}\ .$$ Here $\Delta z$ goes to $0$ "from all directions". If $z=t\in{\mathbb R}$ happens to be a real point then the complex derivative $f'(z)$ coincides of course with the $f'(t)$ defined in $(1)$.


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