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Today we had an online-test and one of the question was whether the function $z = x+iy \mapsto x$ is differentiable in $0 \in \mathbb{C}$. I thought I'd check it using our definition of complex differentiability and I came to the conclusion, that $\lim_{z \to 0} \frac{\Re(z)}{z}=1.$ I even checked it via wolframalpha afterwards and it told me so as well. However, this was considered to be wrong, because apparently the Cauchy-Riemann equations are not fulfilled, that is to say $1 = \partial_x \Re f(0) \neq \partial_y \Im f(0)=0.$ First, I checked wikipedia to find out what these equations looked like and I found that indeed, "a complex function is differentiable in a point" is equivalent to "a complex function fulfils the Cauchy-Riemann equations". This seems inconsistent to me. I don't see how the limit does not exist. Why does it not exist?

Thanks for any answer in advance.

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    The first limit you posted is not 1, it actually doesn't e$x$ist. You can prove it by using two paths, or you can also conjugate the denominator and see why the two (real and imaginary) limits don't e$x$ist. That's a pretty standard 2-varia$b$les limit.2011-04-11

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As others have hinted, I think that the point you're missing is that for such a limit to exist, it has to exist and have the same value regardless of how the variable $z$ approaches 0. The definitions are consistent :-) It's your assertion that $\lim_{z\to 0} \Re(z)/z = 1$ that is false. Consider $z=0+it$ and let $t$ tend to $0$.

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    Yes, because $Re(0+it)=0$. Btw, I think computer programs like Wolphram Alpha generally have a hard time with multidimensional limits. I suspect they just take a particular path, or treat it as a repeated limit (while switching the order is often not allowed).2011-04-11
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this function is linear and its derivative $\mathbb{R}^2\to\mathbb{R}$ is $\Big( \begin{array}{cc} 1&0\\ 0&0\\ \end{array} \Big) $ which is not a complex number, ie not of the form $\Big( \begin{array}{cc} a&-b\\ b&a\\ \end{array} \Big) $