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How can I immediately demonstrate that $z+\overline z$ is not an analytic function?

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closed as off-topic by BLAZE, 6005, T. Bongers, Silvia Ghinassi, colormegone Jan 26 '16 at 5:25

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    $\begingroup$ Because sums of analytic functions are analytic. $\endgroup$ – T. Bongers Jan 20 '16 at 22:39
  • $\begingroup$ @T.Bongers so basically it means that because of that $\overline z$ is not analytic the sum is not? and $\overline z$ is not analytic because of CR don't hold? $\endgroup$ – Sijaan Hallak Jan 20 '16 at 22:43
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    $\begingroup$ @SijaanHallak T. Bongers meant that if $z + \bar z$ was analytic then also $z + \bar z + (-z) = \bar z$ was analytic, which is not. $\endgroup$ – Ant Jan 20 '16 at 22:59
  • $\begingroup$ @Ant I see that! Thanks $\endgroup$ – Sijaan Hallak Jan 20 '16 at 23:01
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Cauchy-Riemann equations don't hold. So no. It's not analytic.

$$z=u+iv=2x+0i$$ $$\frac{\partial u}{\partial x}=2\neq \frac{\partial v}{\partial y}=0$$

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  • $\begingroup$ Thanks! I needed a simple answer like this ;) $\endgroup$ – Sijaan Hallak Jan 20 '16 at 22:59
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$z+\overline{z}$ is a real non constant function of z. The only real valued functions on $\mathbb{C}$ that are analytic are constant (open mapping propty).

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    $\begingroup$ Thanks!! but I still didnt learn what open mapping propty is. Going to read about it now $\endgroup$ – Sijaan Hallak Jan 20 '16 at 22:59
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Your function is equal to $2x$ in the real axis. By analytic continuation, it should be $2z$ in the whole complex line. But this is clearly not true.

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The zeros of an analytic function are isolated, but the zeros of $z+\overline{z}$ is the whole imaginary axis.

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Note that a complex function is analytic if and only if it is holomorphic. The function $f(z)=z+\bar{z}=2Re(z)$ , but the function $g(z)=Re(z)$ is not holomorphic.

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  • $\begingroup$ Up until this moment I thought holomorphic means analytic!! Thanks $\endgroup$ – Sijaan Hallak Jan 20 '16 at 23:00
  • $\begingroup$ @SijaanHallak which indeed it is :) Did you misunderstand something? $\endgroup$ – Ant Jan 20 '16 at 23:01
  • $\begingroup$ @Ant I've been taught that holomorphic is a different meaning of analytic..is that right? $\endgroup$ – Sijaan Hallak Jan 20 '16 at 23:02
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    $\begingroup$ the definitions are different, but one of the most important theorems in complex analysis says that a function is holomorphic iff and only if is analytics (which is want Vincenzo Zaccaro wrote) (en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions) $\endgroup$ – Ant Jan 20 '16 at 23:05
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    $\begingroup$ @Ant Yeah, I told you I've been taught wrong maybe because we don't need it in our course.. Thanks alot really $\endgroup$ – Sijaan Hallak Jan 20 '16 at 23:08
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One equivalent way of defining analytic is that

$$\frac{\partial f}{\partial \bar{z}} =0 $$

or

$$\frac{\partial f}{\partial z} = \frac{\partial f}{\partial x}. $$

So since

$$\frac{\partial (z+\bar{z})}{\partial \bar{z}} =1, $$ it's not analytic. FYI these are some good properties to remember for exams.

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