Example 5.13 goes like this.
Let $A=C(\Gamma)$, where $\Gamma=\{z: |z|=1\}$, let $B$ be the image of the disc algebra $A(\Delta)$ under the isometric isomorphism $f\rightarrow f\restriction_\Gamma$ and $u(z)=z$. Then $Sp(A,u)=\Gamma, Sp(B,u)=\Delta.$
Can any one tell me why $Sp(B,u)=\Delta$?
If $A$ is a commutative, unital Banach algebra, and $a\in A$, then $\mathrm{Sp}(A,a)=\hat a(M_A)$, where $\hat a$ denotes the Gelfand transform of $a$ and $M_A$ denotes the maximal ideal space (or character space) of $A$.
The maximal ideal space of the disk algebra $B$ (following your notation) can be identified with the closed unit disk, and so $\mathrm{Sp}(B,u)=\hat u(M_A)$. But the Gelfand transform of $u$ is just the function $z\mapsto z$ on the defined on the closed unit disk, thus the spectrum is $\Delta$.