Complex Constant and Convergent Power Series

Question

Suppose that the function $f$ is defined by a convergent power series and suppose that $f (z + w) = f (z) f (w)$ for all complex $z$, $w$.

(a) Prove directly from this assumption that there is a constant $a$ $∈$ $\mathbb{C}$ such that $f' (z) = a f (z)$, all $z$. (In fact, $a = f '(0)$.)

(b) Use (a) to prove that $f (z) = $$\sum_{0}^\infty {(az)^n/n!}$

Answer 1

Hints:

Well, (b) follows trivially from (a), and for (a) you should try some continuity\derivability argument perhaps, say:

For constant $\,w\,$ and variable $\,z\,$:

$$f(z+w)=f(z)f(w)\implies f'(z+w)=\frac{d}{dz}(f(z+w))=f(w)f'(z)$$

Putting now $\,w=0\,$ we get the boring differential equation

$$f'(z)=f(0)f'(z)\implies f(0)=1\;\;\text{unless}\;\;f'(z)=0\,\,\forall\,z\in\Bbb C\ldots$$

Answer 2

A more direct aproach for a) would be to write $$f^\prime(z) = \lim_{h\to0} \frac{f(z+h)-f(z)}{h} = \lim_{h\to0} \frac{f(z)f(h)-f(z)}{h} = f(z)\lim_{h\to0} \frac{f(h)-1}{h}.$$

And as Sharkos mentioned arguing that this limit does in fact exist since $f$ is a convergent power series.