A Functional equation in Complex variables 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 ....
 A: 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.
