Majorise the modulus of a function Let's consider the following holomorphic function $f(z)=e^{-z^\gamma}$ with $0\le \gamma <1 $ where the power is the principal power based on the principal Logarithm
Let $H= \{z\in \mathbb{C} , Re(z) > 0 \} $
f is holomorphic on H and continuous on $\overline{H}$.
I am trying to majorise $|f(z)|$ in function of $\gamma$
What I did:
Following the maximum modulus theorem, for $ z \in \overline{H}, |f(z)| $ is reaching its maximum in $\partial H = \{z\in \mathbb{C} , \operatorname{Re}(z) = 0 \} $
So when I calculate $|f(z)|$ I find : 
$|f(z)|=e^{-2\cos(\gamma \operatorname{Arg}(z))e^{\gamma \ln |z|}}$.
But how to go from there and majorise  $|f(z)|$  in function of $\gamma$ knowing that 
I found that this function reaches maximum at $z=0$ which means  $\max(|f(z)|)=1$ ?
Any help much appreciated.
 A: The maximum modulus theorem as I know it applies to functions on a bounded domain.
For example, $f(z)=\sin(iz)$ is analytic in the whole plane, but the maximum modulus of $f|_H$ is not found at $\partial H$ (where $f$ is bounded).
There is no maximum modulus at all.
You can always add infinity to the complex plane, but then the question of continuity up to the boundary becomes typically harder.
In your specific case I would just estimate by hand.
A point $z\in H$ is of the form $re^{i\phi}$ for $r>0$ and $\phi\in(-\pi/2,\pi/2)$.
For these points $z^\gamma=r^\gamma e^{i\phi\gamma}$.
(This is true in $H$ for the typical choice of logarithms when $\gamma\in(0,1]$. I assume you have chosen logarithms so that it is real on the positive real axis and analytic in $H$. Then this identity is indeed true.)
That is, taking the power $\gamma$ maps $H$ into a sector with $r>0$ and $\phi\in(-\gamma\pi/2,\gamma\pi/2)$, which happens to be a subset of $H$.
The question is then about estimating $e^{-z}$ in this sector.
This sector is a subset of $H$, so the supremum of $|f|$ in $H$ is bounded by that of $|e^{-z}|$ in $H$.
That is famously $1$ (and achieved at the origin).
If you want a finer control, we have
$$
|e^{-z^\gamma}|
=
|e^{-r^\gamma e^{i\phi\gamma}}|
=
e^{-r^\gamma\cos(\phi\gamma)}.
$$
