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If I have a function $f(z)$ defined on a domain $D$ in the complex plane that is not continuous at a point $z_0 \in D$, can $f$ be analytic in the region $D$? or I guess another way to phrase this would be can $f$ have a derivative at $z_0$?

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    $\begingroup$ No, because having a derivative there implies continuity there! $\endgroup$ – Shashi Jan 28 '18 at 21:09
  • $\begingroup$ Ok thank you! I was not sure if this carried from real analysis $\endgroup$ – Abrb Jan 28 '18 at 21:11
  • $\begingroup$ @Abrb Identify $\mathbb{C}$ with $\mathbb{R}^2$ math.stackexchange.com/questions/1187000/… $\endgroup$ – mysatellite Jan 28 '18 at 21:13
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$\ f\ is\ analytic \ on \ region \ D \Rightarrow\ f \ is \ continuous \ on \ D $. Therefore $\ f $ cannot be analytic on D if there's a point in D in which $\ f$ is not continuous.

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Generally speaking, you can always rewrite f(z) as: u(x,y) + i*'v(x,y), and see (at any point z = x + iy you like) if: du/dx +idv/dx = du/dy - i*dv/dy. If yes then it is analytic at that point.

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  • $\begingroup$ if you take $f(z)=log(z)$ how would you prove that it doesn't satisfy CREs on the negative real axis? $\endgroup$ – Abrb Jan 28 '18 at 21:59
  • $\begingroup$ Simple: log(Z) = log(abs(Z)) + j*atan2(im(Z) / real(Z)) you now substitude u with the first term and v with the second term, and apply my formula from above. $\endgroup$ – Nadav Carmel Apr 29 '18 at 12:23

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