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I know of some examples (ie: re z, Im z, z bar, etc) that are harmonic but not analytic but I am not sure why that is the case. Can someone explain?

A.Vall
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    Do you mean complex analytic? Not all harmonic functions are complex analytic because satisfying the Cauchy-Riemann equations is stronger than satisfying Laplace's equation. For example, there exist purely real nonconstant functions which are harmonic on the whole plane, whereas no such function can be complex analytic. All harmonic functions are real analytic, however. – Vik78 Apr 18 '17 at 19:13
  • Check out this link. – Martin Apr 18 '17 at 19:36
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    This is not too much different from the question "Why aren't all continuous functions differentiable? I have seen some examples of this, but I am not sure why that is the case." – zhw. Apr 18 '17 at 21:15
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    $f(z)=f(x+iy)$ is complex analytic if for every $z_0 = x_0+iy_0$ there exists $r$ such that for $|z-z_0| < r$ : $$f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!} (z-z_0)^n$$ $f(z)$ is real analytic if for $|z-z_0| < r$ : $$f(z) = \sum_{n=0}^\infty \sum_{m=0}^\infty\frac{\partial_x^n \partial_y^m f(z_0)}{n!} (x-x_0)^n(y-y_0)^n$$ harmonic means real-analytic and $\nabla^2 f = 0$ – reuns Apr 19 '17 at 00:00
  • It might be equivalent to say that a real (complex) analytic function can be locally expressed as power (Laurent) series? – MathArt Dec 13 '23 at 13:37

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