Show that there exist a unique function $f(z)$, define as the series $f(z)=\sum_{n=0}^{\infty}{a_{n}z^{n}}$ Show that there exist a unique function $f(z)$, define as the series $$f(z)=\sum_{n=0}^{\infty}{a_{n}z^{n}}$$ with positive radii convergence, such that $f(0)=2$ and for all $z\in\mathbb{C}$, $f'(z)=f(z)-1$
My approach: If $f'(z)=f(z)-1$, then  $$\sum_{n=1}^{\infty}{a_{n}z^{n}}=\sum_{n=0}^{\infty}{a_{n}z^{n}}-1\implies a_{0}=1$$
But $f(0)=a_{0}=2$, so I don't see how prove this exercise or maybe it has some mistake. 
 A: The series for $f'(z)$ isn't $\sum_{n=1}^{\infty}a_n z^n$, it's $$f'(z)=\sum_{n=0}^{\infty}a_n(z^n)'=\sum_{n=1}^{\infty} na_n z^{n-1}=\sum_{n=0}^{\infty}(n+1)a_{n+1}z^n.$$
Matching coefficients gives you $(n+1)a_{n+1} = a_n-\delta_n$, or $a_{n+1} = (a_n - \delta_n) / (n+1)$ for $n\ge 0$, and $a_0=2$ by the boundary condition $f(0)=2$.  Hence $a_1=1$, and then $a_2=1/2$, $a_3=1/3!$, etc., giving $$f(x)=1+\sum_{n=0}^{\infty}x^n/n!=e^x+1.$$  (For which, indeed, $f'(x)=e^x=f(x)-1$ and $f(0)=2$.)
A: Your mistake is:
$\sum_{n=1}^{\infty}{a_{n}z^{n}}=\sum_{n=0}^{\infty}{a_{n}z^{n}}-1$.
From $f'(z)=f(z)-1$ we get
$\sum_{n=1}^{\infty}{na_{n}z^{n-1}}=\sum_{n=0}^{\infty}{a_{n}z^{n}}-1$.
A: Firstly, treat it as a formal ODE problem: $\frac{dy}{dx}=y-1$ and $y(0)=2$. By
separation of variables, we immediately have $y=e^{x}+1$. Then we
argue it rigorously as follows:


*

*Such function $f$ exists: Define $f(z)=\exp(z)+1$. Then $f$
is an entire function and satisfies $f(0)=2$ and $f'(z)=f(z)-1$.

*Such function $f$ is unique: Suppose that $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$
that satisfies the conditions, with radius of convergence $r\in(0,\infty]$.
$f(0)=2\Rightarrow a_{0}=2$. By induction, we have $f^{(n)}(z)=f^{(n-1)}(z)=\cdots=f'(z)=f(z)-1$
for any $n\geq1$ and $|z|<r$. By direct differentiation, we have $f^{(n)}(z)=a_{n}n!+(\cdots)z+(\cdots)z^{2}+\cdots$.
Put $z=0$, we then have $a_{n}n!=f^{(n)}(0)=f(0)-1=1$. Therefore,
$a_{n}=\frac{1}{n!}$ for $n\geq1$. Therefore, $f(z)=2+z+\frac{1}{2!}z^{2}+\cdots=\exp(z)+1$.
The shows that $f$ is uniquely determined. Moreover, the radius of
convergence is $r=\infty$.
