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I saw Ahlfors's book Complex Analysis. It mentioned that analytic function $f(z)$ can be derived from a given real part $u(x,y)$, where $x$ and $y$ are real.

It said that $$ u(x,y)=\frac{1}{2}[f(x+iy)+\bar{f}(x-iy)]. \tag{1} $$

However, it mentioned that it is 'reasonable' that (1) holds even when $x$ and $y$ are 'complex'. Why?

I think that, if $x$ and $y$ are real, then real part $u(x,y)$ should be written down by $$ u(x,y)=\frac{1}{2}[f(z)+\bar{f}(\bar{z})], \tag{2} $$ where $z=x+iy$.

Hence, if $x$ and $y$ are complex, (2) should be equal to $$ u(x,y)=\frac{1}{2}[f(x+iy)+\bar{f}(\bar{x}-i\bar{y})]. \tag{3} $$ It confused me for a long time. Please help me.

Thanks!

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    Try some examples to see what is happening. $u$ has to be harmonic, so try for example $u(x,y) = x^2-y^2$ which corresponds to $f(z) = z^2$.2011-07-21
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    @ GEdgar. Thanks for your comment. I try your example. Assume that (1) holds when $x,y$ are complex. Then right-hand side of (1) is $1/2[(x+iy)^2+\overline{(x-iy)^2}]=1/2(x^2-y^2+2i\,x\,y)+1/2(\overline{x^2}-\overline{y^2}+2i\,\bar{x}\,\bar{y})$, which is not equal to the left-hand side $x^2-y^2$. What's wrong with my argument?2011-07-21
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    The function $u$ is defined on $\mathbb{R}\times\mathbb{R}$ hence the assertion that (1) holds for $x$ and $y$ complex is mysterious to me.2011-07-21
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    @Didier: As Ahlfors remarks in this derivation, for this method to work $u$ must make sense for complex values. For example $u$ could be a rational function of $x$ and $y$. Ahlfors 2nd edition, Chapter 2, the end of section 1.2.2011-07-21
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    @GEdgar: Thanks for the explanation. *To be a rational function of $x$ and $y$* is a well defined condition but *to make sense for complex values* is not. For example $u(x,y)=0$ for every real values $x$ and $y$ coincide with $U$ defined by $U(z_1,z_2)=z_1-\bar z_1$ for every complex values $z_1$ and $z_2$. One sees that $u$ *makes sense for complex values* in a lot of different ways (although $U$ is probably not the one to have in mind here...).2011-07-21
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    Of course $u$ must be harmonic, hence $C^\infty$. Maybe if it is analytic in two variables $x,y$ that will do. For example, try $u(x,y) = e^x\cos(y)$.2011-07-21

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