Find analytic Functions such that $f'(z)=-2f(z)$ 
Find all analytic functions $f:\mathbb{C} \longrightarrow \mathbb{C}$ such that
  $$f'(z)=-2f(z),~z \in \mathbb{C}$$ and
  $$f(0)+f'(0)=1$$

the only thing that majorly concerns me is making sure my answer is unique. as i'm sure it's possible that there may be a class of functions anyway
the preferred method i believe is to expand these as power series such as
$$\sum_{n=0}^{\infty}na_nz^{n-1} = -2\sum_{n=0}^{\infty}a_nz^{n}$$
do a bit of rearranging and then use the fact that co-effecient of power series are uniquely defined.
i'm sure theres a bit of rearranging magic to be done as considering the co-effecients at the moment gives
$$na_n = -2 a_n$$ which of course only happens at $n=-2$ apparently, but we have it that $n\geq0$ so that doesn't make sense. i can either presume there to be no solution or try the aforementioned rearrangement magic. by all means if you have a hint for how to solve this using the power series i'm all ears.
anyway, the method i used.
we have
$$f'(z)=-2f(z)$$ and these are meant to be analytic, so they're differentiable and integratable. so considering it to be a first order homogeneous ODE we solve using integrating factor
$$f'(z)+2f(z)=0 \implies$$
$$e^{2z}f'(z)+2e^{2z} = 0 \implies$$
$$\frac{d}{dz}\left(e^{2z}f(z)\right) = 0 \implies$$
$$e^{2z}f(z) = C \implies$$
$$f(z) = Ce^{-2z}$$
Differentiating and using the initial conditions given above we have
$$f'(z) = -2Ce^{-2z}$$ and
$$f'(0)+f(0)=1 \implies -2Ce^{0}+Ce^{0} = 1 \implies$$
$$-C=1 \implies C = -1$$
so
$$f(z) = -e^{-2z}$$ which indeed satisfies the requirements of both being analytic and the initial conditions.
now, with ODE's we use Initial conditions in order to guarantee uniqueness of solutions but assuming that $$f(z) = -e^{-2z} + D$$ is also a solution, plugging this into the above we obviously get $D = 0$
but.... is this function unique? and is this a valid arguement
Cheers for the help.
Incidently i've just figured out how to do it via power series. Consider a generating function 
$$f(z) \leftrightarrow a_0,a_1,a_2,a_3,...a_n$$
then
$$2f(z) \leftrightarrow 2a_0,2a_1,2a_3,...,2a_n$$
and $$f'(z) \leftrightarrow a_1,2a_2,3a_3,...,n a_n$$
then for the equation
$$f'(z)+2f(z) = 0$$ we collate co-effecients and get
$$a_1 + 2a_0 = 0$$
$$2a_1+2a_2 = 0$$
etc...
Rearranging we get
$$a_1 = -2a_0$$
$$a_2 = -a_1$$
$$a_3 = -\frac{2}{3}a_2$$
... etc
back substituting we get
$$f(z) \leftrightarrow a_0,-2a_0,2a_0,-\frac{4}{3}a_0,\frac{4}{6}a_0,...,\frac{(-2)^{n}}{n!}$$
this implies then that
$$f(z) = a_0\sum_{n=0}^{\infty}\frac{(-2)^{n}}{n!}z^n$$
which is agrees with the above, ie$ f(z) = a_0 e^{-2z} $
using initial conditions again gives us $a_0$ = -1 and so $$f(z) = -e^{-2z}$$ so i was correct i believe. if someone could confirm, i'd greatly appreciate it.
 A: You do not need to use theroems about unicity of solutions of ODEs in order to prove that your solution is the only one. If $f(z)$ is such that $f'(z)=-2f(z)$, let $g(z)=f(z)e^{2z}$. Then $g'(z)=f'(z)e^{2z}+2f(z)e^{2z}=0$. So, $g$ is constant. In other words, the only functions $f$ such that $f'(z)=-2f(z)$ are those of the form $f(z)=Ke^{2z}$. And the only $K$ for which you have $f(0)+f'(0)=1$ too is $K=-1$ (as you know).
A: $$f'(z)+2f(z)=0,\forall z\in \Bbb C$$$$\implies e^{2z}f'(z)+2e^{2z}f(z)=0,\forall z\in \Bbb C$$$$\implies\frac{d}{dz}\bigg(e^{2z}f(z)\bigg)=0,\forall z\in \Bbb C$$$$\implies z\mapsto e^{2z}f(z),z\in \Bbb C\text{ is constant function as }\Bbb C \text{ is path-connected.}$$ Say, $e^{2z}f(z)=c,\forall z\in \Bbb C$ for some $c\in \Bbb C$.  Therefore, $f(z)=ce^{-2z},\forall z\in \Bbb C$. Now, $f'(0)+f(0)=1\implies -2c+c=1\implies f(z)=-e^{-2z},\forall z\in \Bbb C.$
Note the idea of of solving these. We know the integrating factor of the differential equation, $y'+p(x)y=0$ is $e^{\int p(x)dx}$ i.e. $\frac{d}{dx}\big(e^{\int p(x)dx}y\big)=0$.
