Let $f$ be a real valued harmonic function on $C,$ then Claim $g= \frac {\partial f}{\partial x} - i \frac {\partial f}{\partial y} $ as holomorphic and $h= \frac {\partial f}{\partial x} + i \frac {\partial f}{\partial y} $ as need not be holomorphic.

My Attempt

To prove g as holomorphic, we need to satisfy two conditions

  1. C.R Eqns : $u_x = \dfrac {\partial ^2 f} {\partial ^2 x} = v_y = -\dfrac {\partial ^2 f} {\partial ^2 y}$ ($\because$ f is homomorphic )

  2. $u_x, u_y, v_x, v_y$ exists and continous.

Here existence is guaranteed BUT HOW TO VERIFY THE CONTINUITY OF $u_x, u_y, v_x, v_y$

  • $\begingroup$ Calculate $\frac{\partial g}{\partial z}$. Use chain rule. $\endgroup$ – mlainz Oct 26 '15 at 2:44
  • $\begingroup$ What makes you think that the partial derivatives are both identically zero? $\endgroup$ – Giovanni Oct 26 '15 at 2:45
  • $\begingroup$ BTW, if it is identically zero, then it is holomorphic. $\endgroup$ – user99914 Oct 26 '15 at 2:46
  • $\begingroup$ Actually, it was a Multiple Choice question. The option identically zero is incorrect. $\endgroup$ – Rising Star Oct 26 '15 at 2:49
  • $\begingroup$ I mean derivative becomes zero if its a real valued function of a complex variable. $\endgroup$ – Rising Star Oct 26 '15 at 2:52

Since $f$ is harmonic, $f \in C^2$ by definition, so your function is automatically smooth enough to Cauchy-Riemann's equations (in the "other direction".)

  • $\begingroup$ Ya, existence of $ \dfrac {\partial ^2 f} {\partial ^2 x} $ is guaranteed but what abt continuity ? $\endgroup$ – Rising Star Oct 27 '15 at 13:50
  • 1
    $\begingroup$ No, $C^2$ means that all second-order partial derivatives exist and are continuous. $\endgroup$ – mrf Oct 27 '15 at 14:27
  • $\begingroup$ @mrf Sir Harmonic function need not satify Cauchy Reimann equation , So how can we talk about holomorphicity ? $\endgroup$ – MathLover Sep 21 '18 at 16:48

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