Let $f:\mathbb{C}\to\mathbb{C}$ be given by $z\mapsto\Re\left(z\right)$ . Where is $f$ differentiable? $f$ is definitely differentiable on $\mathbb{R}$ because on these values $f\left(z\right)=z$ . As for $\mathbb{C}\backslash\mathbb{R}$ , I am not sure. I am guessing it's not, but I can't prove it. The way I'm trying to show it is by constructing a sequence such that $z_{n}\to z$ but $\frac{\Re\left(z_{n}\right)-\Re\left(z\right)}{z_{n}-z}$ does not converge. Any help in showing this would be appreciated.
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1$\begingroup$ What do you mean by "differentiable"? Real differentiable or complex differentiable? $\endgroup$– Giuseppe NegroOct 28, 2012 at 18:37
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$\begingroup$ I mean complex differentiable. I already see a mistake in the above argument for differentiability on $\mathbb{R}$, thanks to Marvis. $\endgroup$– GaloisOct 28, 2012 at 18:41
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$\begingroup$ Another approach: it is real-differentiable but it does not satisfy the Cauchy-Riemann conditions at any point. $\endgroup$– ManzanoOct 28, 2012 at 19:56
3 Answers
Consider $z \in \mathbb{C}$. Approach $z$ along $z + ih$ and $z + h$, where $h \in \mathbb{R}$ and see what happens to the limit $$\dfrac{f(z+h)-f(z)}{h} \, \,\,\,\,\, \text{ and }\dfrac{f(z+ih)-f(z)}{ih}$$
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2$\begingroup$ I believe f is not differentiable anywhere and this is my take: For $h\in\mathbb{R}$ $$f'\left(z\right)=\lim_{h\to0}\frac{\Re\left(z+h\right)-\Re\left(z\right)}{h}=\lim_{h\to0}\frac{\Re\left(z\right)+\Re\left(h\right)-\Re\left(z\right)}{h}=\lim_{h\to0}\frac{h}{h}=1$$ $$f'\left(z\right)=\lim_{h\to0}\frac{\Re\left(z+ih\right)-\Re\left(z\right)}{h}=\lim_{h\to0}\frac{\Re\left(z\right)+\Re\left(ih\right)-\Re\left(z\right)}{h}=\lim_{h\to0}\frac{0}{h}=0,$$ and hence $f'\left(z\right)$ cannot exist. $\endgroup$– GaloisOct 28, 2012 at 18:47
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1$\begingroup$ @Galois Exactly. $\endgroup$– user17762Oct 28, 2012 at 19:15
Cauchy-Riemann equations are satisfied nowhere in the complex plane. Therefore, it is nowhere (complex) differentiable (holomorphic).
Using the definition (as in the answer by Marvis) is what one should do when learning the subject. After a while, it becomes easier to do the following: write the Cauchy-Riemann equation is $\frac{\partial}{\partial \bar z}f=0 $ and observe that $f(z)=\frac{1}{2}(z+\bar z)$ has $\frac{\partial}{\partial \bar z}f=\frac12 $ instead of $0$.