proving an element is unitary in a C* algebra Let $p,q$ be projections in a unital C*-algebra $A$ and let $\tilde{A}$ be the unitization.  I'd like to show that if $p\sim_u q$ (ie $q=zpz^*$ for $z$ unitary in $\tilde A$), then $q=upu^*$ for $u$ unitary in $A$.
The book proceeds as follows:  Set $1_\tilde A$ as the unit in $\tilde A$ and $1_A$ the unit in $A$.  Let $f=1_\tilde A-1_A$.  Then $\tilde A=A+\mathbb{C}f$.  Assuming that $q=zpz^*$ for $z$ unitary in $\tilde A$, then $z=u+\alpha f$ for $\alpha\in\mathbb{C}$ and $u\in A$.  It claims that $u$ is unitary from a calculation.  This is what I cannot figure out.
I want to show that $uu^*=u^*u=1_A$.  Solving for $u$ and multiplying, we get:
$uu^*=(z-\alpha f)(z^*-\bar{\alpha}f)=1_\tilde A-\bar{\alpha}fz-\alpha fz^*-|
\alpha|^2f^2$.
I think that $f^*=f$, but I need to know that the involution on $A^*$ agrees with the involution on $A$.  I think this is true but I am not sure.  I know there is exactly one norm that allows $\tilde A$ to be a $C^*$ algebra but not sure if the involution is extended.  Assuming that's true, we also can see that $f^2=f$, and so the problems reduces to showing:
$-\bar{\alpha}fz-\alpha fz^*-|
\alpha|^2f=-f=1_A-1_\tilde A$.
I don't know how to continue.  I've tried factoring and the like, but my main problems are that the $\alpha$ and $\bar\alpha$ in addition to the $z$ and $z^*$ are separated by addition instead of multiplication so I cannot simplify them.
 A: We have $\tilde{A} = A \oplus \mathbb{C}$, where $\tilde{A}$ is a normed space by the formula $$||(a,\lambda)|| = \sup\limits_{b \in A} ||ab + \lambda b||$$ and a $\ast$-algebra by the formula $(a,\lambda)^{\ast} := (a^{\ast},\overline{\lambda})$.  Identify $A$ as a subalgebra as $a \mapsto (a,0)$.  So the involution is preserved.  The identity $1_{\tilde A}$ of $\tilde{A}$ is $(0,1)$.  So your $f$ is $$1_{\tilde A} - 1_A= (0,1) - (1_A,0) = (-1_A,1)$$ Assume $z = (g,\alpha)$ is unitary in $\tilde A$ for some $g \in A, \alpha \in \mathbb{C}$.  Letting $u := g + \alpha1_A$, we have $$u + \alpha f = (g+\alpha1_A, 0) + (-\alpha 1_A, \alpha ) = (g, \alpha) = z$$ Since $z$ is unitary in $\tilde A$, $$(0,1) = 1_{\tilde A} = (g,\alpha)(g^{\ast}, \overline \alpha) = (gg^{\ast} + \alpha g^{\ast} + \overline \alpha g, |\alpha|^2)$$ So $|\alpha|^2 = 1$ and $gg^{\ast} + \alpha g^{\ast} + \overline \alpha g = 0$.  We want to check that $u = g + \alpha 1_A$ is unitary in $A$: $$uu^{\ast} = (g + \alpha 1_A)(g^{\ast} + \overline{\alpha} 1_A) = gg^{\ast} + \alpha g^{\ast} + \overline \alpha g + |\alpha|^2 1_A = 0 + |\alpha|^2 1_A = 1_A $$
