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I'm considering a new proof of Euler's formula, but I'm not confident if my method works.

If $f(x+iy)= \cos(x)+i \sin(x)$, then we have $f_x/f=1$. Does it follow that $f(x+iy)=C \exp(x+ix)$ ?

Since $f\overline{f}=1$, we could infer that $f(x+iy)= \exp(ix)$.

EDIT: It was a typo to write $f_x/f=1$. Of course, $f_x/f=i$ and thus we just consider a function of $x$.

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    $\begingroup$ LaTeX tip: Use \cos, \sin, and \exp for operators typeset in roman with nice spacing. $\endgroup$ Nov 29, 2023 at 3:30
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    $\begingroup$ What is $f_x$? Why does $y$ disappear? $\endgroup$
    – angryavian
    Nov 29, 2023 at 3:59

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The spirit of your argument looks okay, but I don’t see the need for $y$. Instead, define $f:\mathbb{R}\to\mathbb{C}$, $f(x) = \cos x + i\sin x$. Then $f’/f = (\log f)’$ is a logarithmic derivative, and we have

$$ (\log f)’ = i\quad \Longrightarrow\quad f(x) = Ce^{ix}.$$

A quick comparison with $f(0)$ then shows that $C=1$, and so $f(x)=e^{ix}$ as desired.

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    $\begingroup$ Thanks for answering. That looks like a nice way to derive the identity. $\endgroup$
    – Hulkster
    Nov 29, 2023 at 4:25

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