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I am having difficulties in the technical details of proving that, for all $z_1, z_2 \in \mathbb C$, $$e^{z_1}e^{z_2}=e^{z_1+z_2}$$ from the definition $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}.$$ I tried to prove from right to left, the idea is to show that the $N$-th partial sum $$\sum_{n=0}^N \frac{(z_1+z_2)^n}{n!} = f_N(z_1)f_N(z_2)$$ for some $\{f_n\}$ such that $\lim_{n\to \infty} f_n(z) = e^z$. Then, by the uniqueness of limit and the algebraic property of limit, the theorem can be deduced.

However, I realize that it is the case that the $N$-th partial sum is actually some former terms in $\left(\sum_{n=0}^N z_1^n/n!\right)\left(\sum_{n=0}^N z_2^n/n!\right)$. So my question is, how can I handle this difficulty?

Kenta S
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Ryan Hu
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    Compute the Cauchy Product $(\sum_{n=0}^\infty \frac{z_1^n}{n!})(\sum_{n=0}^\infty \frac{z_2^n}{n!}).$ – Fred Sep 07 '22 at 10:34
  • As these are holomorphic functions doesn't it suffice to prove it for real positive $z_1, z_2$? Then if you look at the $2N$-th partial sum of the RHS it gets trapped between the product of the $N$ partial sums of $e^{z_1}, e^{z_2}$ and the $2N$-th partial sums of $e^{z_1}, e^{z_2}$. – ancient mathematician Sep 07 '22 at 11:13

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There's no reason to expect the partial sum to factor like that. You can do the series expansion in the other direction, as Fred suggests in the comments, but it's pretty tedious and annoying. A much cleaner approach is to argue as follows: first show from the series expansion that $e^z$ is the unique function satisfying the differential equation $f'(z) = f(z)$ with initial condition $f(0) = 0$, which follows from standard existence and uniqueness theorems for ODEs.

Next, consider the same differential equation but with initial condition $f(0) = e^w$; the same existence and uniqueness theorems give that the solution is unique, and $e^{z+w}$ and $e^z e^w$ are both solutions. So they must be equal!

Qiaochu Yuan
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