Find a holomorphic function

Question

Find a holomorphic function $f(z,\bar{z})$ such that $Argf(z)=xy \;mod\;2\pi$. I'm trying to apply CRE, i.e if $f(z)=u(x,y)+iv(x,y)$ , and by condition we have $tgArgf(z)=\frac{u}{v}$ i.e $u=vtgArgf(z)$. Then use CRE $\frac{\partial{u}}{\partial{x}}=\frac{\partial{v}}{\partial{x}}tg(xy)+\frac{vy}{cos^2(xy)}=\frac{\partial{v}}{\partial{y}}$ and $\frac{\partial{u}}{\partial{y}}=\frac{\partial{v}}{\partial{y}}tg(xy)+\frac{vx}{cos^2(xy)}=-\frac{\partial{v}}{\partial{x}}$. Then if we sum right-hand sides of this equations, we get $\frac{\partial{v}}{\partial{y}}-\frac{\partial{v}}{\partial{x}}=(\frac{\partial{v}}{\partial{y}}-\frac{\partial{v}}{\partial{x}})tg(xy)$. Thus tg(xy)=1? So, in this case we have u=v and then $f(z)$ is a constant?

Answer 1

If it's holomorphic, it's a function of $z$ alone, because the Cauchy-Riemann equations mean $\displaystyle\frac{\partial f}{\partial \overline{z}}=0$. But what do you think of $e^{z^2/2}$? Your equation $Argf(z)=xy \;mod\;2\pi$ implies a factor $e^{ixy}$, and $ixy$ is clearly the imaginary part of $z^2/2=(x^2-y^2)/2+ixy$.