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

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.

• Both are equivalent to: $0$ is connected to $\infty$ outside $U$. Feb 7, 2016 at 18:26
• I swear this is a duplicate Feb 10, 2016 at 17:42

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 $V$.

• In the first answer "first we replace $\gamma$ by a closed polygon?" How to get the closed polygon?
– 2016
May 23, 2016 at 5:37
• 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. May 23, 2016 at 5:58