Let $f$ be an analytic function on open unit disc $D$ whose image is contained in $\mathbb{C} \setminus (-\infty, 0]$.
Since $f$ is always nonzero, we can define the logarithm of $f$ suitably and so is its sqaureroot.
Now my question is that
- Does there exist a analytic function $g$ on $D$ such that $\Re(g(z)) \ge 0$ and $g^2 = f$? for all $z$ in $D$
- Does there exist a analytic function $g$ on $D$ such that $\Re(g(z)) \le 0$ and $g^2 = f$? for all $z$ in $D$
For constant maps it is obviously true but is it true in general?
Best Answer
In $U=\mathbb C\setminus (-\infty,0]$ the analytic square roots are: $$ s_1(r\mathrm{e}^{i\vartheta})=r^{½}\mathrm{e}^{i\vartheta/2},\quad s_2(r\mathrm{e}^{i\vartheta})=-r^{½}\mathrm{e}^{i\vartheta/2}, $$ where $r>0$ and $\vartheta\in(-\pi,\pi)$. So $$ s_1[U]=\{r\mathrm{e}^{i\vartheta}: -\pi/2<\vartheta<\pi/2\}=\{x+iy\in\mathbb C: x>0\}, \\ s_2[U]=\{r\mathrm{e}^{i\vartheta}: \pi/2<\vartheta<3\pi/2\}=\{x+iy\in\mathbb C: x<0\}. $$ So indeed, both a. and b. hold, since the square roots of $f$ are $$ s_1\circ f \quad\text{and}\quad s_2\circ f. $$