Fall 2009 - Problem 8

Solution.

1. Let $z_1, z_2 \in \Omega$ be distinct. Since $\Omega$ is convex, the line $\func{\gamma}{\br{0,1}}{\Omega}$, $t \mapsto tz_1 + \p{1 - t}z_2$, is contained entirely within $\Omega$.

   Write $f = u + iv$, where $u = \Re\p{f}$ and $v = \Im\p{f}$. Since $f$ is holomorphic, it satisfies the Cauchy-Riemann equations; in particular,

   $$f'\p{z} = \pder{f}{x}\p{x} = \pder{u}{x}\p{z} + i\pder{v}{x}\p{z}.$$

   Thus, by the fundamental theorem of calculus,

   $$\begin{aligned} f\p{z_1} - f\p{z_2} = \int_\gamma f'\p{z} \,\diff{z} &= \int_\gamma \pder{f}{x}\p{z} \,\diff{z} \\ &= \int_\gamma \pder{u}{x}\p{z} + i\pder{v}{x}\p{z} \,\diff{z} \\ &= \int_\gamma \pder{u}{x}\p{z} \,\diff{z} + i\int_\gamma \pder{v}{x}\p{z} \,\diff{z} \\ \implies \Re\p{f\p{z_1} - f\p{z_2}} &= \int_\gamma \pder{u}{x}\p{z} \,\diff{z} \\ &> 0. \end{aligned}$$

   Indeed, $z_1 \neq z_2$, so $\gamma$ has positive length, and $\pder{u}{x} > 0$ and is continuous, so it is bounded below by some $\delta > 0$ on a small segment of $\gamma$ with positive length. Hence, the real part of $f\p{z_1} - f\p{z_2}$ never vanishes, so $f\p{z_1} \neq f\p{z_2}$, so $f$ is injective.

2. Consider $\Omega = \C \setminus \poc{-\infty, 0}$ and $g\p{z} = z^{1/2}$, which is holomorphic on $\Omega$ since it is simply connected and avoids $0$. Here, we choose the branch chosen such that $g\p{x} > 0$ on $\p{0, \infty}$. Set $f\p{z} = \br{g\p{z}}^3$ so that for $z \in \p{0, \infty}$,

   $$f'\p{z} = \frac{3}{2}g\p{z}.$$

   This branch of $g\p{z}$ maps to the right-half plane, and so $\Re\p{f'\p{z}} > 0$, but $f$ is not injective, e.g., $f\p{1} = 1$, and

   $$f\p{e^{i4\pi/3}} = \br{g\p{e^{i4\pi/3}}}^3 = 1.$$