Proof of exponential property using Analytic continuation

Question

The problem is to prove $e^{z+w} = e^z e^w$ for all $z, w \in \mathbb{C}$, using analytic continuation.

Can someone please explain how to do this?

Answer 1

• Assume $w \in \mathbb{R}$

  Let $f(z) = e^{z+w}$ and $ g(z) = e^z e^w$ both entire (analytic on the whole complex plane, by the definition $e^z:=\sum_{k=0}^\infty\frac{z^k}{k!}$) .

  As you know $\forall x \in \mathbb{R}, f(x) = g(x)$.

  It means that $f(z)-g(z)$ is an entire function with some non-isolated zeros on the real line,

  hence it is identically zero and $\forall z \in \mathbb{C}, f(z) = g(z)$.

• Repeat the same process with $w$

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The analytic continuation works if we know that $f(z) = g(z)$ for $z \in U$ some open, for example the unit disk.

But the real line is not open in the complex topology, so it is not obvious how to use the analytic continuation.

Maybe you want to use some theorem about the complex logarithm for $|z-1 | < 1$ ?