Let $A$ be a unital $C^*$-algebra and $a\in A$ be a self-adjoint unitary, i.e. $a=a^*$ and $a^2 = 1$. Is it true that $-i\exp(i\pi a/2) = a?$
Attempt: Look what happens for $A= \mathbb{C}$. The element $a$ with $a=a^*$ and $a^2=1$ are $a\in \{1,-1\}$. For these values, the identity holds.
Hence, it may hold for general $C^*$-algebras. We calculate
$$\exp(i\pi a/2) = \exp(i\pi/2) + \exp(a) = i + \exp(a)$$
and $$\exp(a) = 1 + a + a^2/2 + a^3/3! + a^4/4! + \dots $$
$$=(1 + 1/2 + 1/4!+\dots) + a(1 + 1/3! + 1/5! + \dots)$$ $$= (1+e^2)/(2e) + a(e^2-1)/(2e)$$
but unless I'm missing something this does not further simplify.
Any help?
Let $z \in\mathbb C$ and $a\in A$ with $a^* =a$ and $a^2 = 1$. Then : \begin{align} \exp(za) &= 1 + za + \frac{1}{2}(za)^2 + \ldots + \frac{1}{n!} (za)^n + \ldots \\ &= \left(1 + \frac{z^2}{2}+ \ldots + \frac{z^{2n}}{(2n)!} + \ldots\right) + a \left( z + \frac{z^3}{3}+\ldots + \frac{z^{2n+1}}{(2n+1)!}+\ldots\right) \\ &= \cosh z +a\sinh z \end{align}
In your case, $z = \frac{i\pi}{2}$ so we have $\cosh z = 0$ and $\sinh z = i$ so indeed : $$\exp\left(\frac{i\pi}{2}a\right) = ia$$