# Show that there exists a $g(z)\in O(Ω)$ so that $e^{g(x)}\equiv f(x)$ where $f:Ω\rightarrow C$ ,Ω is convex-open set, $f(z)\in O(Ω)$ and $f(z)\neq 0$

Show that there exists a $g(z)\in O(Ω)$ so that $e^{g(x)}\equiv f(x)$ where $f:Ω\rightarrow C$ ,Ω is convex-open set, $f(z)\in O(Ω)$ and $f(z)\neq 0$

Any ideas? I thought abut working using series since they are both Holomorphic in the same set and Ω is open and convex. But nothing good so far

• integrate $f'(z)/f(z)$ – Lord Shark the Unknown Sep 18 '17 at 17:41
• What's $O(.)$.? – Nosrati Sep 18 '17 at 17:51
• $O(Ω)$ means that its holomorphic to $Ω$ set – Pookaros Sep 18 '17 at 21:49

## 1 Answer

I assume that $O(\Omega)$ is the set of holomorphic functions on $\Omega$, which many also denote as $H(\Omega)$.

The requisite such $g(z)$ may be found constructively as follows:

Pick $z_0 \in \Omega$, and for any $z \in \Omega$ define the path

$\gamma_z(t) = (1 - t) z_0 + tz, \tag 1$

where $t \in [0, 1]$; note that $\gamma_z(t) \subset \Omega$, since $\Omega$ is convex, and since

$z \in \Omega \Longrightarrow f(z) \ne 0, \tag 2$

we may define

$F(z) = \displaystyle \int_{z_0,\; \gamma_z(t)}^z \dfrac{f'(s)}{f(s)} ds, \tag 3$

that is, $F(z)$ is the integral of $f'(z)/f(z)$ taken along $\gamma_z(t)$. By virtue of (2), both $(f(z))^{-1} \in O(\Omega), f'(z) \in O(\Omega)$; thus so is $(f(z))^{-1}f'(z) \in O(\Omega)$ as is $F(z) \in O(\Omega)$; so also $e^{\pm F(z)} \in O(\Omega)$ and we have

$(f(z)e^{-F(z)})' = f'(z)e^{-F(z)} + f(z) (-F(z))'e^{-F(z)}; \tag 4$

now

$F'(z) = (\displaystyle \int_{z_0,\; \gamma_z(t)}^z \dfrac{f'(s)}{f(s)} ds)' = \dfrac{f'(z)}{f(z)}, \tag 5$

so (4) yields

$(f(z)e^{-F(z)})' = f'(z)e^{-F(z)} + f(z) (-F(z))'e^{-F(z)}$ $= f'(z)e^{-F(z)} + f(z) (-\dfrac{f'(z)}{f(z)})e^{-F(z)} = f'(z)e^{-F(z)} - f'(z)e^{-F(z)} = 0; \tag 6$

whence

$f(z) e^{-F(z)} = c \in \Bbb C, \tag 7$

$c$ constant; then

$f(z) = ce^{F(z)}, \tag 8$

and we can evaluate $c$ by setting $z = z_0$, using

$F(z_0) = \displaystyle \int_{z_0,\; \gamma_{z_0}(t)}^{z_0} \dfrac{f'(s)}{f(s)} ds = 0, \tag 9$

obtaining

$c = f(z_0); \tag {10}$

thus

$f(z) = f(z_0)e^{F(z)}. \tag{11}$

Now $f(z_0) \ne 0$ allows us to write

$f(z_0) = r_0e^{i\theta_0}, \tag{12}$

where

$r_0 = \vert f(z_0) \vert > 0, \tag{13}$

and $\theta_0 \in [0, 2\pi)$ is the unique value in this range such that

$\cos \theta_0 = \Re[\dfrac{f(z_0)}{r_0}] \tag{14}$

and

$\sin \theta_0 = \Im[\dfrac{f(z_0)}{r_0}]; \tag{15}$

with these choices we may write

$f(z_0) = r_0e^{i\theta_0} = e^{\ln r_0} e^{i\theta_0} = e^{\ln r_0 + i\theta_0}; \tag{16}$

thus, from (11) and (16),

$f(z) = f(z_0)e^{F(z)} = e^{\ln r_0 + i\theta_0}e^{F(z)} = e^{\ln r_0 + i\theta_0 + F(z)}; \tag{17}$

now setting

$g(z) = \ln r_0 + i\theta_0 + F(z), \tag{18}$

we have

$f(z) = e^{g(z)}; \tag{19}$

furthermore $g(z) \in O(\Omega)$ by virtue of the fact that $F(z) \in O(\Omega)$; this establishes the existence of the requisite $g(z) \in O(\Omega)$. Note that $g(z)$ is not unique, indeed we have

$e^{g(z) + 2n \pi i} = e^{g(z)}e^{2n \pi i} = e^{g(z)} \cdot 1 = e^{g(z)} = f(z), \tag{20}$

for any $n \in \Bbb Z$.

The methods and technique used here are similar to those found in the answers to this and related questions.

Note on the Convexity of $\Omega$: In the preceding discussion, I have specified the path integral defining $F(z)$ be taken along the particular path $\gamma_z(t)$ given in (1), thus exploiting the convexity of $\Omega$, which ensures $\gamma_z(t) \subset \Omega$, since it is a line segment joining two points of $\Omega$. In fact, the hypothesis that $\Omega$ is convex allows us to simply construct a homotopy 'twixt any two paths $\gamma_1, \gamma_2: [0, 1] \to \Omega$; we simply join the corresponding points on each curve (that is, points $\gamma_1(t)$, $\gamma_2(t)$ for $t \in [0, 1]$) by a line segment, thus:

$H(t, s) = (1 - s) \gamma_1(t) + s\gamma_2(t), \tag{21}$

where $s \in [0, 1]$; as $0 \to s \to 1$, $H(t, s)$ continuously carries $\gamma_1(t)$ to $\gamma_2(t)$; the convexity of $\Omega$ implies that the image of $H(t, s)$ lies entirely in $\Omega$; thus any two paths in $\Omega$ are homotopic in $\Omega$; in particular, taking $\gamma_1(t)$ to be a closed loop and $\gamma_2(t)$ to be a constant path, we see that every loop in $\Omega$ is null homotopic, that is, $\Omega$ is simply connected. This is in fact the usual criterion imposed when seeking the complex logarithm, that is, a function $g(z)$ such that $e^{g(z)} = f(z)$, of a holomorphic function $f(z)$ which never takes the value $0$ on $\Omega$. But in the present problem, we are able to avoid explicit mention of this essential topological property (simple connectedness) by exploiting the convexity of $\Omega$, which also allows all the necessary path integrals to be taken along line segments, as in (1), (3). This is a considerable conceptual simplification, and allows the proof(s) to proceed without the full apparatus of simple connectedness, making it that much more accessible to those less familiar with the topology of the complex plane $\Bbb C$. End of Note.