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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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    Another approach: it is real-differentiable but it does not satisfy the Cauchy-Riemann conditions at any point.2012-10-28

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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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    @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 $0$.