# Derived analytic function from real part

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 $f(z) = z^2$. –  GEdgar Jul 21 '11 at 0:40
@ 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 $1/2[(x+iy)^2+\overline{(x-iy)^2}]=1/2(x^2-y^2+2i\,x\,y)+1/2(\overline{x^2}-\ove‌​rline{y^2}+2i\,\bar{x}\,\bar{y})$, which is not equal to the left-hand side $x^2-y^2$. What's wrong with my argument? –  Paul Jul 21 '11 at 15:27
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. –  Did Jul 21 '11 at 16:33
@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. –  GEdgar Jul 21 '11 at 17:04
@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...). –  Did Jul 21 '11 at 17:12

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 $y$.
Sorry it still confused me. As you said that if $\bar{f}(z):=\overline{f(\bar{z})}$, then $\bar{f}(z)=z^2$. Why did you take $z=x-iy$ into that? $z$ should be $x+iy$? Thanks! –  Paul Jul 22 '11 at 1:17
The formula (1) has $x-iy$ in it. That is why I used it. Here, $x,y,z$ are all just variables, not assumed to be related to each other. –  GEdgar Jul 22 '11 at 15:12