- Supppose \(F ( z )\) is holomorphic near \(z = z _ { 0 }\) and \(F ( z _ { 0 } ) = F ^ { \prime } ( z _ { 0 } ) = 0\) , while \(F ^ { \prime \prime } ( z _ { 0 } ) \neq 0\) . Show that there are two curves \(\Gamma _ { 1 }\) and \(\Gamma _ { 2 }\) that pass through \(z_0\), are orthogonal at \(z _ { 0 }\) , and so that \(F\) restricted to \(\Gamma _ { 1 }\) is real and has a minimum at \(z _ { \mathrm { 0 } }\) , while \(F\) restricted to \(\Gamma _ { 2 }\) is also real but has a maximum at \(z _ { 0 }\)
[Hint: Write \(F ( z ) = ( g ( z ) ) ^ { 2 }\) for \(z \ \mathrm { n e a r } \ z _ { 0 }\) , and consider the mapping \(z \mapsto g ( z )\) and its inverse.]
Solution
Since \(F\) has a zero of order exactly two at \(z_0\), there is a holomorphic function \(h\) near \(z_0\) such that \[ F(z)=(z-z_0)^2h(z),\qquad h(z_0)=\frac{F''(z_0)}2\ne0. \] After shrinking the neighborhood, \(h\) has no zeros and therefore admits a holomorphic square root \(q\). Set \[ g(z)=(z-z_0)q(z). \] Then \(F=g^2\), \(g(z_0)=0\), and \(g'(z_0)=q(z_0)\ne0\). By the holomorphic inverse-function theorem, after shrinking once more, \(g\) is biholomorphic from a neighborhood \(U\) of \(z_0\) onto a disc \(V\) centered at \(0\).
Let \[ \Gamma_1=g^{-1}(V\cap\mathbb R), \qquad \Gamma_2=g^{-1}(V\cap i\mathbb R). \] These are smooth curves through \(z_0\). On \(\Gamma_1\), write \(g(z)=t\in\mathbb R\); then \[ F(z)=t^2\ge0=F(z_0), \] so \(F\) is real there and has a strict local minimum at \(z_0\). On \(\Gamma_2\), write \(g(z)=it\) with \(t\in\mathbb R\); then \[ F(z)=-t^2\le0=F(z_0), \] so \(F\) is real there and has a strict local maximum at \(z_0\).
Finally, a conformal map preserves angles. The real and imaginary axes in the \(g\)-plane meet orthogonally at \(0\), and \(g^{-1}\) is conformal at \(0\) because \((g^{-1})'(0)\ne0\). Hence \(\Gamma_1\) and \(\Gamma_2\) meet orthogonally at \(z_0\).