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Show that if there is a branch of $\sqrt{z}$ on an open set $U$ with $0 \notin U,$ then there is also a branch of $arg$ $z.$

I am unable to proceed any further in this and any help in this regard would be greatly appreciated.

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    $\begingroup$ Both are equivalent to: $0$ is connected to $\infty$ outside $U$. $\endgroup$ – Hagen von Eitzen Feb 7 '16 at 18:26
  • $\begingroup$ I swear this is a duplicate $\endgroup$ – mercio Feb 10 '16 at 17:42
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A natural way is proving the ``obvious'' fact that if an open set $U\subset\mathbb{C}$ contains a closed curve $\gamma$ with non-zero winding number around $0$ then $U$ contains a closed curve $\gamma_1$ with winding number $1$ around $0$. To do this, first we replace $\gamma$ by a closed polygon. After some perturbation the polygon has no multiple segments, just finitely many self-intersections. At the self-intersections the polygon can be split into finitely many simply closed polygons. From the sum of the winding numbers of the small polygons, at least one of them has non-zero winding number. But, due to Jordan's curve theorem, the winding number of a simply closed polygon can be only $0$ or $\pm 1$.


Another proof: Suppose that there is a continuous branch of $\sqrt{z}$ on $U$. Let $V=\{\sqrt{z}:z\in U\}$. We will prove that $\log z$ has a holomorphic branch on $V$; this provides a branch of $2\log\sqrt{z}$ on $U$.

Notice that the sets $V$ and $-V$ are disjoint.

For the existence of $\log z$ on $V$ it suffices if every closed polygon in $V$ has zero winding number around $0$. Take an arbitrary closed polygon $\gamma\subset V$, and for every $z\in V\subset \gamma$, let $n_\gamma(z) =\frac1{2\pi i}\int_{w\in\gamma}\frac{\mathrm{d}w}{w-z}$ be the winding number of $\gamma$ around the point $z$. We want to prove $n_\gamma(0)=0$.

Consider the polygon $-\gamma$. As $\gamma\subset V$ and $-\gamma\subset (-V)$ and the sets $V$ and $-V$ are disjoint, the curves $\gamma$ and $-\gamma$ are disjoint, too. Let $z_1\in(-\gamma)$ and $z_2\in(-\gamma)$ be two points of $-\gamma$ with minimum and maximum distance from $0$. In the open set $\mathbb{C}\setminus(\{0\}\cup \gamma)$ the points $0$ and $z_1$ are connected by a line segment; the points $z_1$ and $z_2$ are connected by the curve $-\gamma$; finally $z_2$ and $\infty$ are connected by a ray. Therefore, the points $0,z_1,z_2,\infty$ are in the same component of $\overline{\mathbb{C}}\setminus(\{0\}\cup\gamma)$, so $$ n_\gamma(0) = n_\gamma(z_1) = n_\gamma(z_2) = n_\gamma(\infty) = 0. $$

Hence, every closed polygon $\gamma\subset V$ has winding number $0$ around $0$, so there is a holomorphic branch of $\log z$ on $V$.

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  • $\begingroup$ In the first answer "first we replace $\gamma$ by a closed polygon?" How to get the closed polygon? $\endgroup$ – 2016 May 23 '16 at 5:37
  • $\begingroup$ The curve $\gamma$ lies in an open domain, so the distance between $\gamma$ and $\mathbb{C}\setminus{U}$ is positive. We can choose finitely many sampling points along the curve. $\endgroup$ – user141614 May 23 '16 at 5:58

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