Question

Why isn't the complex conjugate of z complex differentiable?

Answer 1

Because the complex limit [f(z+h)-f(z)]/h doesn't converge, where f(z)=z^(\*). For h=x+iy, this expression simplifies to (x-iy)/(x+yi). Notice that this does approach different values, depending on how x and y approach 0

Equivalently if you already know what the Riemann cauchy differential equations are, it's easy to show that f doesn't fulfil them and thus is not complex differentiable

Answer 2

f(z) = z* is certainly differentiable when viewed as a function from R^2 to R^(2). But complex differentiability is a stronger condition: it means that locally, the function looks like multiplication by a complex number. But geometrically z* is a reflection, while multiplication by a complex number is a rotation and dilatation - no complex number will give you a reflection.

However, compared to uglier functions, the conjugate is a bit of a special case - it's almost complex differentiable, it just kind of fails for an almost trivial reason. And this is why the concept of an antiholomorphic function exists: a function which is complex differentiable with respect to to z*.