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For those of you wondering why on Earth such a simple function wouldn't be well-behaved, it can help to think about what you can do with the conjugate. In part

For those of you wondering why on Earth such a simple function wouldn't be well-behaved, it can help to think about what you can do with the conjugate. In particular, since sums of holomorphic functions are holomorphic, the sum f(z) = z + z* of the identity function g(z) = z and conjugate function h(z) = z* would need to be holomorphic. But z + z* cancels out the imaginary part of z and therefore maps the complex plane to the real line - and it should be obvious that a map from C to R cannot possibly preserve the properties of C in any intuitive way. Roughly speaking, the conjugate fails to be holomorphic because it maps C to its mirror image, rather than C itself, in much the same way that reflecting a circle is different from rotating it.

bobbylarrybobby 2y ago
Constant functions are holomorphic without preserving the properties of C The issue is that the Cauchy Riemann equations don't hold. (You would need 1==-1)
- rachofsunshine 2y ago
  Picard's theorem tells us constant functions are the only example of that, though (more or less because 0 == -0 [1] and that's only true for 0). [1] I sincerely apologize to anyone who works with floats for this statement.
  - eapriv 2y ago
    No need to apologize, 0 == -0 even if we are talking about floats.
amelius 2y ago
Is there an analytic function that approximates the conjugate, and can this approximation be made better and better by making the function more complicated? Edit: no, it can't be done because the Cauchy-Riemann equations make no sense here.
- gunnihinn 2y ago
  For any smooth function (like the conjugate) it makes sense to ask whether there exist holomorphic functions that approximate it arbitrarily well. However: Suppose that for every n > 0 there exists a holomorphic function f_n such that |f_n(z) - z| < 1/n for all z. Then |f_n(z)| <= |f_n(z) - z| + |z*| = |z| + 1/n by the triangle inequality. A consequence of Liouville's theorem is that any entire holomorphic function with polynomial growth is a polynomial; here in particular we would need to have f_n(z) = a_n z + b_n for some complex numbers a_n and b_n. For real x we would have |(a_n - 1)x + b_n| < 1/n for all x, so a_n = 1. For imaginary iy we would have |(a_n + 1)iy + b_n| < 1/n for all y, so a_n = -1, which is a contradiction. In fact, if a sequence of holomorphic functions converges uniformely on compact sets, the limit is itself holomorphic because of Cauchy's theorem.

bobbylarrybobby 2y ago

Constant functions are holomorphic without preserving the properties of C The issue is that the Cauchy Riemann equations don't hold. (You would need 1==-1)

rachofsunshine 2y ago
Picard's theorem tells us constant functions are the only example of that, though (more or less because 0 == -0 [1] and that's only true for 0). [1] I sincerely apologize to anyone who works with floats for this statement.
- eapriv 2y ago
  No need to apologize, 0 == -0 even if we are talking about floats.

amelius 2y ago

Is there an analytic function that approximates the conjugate, and can this approximation be made better and better by making the function more complicated? Edit: no, it can't be done because the Cauchy-Riemann equations make no sense here.

gunnihinn 2y ago
For any smooth function (like the conjugate) it makes sense to ask whether there exist holomorphic functions that approximate it arbitrarily well. However: Suppose that for every n > 0 there exists a holomorphic function f_n such that |f_n(z) - z| < 1/n for all z. Then |f_n(z)| <= |f_n(z) - z| + |z*| = |z| + 1/n by the triangle inequality. A consequence of Liouville's theorem is that any entire holomorphic function with polynomial growth is a polynomial; here in particular we would need to have f_n(z) = a_n z + b_n for some complex numbers a_n and b_n. For real x we would have |(a_n - 1)x + b_n| < 1/n for all x, so a_n = 1. For imaginary iy we would have |(a_n + 1)iy + b_n| < 1/n for all y, so a_n = -1, which is a contradiction. In fact, if a sequence of holomorphic functions converges uniformely on compact sets, the limit is itself holomorphic because of Cauchy's theorem.