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The following is an exercise from Gamelin VIII.5.4

Suppose that $f(z)$ is analytic at $z_0$. Show that if the set of $z$ such that $\text{Re } f(z) = \text{Re } f(z_0)$ consists of just one curve passing through $z_0$, then $f'(z_0) \ne 0$. Show also that if the set of $z$ such that $|f(z)| = |f(z_0)|$ consists of just one curve passing through $z_0$, then $f'(z_0) \ne 0$.

I would like to apply the following Theorem:

let $f(z)$ be analytic at $z_0$. Suppose $z_0$ is a critical point of order $m-1$ for $f(z)$ with critical value $f(z_0) = w_0$. Let $\rho > 0$ satisfy $f(z) \ne w_0$ for $0 < |z-z_0| \le \rho$ and $\delta > 0$ satisfy $|f(z) - w_0| \ge \delta$ for $|z - z_0| = \rho$. Then for each $w$ such that $0 < |w - w_0| < \delta$, the equation $f(z) = w$ has exactly $m$ distinct solutions $z_1 (w) , \dots, z_m(w)$ in the disk of radius $\rho$ centered at $z_0$, which can be chosen to be analytic on a slit disk.

  1. Suppose $z_0$ is a critical point of order $m-1$. I wish to show $m = 1$. The theorem says that for each value $w \ne f(z_0)$ in a neighborhood of $f(z_0)$, there are $m$ solutions to $f(z) = w$, where $z\ne z_0$ is in a neighborhood of $z_0$. I want to say at this point that taking the points $z$ closer to $z_0$, these will correspond to $m$ curves passing through $z_0$ and so by assumption, $m = 1$. Is this the right approach?

  2. I am hesitant on the above approach since nowhere did I use the fact about it being true for real part only. So it seems, if the above works, then this follows by almost the exact same argument.

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Let $D$ be a small disk centered at $z_0$ which does not contain any zeros of $f(z)-f(z_0)$ other than $z_0$ and $C=\partial D$. The curve $ \operatorname{Re}\, f(z)=\operatorname{Re}\, f(z_0)$, say $l$, divides $D$ into just two parts $D^+$ and $D^-$ by the assumption.
We may assume that $\operatorname{Re}\, f(z)>\operatorname{Re}\, f(z_0)$ in $D^+$ and $\operatorname{Re}\, f(z)<\operatorname{Re}\, f(z_0)$ in $D^-$. Also $l$ divides $C$ into $C^+$ and $C^-$. We note that the image $f(C^+)$ of $C^+$ lies in the half-plane $\operatorname{Re}\,w>\operatorname{Re}\, w_0$ and $f(C^-)$ lies in $\operatorname{Re}\,w<\operatorname{Re}\, w_0$. See illustration in Fig.1 below. enter image description here

Thus $f(C)$ can go around $w_0$ only once, and hence $$ \int_{C} d\arg (f(z)-f(z_0))=2\pi.$$

If we suppose that $ f(z)-f(z_0)=(z-z_0)^n\cdot \varphi (z)$ with $\varphi (z)\ne 0 $ in a neighborhood of $z_0$, then $$ \int_{C} d\arg (f(z)-f(z_0))=2n\pi.$$ Hence we have $n=1$ which implies $f^\prime(z_0)\ne 0$.

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  • $\begingroup$ Thank you for your beautiful explanation and illustration! I have only one question: why may we assume that the difference of the real parts is positive on one side and negative on the other side? $\endgroup$
    – Future
    Commented Feb 22, 2016 at 14:26
  • $\begingroup$ Suppose that, for instance, $u=\operatorname{Re}f(z)-\operatorname{Re}f(z_0)>0$ on both sides. Then $u\ge 0$ on $D$ and $u=0$ on $l$. The maximum principle for harmonic functions yields that $u$ is a constant. Therefore $u$ is positive on one side and negative on the other side. $\endgroup$
    – ts375_zk26
    Commented Feb 22, 2016 at 22:39

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