= Solution
\b[Logarithms on sufficiently deep units.] Suppose $\operatorname{char}K=0$, let $\ell$ be the <residue characteristic>, and put $e=v(\ell)$. For every integer $r>e/(\ell-1)$ the <p-adic logarithm> and <p-adic exponential function>, evaluated in $K$, converge on the required domains:
$$
\log(1+x)=\sum_{m\geq1}\frac{(-1)^{m+1}x^m}{m},\qquad
\exp(y)=\sum_{m\geq0}\frac{y^m}{m!}.
$$
The bounds $v(m!)=e\,v_\ell(m!)\leq e(m-1)/(\ell-1)$ and $v(m)\leq e(m-1)/(\ell-1)$ show that, for $x,y\in\pi^r\mathcal O_K$, every term beyond the linear one has <valuation> strictly greater than $v(x)$ or $v(y)$. The series therefore converge and preserve these lattices; in particular $\exp(y)\in1+\pi^r\mathcal O_K$. Their formal composition and addition identities are valid by convergence, giving inverse continuous group homomorphisms
$$
\log:1+\pi^r\mathcal O_K\xrightarrow{\;\cong\;}\pi^r\mathcal O_K,
\qquad \exp:\pi^r\mathcal O_K\xrightarrow{\;\cong\;}1+\pi^r\mathcal O_K.
$$
Thus the <logarithm isomorphism on deep principal units> is
$$
\boxed{1+\pi^r\mathcal O_K\cong(\mathcal O_K,+),
\qquad u\longmapsto\pi^{-r}\log u,\quad r>\frac e{\ell-1}.}
$$
The right-hand group is torsion-free.
\b[Roots of unity in $\mathbb Q_p$.] For odd $p$, take $\ell=p,e=1,r=1$. The <principal unit> group $1+p\mathbb Z_p$ is torsion-free by the <logarithm isomorphism on deep principal units>. The <Teichmuller representative> splitting leaves precisely the $p-1$ roots in $\mu_{p-1}$.
For $p=2$, take $r=2$. Each odd unit is uniquely $\pm u$ with $u\in1+4\mathbb Z_2$, since its residue modulo four is either one or minus one. This latter group is torsion-free. Consequently
$$
\boxed{|\mu(\mathbb Q_p)|=
\begin{cases}p-1,&p\text{ odd},\\2,&p=2,\end{cases}
\qquad \mu(\mathbb Q_2)=\{1,-1\}.}
$$
This computes the <roots of unity in the p-adic numbers>.
\b[The units of $\mathbb Q_2(i)$.] Set $\pi=1-i$. Its polynomial $T^2-2T+2$ is <Eisenstein>, so
$$
\mathcal O_K=\mathbb Z_2[i],\qquad v(\pi)=1,\quad v(2)=2,\quad k=\mathbb F_2.
$$
The <Eisenstein polynomial> facts justifying this are stated in the next solution. Write $U_s=1+\pi^s\mathcal O_K$. Reduction gives $\mathcal O_K^\times=U_1$, and the <successive quotients of principal-unit groups>
$$
U_s/U_{s+1}\cong(k,+),\qquad
1+\pi^s b\longmapsto\bar b
$$
have order two for $s\geq1$. Hence $[U_1:U_3]=4$. The four roots in $\mu_4=\{1,i,-1,-i\}$ have distinct images modulo $U_3$, since
$$
v(i-1)=v(-i-1)=1,\qquad v(-1-1)=2.
$$
They exhaust the quotient and intersect $U_3$ trivially. Since $r=3>2/(2-1)$, the <logarithm isomorphism on deep principal units> gives $U_3\cong(\mathcal O_K,+)$. Therefore the <unit decomposition of the 2-adic Gaussian field> is
$$
\boxed{\mathcal O_K^\times=\mu_4\times U_3
\cong\mu_4\times(\mathcal O_K,+).}
$$
In particular these four elements are all its roots of unity.
\b[<Quadratic extensions>.] A nonzero element of $K$ is uniquely a power of $\pi$ times a unit, so $K^\times\cong\pi^{\mathbb Z}\times\mathcal O_K^\times$. Squaring on the decomposition above acts as squaring on $\mu_4$ and multiplication by two on the additive ring $\mathcal O_K=\mathbb Z_2\oplus\mathbb Z_2i$. Thus the <square-class group of the 2-adic Gaussian field> is
$$
\boxed{K^\times/(K^\times)^2
\cong C_2\times(\mu_4/\mu_4^2)\times(\mathcal O_K/2\mathcal O_K)
\cong C_2^4.}
$$
By <quadratic extensions from square classes>, in characteristic different from two the nontrivial <square classes> classify <quadratic extensions> $K(\sqrt a)$: two such extensions are $K$-isomorphic exactly when $a/b$ is a square. There are therefore $\boxed{2^4-1=15}$ <quadratic extensions> up to $K$-isomorphism.
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