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Suppose $F(z)$ is holomorphic near $z=z_0$ and $F(z_0)=F'(z_0)=0$, while $F''(z_0)\neq 0$. 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 an has a minimum at $z_0$, while $F$ restricted to $\Gamma_2$ is also real but has a maximum at $z_0$.

Progress: I know that $F(z)=z^2f(z)$ where $f(z)$ doesn't vanish at $z_0$. Then in a small disk around $z_0$ the function $f$ doesn't vanish. So I can write $F(z)=g(z)^2$ where $g$ is a bijection in a small disk around $z_0$. The hint of this exercise suggests to use this function $g$ and its inverse to get the curves $\Gamma_1$ and $\Gamma_2$, but I don't see how to do it. Any suggestions?

Arctic Char
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Luz
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1 Answers1

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You are almost there. $g$ is a bijection of some neighborhood of $z_0$ to a disk $B_r(0)$, therefore you can define $$ \Gamma_1(t) = g^{-1}(t) \\ \Gamma_2(t) = g^{-1}(it) $$ where $ t \in (-r, r)$. Then $f(\Gamma_1(t)) = t^2$ is real and has a minimum at $t=0$, whereas $f(\Gamma_2(t)) = -t^2$ is real and has a maximum at $t=0$.

$\Gamma_1$ and $\Gamma_2$ are orthogonal at $z_0$ because $g^{-1}$ preserves angles.

Martin R
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