Suppose $\Bbb T$ is the unit circle and $M$ is the C*-algebra of all $2\times 2$ complex matrices. Consider the C*-algebra $A: = C(\Bbb T, M)$. Let $E$ and $F$ be the projections in $A$ given by $$E(e^{i\theta})= \begin{bmatrix}0& 0\\\phantom{}0& \phantom{}1\end{bmatrix},~~~~~~~~~~~~~~~~~~~F(e^{i\theta})= \frac{1}{2}\begin{bmatrix}1-\cos\theta& -\sin\theta\\\phantom{}-\sin\theta& \phantom{}1+\cos\theta\end{bmatrix}$$ $(0\leq \theta\leq 2\pi)$, and let $U$ be the unitary element $(exp i\pi E)(exp i\pi F)$ of $A$. Show that $$U(e^{i\theta})= e^{i\theta} P + e^{-i\theta} Q~~~~~~(0\leq \theta\leq 2\pi)$$ where $P$ and $Q$ are the projections in $A$ given by $$P=\frac{1}{2}\begin{bmatrix}1& -i\\\phantom{}i& \phantom{}1\end{bmatrix},~~~~~~~~~~~~~~~~~~~Q=\frac{1}{2}\begin{bmatrix}1& i\\\phantom{}-i& \phantom{}1\end{bmatrix}$$ Deduce that $U$ is not itsef an exponential unitary.
Please help me. Thanks.
The exponential of an idempotent $R$ is very easy to calculate: if $R^2=R$, then $$ e^{i\pi R}=\sum_{k=0}^\infty\frac{i^k\,\pi^k\,R^k}{k!} =I+\sum_{k=1}^\infty\frac{i^k\,\pi^k\,R}{k!} =I-R+R\left(\sum_{k=0}^\infty\frac{i^k\pi^k}{k!}\right)\\ =I-R+e^{i\pi}R=I-2R. $$ Then $$ U(e^{i\theta})=e^{i\pi E(e^{i\theta})}e^{i\pi F(e^{i\theta})} =(I-2E(e^{i\theta}))(I-2F(e^{i\theta}))\\ =\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}\cos\theta&\sin\theta\\ \sin\theta&-\cos\theta\end{bmatrix} =\begin{bmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{bmatrix}. $$ On the other hand, $$ e^{i\theta}P+e^{-i\theta}Q=\begin{bmatrix}\frac12(e^{i\theta}+e^{-i\theta})&-\frac i2(e^{i\theta}-e^{-i\theta})\\ \frac i2(e^{i\theta}-e^{-i\theta})&\frac12(e^{i\theta}+e^{-i\theta})\end{bmatrix} =\begin{bmatrix} \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta \end{bmatrix}. $$
As for the last part, I don't know what "exponential unitary" means.