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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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    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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The way it is used here, given a function $f$, a new function $\overline{f}$ can be defined by $ \overline{f}\big(z\big) = \overline{f(\overline{z})} . $ For example, if $f$ is a polynomial, change all coefficients to their complex conjugates, but leave the variable alone.

Let's try $f(z) = z^2$ as I suggested above. So $\overline{f}(z) = z^2$. Then $f(x+iy) = (x+iy)^2 = (x^2-y^2)+2ixy$; $\overline{f}(x-iy) = (x-iy)^2=(x^2-y^2)-2ixy$; so $ \frac{1}{2}\big[f(x+iy)+\overline{f}(x-iy)\big] = x^2-y^2 . $ As required, this holds even for complex $x$ and $y$.

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    @ GEdgar: Thanks $f$or your help~2011-07-22