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The following assertion appears plausible to me: Let $f(z,w,u)$ and $g(z,w,u)$ be holomorphic in $z$, $w$, and $u$. Suppose that $f(z_0,w_0,u_0) = g(z_0,w_0,u_0) = 0$ and that $f$ and $g$ are non-degenerate in the sense that none of $f(z,w_0,u_0)$, $f(z_0,w,u_0)$,...,$g(z_0,w_0,u)$ are identically zero. Then there exists $w(u)$ and $z(u)$ defined near $u_0$ such that $f(z(u),w(u),u) = g(z(u),w(u),u) = 0$. Note that I am not imposing any uniqueness or regularity requirements on $z(u)$ and $w(u)$.

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  • $\begingroup$ Taking $f=z+w+u$, $g=z+w+2u$, $z_0=w_0=u_0=0$ gives a counterexample. Is there an extra hypothesis that should be added? $\endgroup$
    – Julian Rosen
    Nov 17, 2012 at 17:37
  • $\begingroup$ Holomorphic implicit function theorem? $\endgroup$
    – Vít Tuček
    Nov 17, 2012 at 17:46
  • $\begingroup$ @Julian Rosen: Thanks, that's indeed a relevant counter-example.If you want to add that as an answer I can accept it. $\endgroup$
    – Yakov Shlapentokh-Rothman
    Nov 17, 2012 at 18:06
  • $\begingroup$ @robot: In the application I have in mind it is not easy to check the hypothesis for the implicit function theorem. $\endgroup$
    – Yakov Shlapentokh-Rothman
    Nov 17, 2012 at 18:06

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In general, $S:=\{(z,w,u):f=g\}$ will have codimension 1. If we arrange for $S$ to be contained entirely in the fiber over $u_0$, then the desired functions $z(u)$, $w(u)$ won't exist.

In particular, if fix $z_0$, $w_0$, find $h(z,w)$ holomorphic with $h(z_0,w_0)=0$, then we can take $u_0=0$, $f(z,w,u)=h(z,w)+u$, $g(z,w,u)=h(z,w)+2u$ (we will need some condition on $h$ so that $f$ and $g$ are non-degenerate).

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