Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
1  
@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
show 1 more comment

1 Answer

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 $y$.

share|improve this answer
    
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
    
@ GEdgar: Thanks for your help~ –  Paul Jul 22 '11 at 20:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.