Let $(, )$ be a harmonic function which is the real part of a holomorphic function $ ()$, so that $$(, )=\frac{1}{2}(f(z)+\overline {f(z)})$$
Argue that $\overline {()} = g(\bar z)$ for some function $$, and hence that:
$(, )=\frac{1}{2}( ( + ) + ( − ))$. Use this to show that it still holds even if $x,y$ are complex. Part 2) Assume that $ (0)$ is real. Show that in this case $(0) = (0) = (0, 0)$, and derive the formula: $$f(z)=2u(\frac{1}{2}z,\frac{1}{2i}z)-2u(0,0)$$
Arguement $\overline {()} = g(\bar z)$
$\overline {f(x+iy)}=u(x,y)-iv(x,y)$ which should be equal to $g(\bar z)=g(x-iy)=p(x,-y)+is(x,-y)$. I can see that if I take $p$ to be the same function as $u$ but multiply all $y$'s by $-1$. To make the imaginary parts the same I should multiply $s$ by $-1$ and $replace $-y$ by $i$. Is this a convincing arguement?
For Complex $x,y$
Take $x=\frac{1}{2}z$ and $y=\frac{1}{2i}z$. Then $$(\frac{1}{2}z, \frac{1}{2i}z)=\frac{1}{2}( (\frac{1}{2}z + \frac{1}{2i}z) + (\frac{1}{2}z − \frac{1}{2i}z))=\frac{1}{2}(f(z)-g(0))$$
Part 2
Why is $f(0)$ real? Can't the fuction simply be $f(x,y)=5x+i5y+10i$ in which case at $(0,0)$ I have $10i$?
After little rearranging I get:
$f(z)=2u(x,y) -\overline {f(z)}$. If $f(0)$ is real, then $g(0)$ doesn't exist, which means that $g(0)=0$, wouldn't it mean that $y=0$ since $y$ is our imaginary part in $z=x+iy$?
I would be gratefull if somebody answered all my questions and checked my working, especially part 2.