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I have been stuck on this problem for a long time : If $f(z)=u(x,y)+iv(x,y)$ , prove that

a. $f(z)=2u(z/2,(-iz)/2) +$ constant

b.$f(z)=2iv(z/2,(-iz)/2) +$ constant

This result seems very interesting in itself ( nothing that I have done comes close to a solution) .Any help would be greatly appreciated ....

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  • $\begingroup$ Is $f$ supposed to be analytic? For a general $f$ this is clearly not true, consider for instance $f(z) = 0 + i (x + y )$, then of course (a) would imply $f(z) =$ constant. which is clearly not true. $\endgroup$ – Rogelio Molina May 2 '15 at 5:30
  • $\begingroup$ Nothing is assumed about f being analytic ....Could you by any chance help me understand the very last part of Mr Chappers proof ?? $\endgroup$ – herashefat May 2 '15 at 5:41
  • $\begingroup$ If $f$ is arbitrary then then claim is false, I just gave a counterexample, this cannot be proven by any means. $\endgroup$ – Rogelio Molina May 2 '15 at 17:27
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Let $z=x+iy$, $w=x-iy$. Inverting these equations, we find that $$ x = \frac{1}{2}(z+w), \qquad y = \frac{1}{2i}(z-w), $$ so, substituting in, $$ f(z) = u\left( \frac{z+w}{2},\frac{z-w}{2i} \right) + i v\left( \frac{z+w}{2}, \frac{z-w}{2i} \right). \tag{1} $$ Setting $z=z_0$ gives $$ f(z_0) = u\left( \frac{z_0+w}{2},\frac{z_0-w}{2i} \right) + i v\left( \frac{z_0+w}{2}, \frac{z_0-w}{2i} \right), $$ and taking the complex conjugate gives $$ \overline{f(z_0)} = u\left( \frac{\bar{z}_0+\bar{w}}{2},\frac{\bar{w}-\bar{z}_0}{2i} \right) - i v\left( \frac{\bar{z}_0+\bar{w}}{2}, \frac{\bar{w}-\bar{z}_0}{2i} \right), $$ and setting $w=\bar{z}$ gives $$ \overline{f(z_0)} = u\left( \frac{z+\bar{z}_0}{2},\frac{z-\bar{z}_0}{2i} \right) - i v\left( \frac{z+\bar{z}_0}{2}, \frac{z-\bar{z}_0}{2i} \right), \tag{2} $$ Now, the left-hand side of (1) is independent of $w$, so we can set it to whatever we like; choose $\bar{z}_0$, which gives $$ f(z) = u\left( \frac{z+\bar{z}_0}{2},\frac{z-\bar{z}_0}{2i} \right) + i v\left( \frac{z+\bar{z}_0}{2}, \frac{z-\bar{z}_0}{2i} \right). \tag{3} $$

Now add and subtract (2) and (3) to get $f(z)$ in terms of just $u$ or $v$. Set $z_0=0$ to get what you want.

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  • $\begingroup$ Does (1) hold only for $w=\bar{z}$? (Line 1, "Let $z=x+iy$, $w=x−iy$.", in other words, "Let $w=\bar{z}$") If yes, then how "the left-hand side of (1) is independent of $w$". If no, then how to obtain it for an arbitrary $w$? (To say, we use only the case $w=\bar{z}$, don't we?) $\endgroup$ – Alexey Burdin May 2 '15 at 1:02
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    $\begingroup$ @AlexeyBurdin The problem with that is what to do when you set $z=z_0$: you have to stop setting $w=\bar{z}_0$ at the same time or you can't get to the result. In complex variables we sometimes treat $z$ and $\bar{z}$ as independent (see writing the C-R equations as "$\partial f/\partial \bar{z} = 0$", but that's quite hard to deal with here. I've never been very happy with this proof, but somehow it does work. $\endgroup$ – Chappers May 2 '15 at 1:17
  • $\begingroup$ Thanks Mr Chappers for your effort for solving this , but I cannot follow some of this , After taking the Complex conjugate why did the second variable change in u and v??? (looks like it was multiplied by negative 1) Sorry if this question is elementary..... $\endgroup$ – herashefat May 2 '15 at 3:20
  • $\begingroup$ $u$ and $v$ are real functions, so you take the complex conjugate inside and apply to of both of their arguments: specifically, it's because $\bar{i}=-i$ that the second argument changes. $\endgroup$ – Chappers May 2 '15 at 3:26
  • $\begingroup$ Thanks again for your comment Ok so I understood your remark ,Now in equation 2 setting w=z (bar) ,why is the bar missing in (2) over z ?? $\endgroup$ – herashefat May 2 '15 at 3:49

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