Let \(P=(1,3-2\sqrt2)\) and define \(f(s,t)=s^3-6st+t^2\). Then \(f(P)=0\).
State the conclusion of the Implicit Function Theorem concerning \(f(s, t) = 0\) when \(f\) is considered a function \(\mathbb{R}^2\to\mathbb{R}\).
State the above conclusion when \(f\) is considered a function \(\mathbb{C}^2\to \mathbb{C}\).
Use the implicit function theorem for a function \(\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}^2\) to prove (b).
Solution
Let \[ P=(s_0,t_0)=(1,3-2\sqrt2). \] Then \(f(P)=0\) and \[ f_t(s,t)=-6s+2t, \qquad f_t(P)=-4\sqrt2\ne0. \]
1. Over \(\mathbb R\), the implicit-function theorem gives neighborhoods \(U\ni s_0\) and \(V\ni t_0\) and a unique \(C^1\) function\[ g:U\to V \]with\[ g(s_0)=t_0, \qquad f(s,g(s))=0. \]Moreover,\[ g'(s)=-\frac{f_s(s,g(s))}{f_t(s,g(s))}. \]
2. Over \(\mathbb C\), regarding the same polynomial as holomorphic on \(\mathbb C^2\), the complex implicit-function theorem gives neighborhoods \(U,V\subset\mathbb C\) and a unique holomorphic map \(g:U\to V\) satisfying the same equations.
3. To derive the complex theorem from the real one, write\[ s=x+iy, \qquad t=u+iv, \qquad f=F_1+iF_2, \]and set\[ F=(F_1,F_2):\mathbb R^2\times\mathbb R^2\to\mathbb R^2. \]The real Jacobian with respect to \((u,v)\) is\[ D_tF= \begin{pmatrix} a&-b\\ b&a \end{pmatrix}, \qquad a+ib=f_t. \]Hence\[ \det D_tF=|f_t|^2. \]At \(P\) this determinant is nonzero, so the real implicit-function theorem gives a unique \(C^1\) map \(g\).
4. Differentiating \(F(s,g(s))=0\) gives\[ Dg=-(D_tF)^{-1}D_sF. \]Both \(D_tF\) and \(D_sF\) are real matrices representing multiplication by complex numbers. Such matrices are closed under multiplication and inversion, so \(Dg\) has the form\[ \begin{pmatrix} \alpha&-\beta\\ \beta&\alpha \end{pmatrix}. \]Thus \(g\) satisfies the Cauchy–Riemann equations. Since \(g\) is \(C^1\), it is holomorphic, proving the complex conclusion from the real theorem.