Prove that if $f$ is holomorphic so that $f'(z)=\alpha f(z)$ then $f(z)=ce^{\alpha z}$ Prove that if $f$ is holomorphic so that $f'(z)=\alpha f(z)$, $\alpha$ being a constant, for every $z \neq 0$ then $f(z)=ce^{\alpha z}$, $c \in \mathbb C$.
So what I tried doing is defining $g(z)=f(z)-ce^{\alpha z}$, and proving $g(z)$ is constant using Cauchy-Riemann. Obviously that didn't work out very well..
Any other hints will be great! 
 A: Multiply both sides by $e^{-\alpha z}$ and, with a little magic, you can make it happen.  I won't spoil things by saying any more.
A: Consider the holomorphic function
$g(z) = e^{-\alpha z} f(z); \tag{1}$
we have
$g'(z) = -\alpha e^{-\alpha z} f(z) + e^{-\alpha z} f'(z)$
$= -\alpha e^{-\alpha z} f(z) + e^{-\alpha z} \alpha f(z) = 0; \tag{2}$
thus
$g(z) = c, \;\; \text{a constant}; \tag{3}$
and thus from (1),
$f(z) = ce^{\alpha z}.  \tag{4}$
QED!!!
Note Added Saturday 25 April 2015 2:01 PM PST:  We can, and perhaps should, say a little more.  If we know the value $f(z_0)$ of $f(z)$ at some point $z_0$, then from (4) we have
$f(z_0) = ce^{\alpha z_0},  \tag{5}$
whence
$c = f(z_0) e^{-\alpha z_0},  \tag{6}$
so we may in fact write (4) as
$f(z) = f(z_0) e^{-\alpha z_0} e^{\alpha z} = f(z_0) e^{\alpha(z - z_0)}. \tag{7}$
Writing $f(z)$ in the form (7) is a help to presenting $f(z)$ in the event that the domain $\Omega$ of $f(z)$ is allowed to have more than one topological component, i.e., in the event that it is not necessarily a connected subset of the complex plane.  For then we may only affirm from $g'(z) = 0$ that $g(z)$ is constant on each connected component of $\Omega$.  If we write
$\Omega = \bigcup_{i \in I} \Omega_i, \tag{8}$
where each $\Omega_i$ is a non-empty, connected open subset of $\Bbb C$ with $\Omega_i \cap \Omega_j = \emptyset$, for $i \ne j$, that is, $\Omega$ is topologically decomposed into the $\Omega_i$, then for each $i \in I$ we may choose $c_i \in \Bbb C$ and have
$g(z) = c_i, \;\; f(z) = c_i e^{\alpha z} \tag{9}$
on $\Omega_i$.  If $z_i \in \Omega_i$, then by (7) we have
$f(z) = f(z_i)e^{\alpha(z - z_i)}, \tag{10}$
$z \in \Omega_i$.  Thus (7) provides a method for "automatically" explicitly specifying $f(z)$ on each component of $\Omega$.
Of course, when we unqualifiedly say, with our OP Reyo, "$f$ is holomorphic . . . for every $z \ne 0$" we generally understand $f(z)$ to be holomorphic on all of $\Bbb C \setminus \{ 0 \}$, which is connected. And indeed the general form of $f(z)$ given in (4) shows it may be extended to an entire function; we can take $z_0 = 0$ so that $f(z) = f(z_0) e^{\alpha z}$ and be done with it.  End of Note.
