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The trouble is that complex conjugation is not holomorphic (analytic) over the complex numbers. It's incredibly useful as an operator, and it appears all over
by andrewla 2y ago
The trouble is that complex conjugation is not holomorphic (analytic) over the complex numbers.
It's incredibly useful as an operator, and it appears all over the place in Hilbert spaces and other complex-probability situations, but fundamentally it defies attempts to use it for analytic purposes (like differential equations or contour integration).
Useful when you need to treat the complex plane as a vector space or are interested in the topology of a complex function, but a pain to deal with in almost any other context.
- rachofsunshine 2y agoFor 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 agoConstant 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 agoPicard'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 agoNo need to apologize, 0 == -0 even if we are talking about floats.
- amelius 2y agoIs 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 agoFor 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.
- Tainnor 2y ago> The trouble is that complex conjugation is not holomorphic (analytic) over the complex numbers. Even more generally, a real-differentiable function is complex differentiable iff its (Wirtinger) derivative with respect to z* is 0.
- kevinventullo 2y agoMaybe worth noting that it is real-analytic.
- aabajian 2y agoDoes the problem boil down to selecting the real part of a complex number? Correct me if I'm wrong, but the conjugate of z is merely 2*Re(z) - z. For example, the conjugate of a + bi is 2*a - (a + bi) = a - bi.
- RandomThoughts3 2y agoThere is no problem. This is just a non holomorphic function of C to C. Differentiable functions are the exception not the norm (no pun intended).