Little question about holomorphic function.

Question

I am working on a task where the following is given:

Let $r > 0$ and $f$ a holomorphic function on $U=B_r(a)$ so that $f(a)=0$ but $f'(a)\neq 0$.

The Solution claims the existence of a holomorphic function $g: U \to \Bbb C$ with $f(z)=(z-a)g(z)$. I don't undestand why $g(a)=a_1$ where $a_1$ is a coefficient of the power series of $f$.

Answer 1

A slightly more convenient approach, imho:

Since $f$ is analytic on a disc centered at $a$, it is exactly equal to a power series of the form $f(z)=\sum_{k=0}^\infty a_k (z-a)^k$. Also, we have that $a_k=\frac{f^{(k)}(a)}{k!}$. In particular, $a_0=0$. So we have

$$\begin{align}f(z)&=\sum_{k=0}^\infty a_k(z-a)^k\\ &=\sum_{k=1}^\infty a_k(z-a)^k\\ &=(z-a)\sum_{k=1}^\infty a_k (z-a)^{k-1}\\ &=(z-a)\sum_{k=0}^\infty a_{k+1}(z-a)^k \end{align}$$

The power series obtained in the last step is $g$. It's obviously analytic, and $g(a)=a_1$.

Answer 2

I don't really know what "believe that $g$ exists" means. But if you have proven that such a $g$ exists, then simply differentiate

$f'(z)=g(z)+(z-a)g'(z)$ and plug in $a$ to get $f'(a)=g(a)$.

Answer 3

Simply define$$g(z)=\begin{cases}\frac{f(z)}{z-a}&\text{ if }z\ne a\\f'(a)&\text{ if }z=a.\end{cases}$$It is clear from the way that $g$ is defined that you always have $g(z)=(z-a)f(z)$ and, if $\sum_{n=1}^\infty a_n(z-a)^n$ is the Taylor series of $F$ centered at $a$, that $g(a)=f'(a)=a_1$. Besides, $g$ is analytic. This is clear in $U\setminus\{a\}$ (it's the quotient of two analytic functions). And, near $a$,$$g(z)=\sum_{n=1}^\infty a_n(z-a)^{n-1}=\sum_{n=0}^\infty a_{n+1}(z-a)^n.$$