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I asked this question in stackexchange, but it flashed and disappeared:

Let $A$ be a C*-algebra and $\exp(x)=\sum_{n=0}\frac{x^n}{n!}$, the usual exponential function from $A$ into $A$. Is it true that if $x\ne y\in A$, $x^*=x$, $y^*=y$, then $\exp(x)\ne\exp(y)$?

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    $\begingroup$ Yes, because $x=\lim_{n\to\infty} n(e^{x/n}-1)$ and $e^{x/n}$ is the unique positive $n$th root of $e^x$. $\endgroup$
    – Christian Remling
    Commented Oct 27, 2014 at 20:36
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    $\begingroup$ In a unital $C^*$-algebra, It induces a bijection betweend self-adjoint element and positive inversible self adjoint element: its inverse is the logarithm. ANd because these two functions are holomorphic one should even be able to say something similar for Banach algebra. $\endgroup$
    – Simon Henry
    Commented Jun 11, 2015 at 15:56

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