= Solution
<Anti-self-adjoint> means $T^*=-T$. Unitary means $T^*T=TT^*=I$, so $T^{-1}=T^*$. For a <unitary operator> $\|Tu\|^2=\langle u,T^*Tu\rangle=\|u\|^2$, and the space is nonzero. Thus \b[$\|T\|=1$].
The <series> for the exponential converges absolutely in the <Banach algebra> $B(V)$, since $\sum_n\|T^n/n!\|\leq e^{\|T\|}$. The adjoint is <norm> continuous and conjugate linear, and its action on these real scalar coefficients gives $(e^T)^*=e^{T^*}$. The absolutely convergent Cauchy product gives
$$
e^Te^{-T}=e^{-T}e^T=I,
$$
because its degree-$n$ coefficient for $n>0$ is $T^n\sum_{j=0}^n(-1)^j/[j!(n-j)!]=0$. If $T^*=-T$, these identities prove \b[$e^T$ is unitary].
If $T$ is invertible, zero is outside the spectrum of $T^{-1}$. For $\mu\ne0$ the factorization
$$
T^{-1}-\mu I=-\mu T^{-1}(T-\mu^{-1}I)
$$
shows that the left side is invertible exactly when the final factor is. Hence
$$
\boxed{\sigma(T^{-1})=\{\lambda^{-1}:\lambda\in\sigma(T)\}.}
$$
For unitary $T$, both $T$ and $T^{-1}=T^*$ have <norm> one. The <norm> bound on each spectrum gives $|\lambda|\leq1$ and $|\lambda^{-1}|\leq1$, forcing \b[$\sigma(T)$ to lie on the unit circle].
The converse fails: on $\mathbb C^2$ with its usual <inner product> take
$$
\boxed{T=\begin{pmatrix}1&1\\0&1\end{pmatrix}.}
$$
Its spectrum is $\{1\}$, but $T^*T=\begin{pmatrix}1&1\\1&2\end{pmatrix}\ne I$, so it is not unitary.
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