What are some examples of functions that are differentiable (everywhere) in $\mathbb{R^2}$, but that are not differentiable in the complex plane? We got an example for homework, $f(z)=2xy$, and I was wondering if there were any others.
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1Almost any $\mathbb R^2 \to \mathbb R^2$ function you come up with randomly will fail to be complex differentiable. Complex differentiability is a pretty stringent condition. – Feb 10 '12 at 05:58
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4Your $f(x,y)=(2xy,0)$ is a special case of the fact that $f(x,y)=(u(x,y),0)$ is never complex differentiable unless $f$ is constant, which can be seen as a special case of the Cauchy-Riemann equations or the open mapping theorem. – Jonas Meyer Feb 10 '12 at 06:01
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2Related: How is $\mathbb{C}$ different than $\mathbb{R}^2$? – Jonas Meyer Feb 10 '12 at 06:04
2 Answers
As some commentors have pointed out, there are many, many other such functions. All that's required is that your function fail to satisfy the Cauchy-Riemann equations -- that is, if $f(z) = u(z) + iv(z)$, the function $f$ will fail to be complex differentiable just in case one of the following fails:
$u_x(z) = v_y(z)$
$u_y(z) = -v_x(z)$.
Here are a few classic examples that are easily seen to be real-differentiable:
$f(z) = \overline z$
$f(z) = |z|$
EDIT: although the latter fails to be real-differentiable on the axes (my mistake).

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2I don't think your second example is $\mathbb{R}$-differentiable everywhere... – Najib Idrissi Feb 10 '12 at 06:43
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$|z|$ is real differentiable at most points on the axes. The only point where it is not differentiable is the origin. – Jonas Meyer Feb 12 '12 at 23:10
A function from plane to plane is complex diffeeentiable if and only if its derivative is an orientation preserving orthogonal linear map. So take any matrix that either has negative determinant, rank 1, or the norm of its columns are not equal, is not complex differentiable anywhere.
Take any diagonal matrix for example with nonequal entries.
Conjugate of z is another example.

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