= Solution
An element $a$ is a \b[<Normal element of a C-star algebra>] if $a^*a=aa^*$. It is a \b[<Unitary element of a C-star algebra>] if $a^*a=aa^*=1$. Its <Banach algebra spectrum> is
$$
\boxed{\sigma_A(a)=\{z\in\mathbb C:a-z1\notin A^\times\},}
$$
where $A^\times$ is the <group of invertible elements of a Banach algebra>.
For normal $a$, the \b[<continuous functional calculus>] is the unique unital isometric star-isomorphism
$$
\Phi_a:C(\sigma_A(a))\longrightarrow C^*(1,a)\subseteq A,
\qquad f\longmapsto f(a),
$$
which sends the coordinate function $z\mapsto z$ to $a$. It preserves sums, products, complex <scalars> and the involution: $\Phi_a(\overline f)=\Phi_a(f)^*$. In particular $\|f(a)\|=\max_{z\in\sigma_A(a)}|f(z)|$. The <spectral mapping theorem> in this setting states
$$
\boxed{\sigma_A(f(a))=f(\sigma_A(a))\quad(f\in C(\sigma_A(a))).}
$$
The common unit and spectral permanence for C-star <subalgebras> permit computing this spectrum in $A$ itself.
For unitary $u$, the C-star identity gives $\|u\|=\|u^*\|=1$. If $|z|>1$, $u-z1=-z(1-u/z)$ has a multiplicative inverse by the <Neumann series>. If $|z|<1$, write $u-z1=u(1-zu^*)$ and apply the same series. Therefore $\sigma_A(u)$ lies on the <unit circle>.
For $|t|=1$, the denominator $1+\overline\lambda t$ is nonzero because $|\lambda|<1$. The function is consequently continuous on the spectrum, and
$$
|t+\lambda|^2=1+|\lambda|^2+\overline\lambda t+\lambda\overline t
=|1+\overline\lambda t|^2.
$$
Thus $|f(t)|=1$ throughout $\sigma_A(u)$, so $\overline f f=f\overline f=1$ there. Applying the star-homomorphism $\Phi_u$ proves
$$
\boxed{f(u)^*f(u)=f(u)f(u)^*=1.}
$$
Hence $f(u)$ is unitary. Explicitly, the same calculus gives
$$
\boxed{f(u)=(u+\lambda1)(1+\overline\lambda u)^{-1},}
$$
with the denominator's inverse also supplied by a norm-convergent Neumann series. This is the <Möbius transform of a unitary element>. The conjugation on $\lambda$ in the denominator is essential to the unit-modulus identity.
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