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Suppose we use the following definition of complex exponentiation: $$ e^z=\lim_{n \to\infty}\left(1+\frac{z}{n}\right)^n \, . $$ For a given real value of $\theta$, let $e^{i\theta}=x+iy$. I want to show that $e^{-i\theta}=x-iy$ without using Euler's formula.

I think that this is true because the algebraic properties of $i$ and $-i$ are identical. This means, roughly, that if we have a valid equation involving $i$, and replace every occurence of $i$ with $-i$, then we still get a valid equation. Here is my attempt to turn it into a rigorous argument. Is it correct?

Since the mapping $a+bi\mapsto a-bi$ is a field automorphism of $\Bbb{C}$, we know that for all $n\in\Bbb{N}$, $$ \left(\overline{1+\frac{i\theta}{n}}\right)^n=\overline{\left(1+\frac{i\theta}{n}\right)^n} \, . $$ Hence, \begin{align} e^{-i\theta} &= \lim_{n \to \infty}\left(1-\frac{i\theta}{n}\right)^n \\[5pt] &= \lim_{n\to\infty}\left(\overline{1+\frac{i\theta}{n}}\right)^n \\[5pt] &= \lim_{n\to\infty}\overline{\left(1+\frac{i\theta}{n}\right)^n} \\[5pt] &= \overline{\lim_{n\to\infty}\left(1+\frac{i\theta}{n}\right)^n} \tag{*}\label{*}\\[5pt] &= \overline{e^{i\theta}} \\[5pt] &= x-iy \, . \end{align} The line $\eqref{*}$ is justified because the function $a+bi\mapsto a-bi$ is continuous.

Joe
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    Yes, your reasoning is fine. If $a(n) \to A$ as $n \to \infty$, then $\overline{a(n))} \to \overline{A}$ as $n \to \infty$. You can just read this off the definitions. – Rob Arthan Aug 14 '21 at 21:21
  • Possible duplicate - https://math.stackexchange.com/q/3997692/443030 ? Otherwise your proof seems correct. – V.S.e.H. Aug 14 '21 at 21:45
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    Very elegant proof. – Sebastiano Aug 14 '21 at 22:50
  • From your definition you can directly prove that $e^{\phi+i\theta} =e^{\phi} (\cos\theta+i\sin \theta) $. And then your identity follows by using $\phi=0$. – Paramanand Singh Aug 23 '21 at 09:30

1 Answers1

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It seems fine, maybe we can also proceed as follows, since

$$e^{i\theta}\cdot e^{-i\theta}=e^0=1$$

we have

$$(x+iy)(a+ib)=ax-by+i(ay+bx)=1$$

and

  • $ay+bx=0 \implies a=-b\frac x y$
  • $ax-by=-b\frac {x^2}y-by=-\frac b y=1 \implies b=-y \quad a=x$
user
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    Hi user. Welcome back to MSE. Thanks for posting this answer. I have one query: I'm not sure how your argument works if $x=0$ or $y=0$, since we have no way of knowing a priori that $e^{-i\theta}$ can still be written in the form $ax+iby$ for some constants $a$ and $b$. – Joe Aug 14 '21 at 21:59
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    Hi @Joe thanks for your welcome! For $x=0$ we obtain $-by^2=1$ as condition which implies $b=-1$ and similarly for $y=0$ it seems to work. Maybe, is there some other properties we are using in this algebraic argument which you are nor assuming among the hypoteses? – user Aug 14 '21 at 22:06
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    What I mean is: suppose that $e^{i\theta}=0+1i$, for instance. Then, if the real part of $e^{-i\theta}$ is non-zero, then there is no way of writing $e^{-i\theta}$ in the form $ax+iby$. Of course, the real part of $e^{-i\theta}$ is zero, but we don't know that yet. – Joe Aug 14 '21 at 22:13
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    @Joe Ah ok now I get your point. I think you are right. I try to edit and use a more general approach. – user Aug 14 '21 at 22:21
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    @Joe Now it should be fixed for the issue you pointed out! – user Aug 14 '21 at 22:25
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    @user Hi, what happen if $b=-y, \text{ and } a=x$? :-) – Sebastiano Aug 14 '21 at 22:52
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    Hi @Sebastiano, we can conclude that $e^{-i\theta}=x-iy$. – user Aug 14 '21 at 22:54
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    I have my mind....fused with the high temperatures in Sicily. – Sebastiano Aug 14 '21 at 22:57