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.
Where is $f(z)=\Re (z)$ differentiable?
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complex-analysis
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1What do you mean by "differentiable"? *Real* differentiable or *complex* differentiable? – 2012-10-28
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0I mean complex differentiable. I already see a mistake in the above argument for differentiability on $\mathbb{R}$, thanks to Marvis. – 2012-10-28
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0Another approach: it is real-differentiable but it does not satisfy the Cauchy-Riemann conditions at any point. – 2012-10-28
3 Answers
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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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2I 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. – 2012-10-28
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1@Galois Exactly. – 2012-10-28
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Cauchy-Riemann equations are satisfied nowhere in the complex plane. Therefore, it is nowhere (complex) differentiable (holomorphic).
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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 $0$.