# Why is $z = x+iy \mapsto x$ not differentiable in $0 \in \mathbb{C}$?

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?

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Think about what happens if you approach $0$ along the $y$-axis. Indeed, taking limits along the $x$ and $y$-axes is precisely how the Cauchy-Riemann equations are derived. –  Santiago Canez Apr 11 '11 at 18:35
What you said about the Cauchy-Riemann equations is not quite 100% accurate. It is true that if a function is complex differentiable at a point, then the C-R equations are satisfied at that point, but not necessarily conversely. If a function has partial derivatives that satisfy the C-R equations at every point of an open subset of the plane, then the function is complex differentiable on that set. (In the case in your question, the function is complex differentiable nowhere.) –  Jonas Meyer Apr 11 '11 at 18:43
The first limit you posted is not 1, it actually doesn't exist. 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 exist. That's a pretty standard 2-variables limit. –  N. S. Apr 11 '11 at 18:57

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 $0$.
For $z = 0 + it$ and $t \to 0$, is $\frac{\Re(z)}{z}=0$? If so, then I believe I understood where I was wrong. –  Huy Apr 11 '11 at 20:58
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). –  wildildildlife Apr 11 '11 at 22:46
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 $$\Big( \begin{array}{cc} a&-b\\ b&a\\ \end{array} \Big)$$