Consider \(f(x,y)=x^2-2y+y^3\). Let \(P=(1,1)\).
Let \(X=\{(x,y)\in\mathbb{R}^2:f(x,y)=0\}\). Prove that there is a neighborhood \(U\) of \(P\) so that \(U\cap X\) is given by \(y=\phi(x)\) for some \(\mathcal{C}^1\) function \(\phi\). Give \(\phi'(x)\).
Now consider the same equation in \(\mathbb{C}^2\) (i.e., \(Y=\{(x,y)\in\mathbb{C}^2:f(x,y)=0\}\)). Prove that the analogous statement holds. What is \(\phi'(x)\) as an \(\mathbb{R}\)-linear map from \(\mathbb{C}\) to \(\mathbb{C}\)? Why is \(\phi\) holomorphic? (Hint: If you consider \(g:\mathbb{C}\to\mathbb{C}\) as a map \(\widetilde{g}:\mathbb{R}^2\to\mathbb{R}^2\), how is holomorphicity of \(g\) characterized by \(D\widetilde{g}\)?)
Solution
We have \(f(1,1)=0\) and \[ f_y(x,y)=-2+3y^2, \qquad f_y(1,1)=1\ne0. \] Thus the real implicit-function theorem gives neighborhoods of \(1\) and a unique \(C^1\) function \(y=\phi(x)\) with \(\phi(1)=1\) and \(f(x,\phi(x))=0\). Differentiating, \[ 2x+(-2+3\phi(x)^2)\phi'(x)=0, \] so \[ \boxed{\phi'(x)=-\frac{2x}{-2+3\phi(x)^2}}. \]
For (b), regard \(f:\mathbb C^2\to\mathbb C\) as a smooth map \(\mathbb R^4\to\mathbb R^2\). Its derivative in the \(y\) variable at \((1,1)\) is multiplication by \(f_y(1,1)=1\), hence is an invertible real-linear map. The real implicit-function theorem therefore gives a \(C^1\) map \(\phi\) from a neighborhood of \(1\) in \(\mathbb C\) to \(\mathbb C\) with the same graph property. Differentiating in a direction \(h\in\mathbb C\) gives \[ D\phi(x)(h)=-\frac{2x}{-2+3\phi(x)^2}\,h. \] Thus \(D\phi(x)\) is multiplication by a complex scalar, so it is \(\mathbb C\)-linear at every point. A \(C^1\) map \(\mathbb C\to\mathbb C\) is holomorphic exactly when its real derivative is complex-linear. Hence \(\phi\) is holomorphic, with complex derivative given by the same displayed scalar.