An exercise from Steins's complex analysis.

Question

This is an exercise from Steins's complex analysis chapter $8$:

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 and has a minimum at $z_0$ ,while $F$ restricted to $\Gamma_2$ is also real but has a maximum at $z_0$.

This hint is also given:

Write $F(z)=(g(z))^2$ for $z$ near $z_0$ , and consider the mapping $z \rightarrow g(z)$ and its inverse.

I really have no idea.

Thanks

the graph of $z^2$ and its inverse $\sqrt z$ are the curves which are orthogonal in $z_0$ ? i need explanation. — user115608, May 29 '14 at 17:14
I meant - first solve the $F(z)=z^2$ case without the hint, then use the hint to reduce general $F$ to this case — user8268, May 29 '14 at 17:37
i consider y- axis .restriction of $z^2$ to this line is real and has maximum at $z=0$ and for other curve the x-axis that has minimum at $z=0$!am i true? — user115608, May 29 '14 at 17:54
In addition: you don't have to invert $z^2$, but you should think about why $g(z)$ is invertible and how to use it in your two paths. — Emrys-Merlin, May 29 '14 at 17:54

score 5 · Answer 1 · edited May 16 '21 at 03:41

Suppose $f(z)=\left(g(z)\right)^2$. ($g(z)$ is well-defined in a disk centered at $z_0$ as $g$ is not constant).

Twice differentiating $f$ gives us:

$\begin{align} &g(z_0)=0\\ &f''(z_0)=2\left[(g'(z_0))^2+g(z_0)g''(z_0)\right]\Rightarrow g'(z_0)\neq 0 \end{align}$

Hence by Exercise 1 (local bijection equivalence) of that chapter, $g$ is conformal in some disk around $z_0$. Therefore, $g^{-1}$ is also conformal. Define these curves:

$\left\{\begin{array}{lc} \gamma_1(t)=t\:;\:t\geq0 & \& &\Gamma_1=g^{-1}(\gamma_1(t))\:;\:t\geq0\\ \gamma_2(t)=it\:;\:t\geq0 & \& &\Gamma_2=g^{-1}(\gamma_2(t))\:;\:t\geq0\\ \end{array}\right.$

$\Longrightarrow\left\{\begin{array}{lc} F(\Gamma_1(t))=t^2\:;\:t\geq0\\ F(\Gamma_2(t))=-t^2\:;\:t\geq0\\ \end{array}\right.$

Now $\Gamma_1$ and $\Gamma_2$ satisfy the problem conditions.

You could also mention that the conformal map preserves the angel, so since the two curves you gave are orthogonal, the curve after the inverse map is also orthogonal. — JacobsonRadical, Apr 05 '18 at 16:07

score 1 · Answer 2 · answered Jun 27 '21 at 17:51

Fardad's answer is correct except for its explanation why $g(z)$ is well-defined. After all, we know $f(z) = z$ does not have a well-defined square root function around 0 as the square root would be multi-valued. Below is how to respond to this objection.

We can write $F(z) = (z-z_0)^2 h(z)$, where $h(z)$ is a nowhere vanishing holomorphic function around $z_0$, because $F(z_0) = F'(z_0) = 0$ and $F''(z_0) \neq 0$.

Since $h(z)$ is nowhere vanishing in a sufficiently small disc around $z_0$, $h^{1/2}(z) \equiv e^{\frac{1}{2}\log(h(z))}$ is well defined. (See Theorem 6.2 in Chapter 3.) Now we can define $g(z) \equiv (z-z_0)h^{1/2}(z)$.

An exercise from Steins's complex analysis.

2 Answers2