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I'm trying to set about proving that given a complex number $z$, the principles that hold for real numbers also hold for complex numbers. Namely,the product of two functions $e^{f_1}$ and $e^{f_2}$ is $e^{f_1+f_2}$.

I do believe I have shown it in my work here: but I'm not sure if I have all the justifications for proving it. Do you feel that I've shown this statement is true?

(I've attached a picture of my work, you can also use the link for a higher resolution version, your choice)!

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3 Answers 3

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Here's a more analysis-based proof, assuming you know $e^z$ is holomorphic $\mathbb C \to \mathbb C$ with derivative $e^z$ (for example via some results on power series), and you've established basic rules of differentiation (chain rule, product rule).

Choose arbitrary $a,b \in \mathbb C$. Consider $F:\mathbb C \to \mathbb C$ given by $F(z) = e^{a+b-z}e^{z}$. Clearly $F$ is holomorphic, and differentiating we find $F'(z) \equiv 0$. So $F$ is constant, thus

$e^a e^b = F(b) = F(0) = e^{a+b}$

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You can use the definition for exponential function: $e^z:=\sum_{n=0}^\infty\frac{z^n}{n!}$ (in fact, Euler's formula can be deduced from this definition and the power series expansion of $\sin$ and $\cos$.) We write: $$ \begin{align*} e^{z_1 + z_2}&= \sum_{n=0}^\infty \frac{(z_1+z_2)^n}{n!}.\\ &=\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}\frac{z_1^kz_2^{n-k}}{n!} \end{align*} $$

where we use the Binomial theorem in the second equality.

Since we know that the series converges absolutely, we can change the order of summation: $$ \begin{align*} e^{z_1+z_2} &= \sum_{k=0}^\infty\sum_{n=k}^\infty\frac{z_1^kz_2^{n-k}}{k!(n-k)!}\\ &= \sum_{k=0}^\infty\sum_{m=0}^\infty\frac{z_1^kz_2^{m}}{k!m!}\\ &=e^{z_1}e^{z_2}. \end{align*} $$

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If you define the complex exponential in terms of Euler's formula, that all $x,y\in\mathbb{R}$ has $e^{x+iy}:=e^x(\cos(y)+i\sin(y))$, then your proof is correct. I would be a bit more clear about the application of the angle sum formula, and use more equals signs for clarity, but the basic proof strategy is valid.

It is worth noting that the complex exponential is usually defined in terms of a power series though, rather than with Euler's formula. Typically Euler's formula is a theorem about the power series, so anyone who starts with the power series definition won't be able to accept your proof unless you first prove that the Euler's formula definition is identical to the power series definition.

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  • $\begingroup$ Thank you @jade for your feedback. I will take your suggestions into account. I also take your point about the complex exponential being represented as a power series. I would typically use the series. Whenever I was problem-solving, I instinctually jumped to Euler's. I will also clarify what identies I used to reduce the problem down. $\endgroup$
    – Aaron
    Mar 30, 2022 at 21:54

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