= Solution
If the real part vanishes identically, the <norm>-difference identity becomes equality, so
$$
\|(I-T)x\|=\|(I+T)x\|\quad\text{for all }x.
$$
Thus the same substitution as above gives $\|Sz\|=\|z\|$ for every $z$. To prove surjectivity as well, observe that both $T$ and $-T$ satisfy the nonnegative-real-part hypothesis. The preceding part makes both $I+T$ and $I-T$ invertible. Their product $S=(I-T)(I+T)^{-1}$ is therefore invertible and onto. It is a surjective <norm>-preserving <linear operator>, so the equivalences in (a) prove
$$
\boxed{S\text{ is unitary}.}
$$
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