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Consider a surface somewhat like bottom of a boat. Imagine cutting a paraboloid (see picture) in half, pulling the halves at a distance and joining the two halves with a matching parabolic cylinder (see picture).

In other words consider $f(z)$ for $z=x+iy$ as follows:

$f(z) = x + 0i$, when $-10 \le x \le 10$ and $y = 0$

$f(z)=x + iy^2$, when $-10 \le x \le 10$ and $y \ne 0$

$f(z)=(x+10)^2 + iy^2$, when $x < -10$

$f(z)=(x-10)^2 + iy^2$, when $x > 10$

Is this surface holomorphic at all points on the segment connecting $z1=10+0i$ and $z2=-10+0i$, (including the end points).

Thanks

Shree
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    Holomorphic functions go from $\mathbf{C}$ (or $\mathbf{C}^n$) to $\mathbf{C}$. This one goes from $\mathbf{R}^2$ to $\mathbf{R}$, so how could it be holomorphic? – Hans Lundmark Jan 01 '17 at 10:07
  • @HansLundmark Thanks. I have made changes to make the function go from C to C. – Shree Jan 02 '17 at 03:41

2 Answers2

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As currently written, your function is not even continuous at points on the real axis between $-10$ and $10$ (except at $z=0$), so it's certainly not complex differentiable.

Due to the identity theorem, attempts to define a complex function "piecewise" will almost never produce a holomorphic function. If any of the "pieces" has a natural continuation into an area covered by a neighboring piece, which differs from that other piece there, then the stitched-together function cannot be holomorphic.

hmakholm left over Monica
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In this case the fibre of $0$ under $f$ (a.k.a. $f^{-1}[\{0\}]$) has a limit point. By the Identity Theorem, $f$ must be zero everywhere if it is holomorphic, but it is not.

Henricus V.
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