Let $\phi$ be holomorphic. Show $R = \{x \in U\;|\; d_x\phi \;\text{is not an isomorphism}\}$ is defined by one holomorphic equation.

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Let $\phi: U \to V$ be holomorphic with $U,V \subseteq \mathbb{C}^n$ open. For context, this is the first exercise in Voisin's "Hodge Theory and Complex Algebraic Geometry". Overall, I'm just confused about what the question is asking. My guess is that I'm supposed to use some higher dimensional version of the Cauchy Riemann equations to derive some simple equation but I'm at a bit of a loss. Can anyone please give me an idea of where to go with this? I'll try to work with the two-dimensional case to see if anything interesting comes of it.

Thank you!

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Recall that the differential of your holomorphic map $d\phi_x$ is a linear map between the tangent spaces of $x$ and $\phi(x)$ (both tangent spaces isomorphic to $\mathbb{C}^n$). A linear map is an isomorphism if and only if its determinant is nonzero. Hence, the equation that determines your set $R$ is $\det(d\phi_x) = 0$. Holomorphicity of this determinant map follows from examining what the entries of your differential are, and the fact that finite products of holomorphic functions are holomorphic.