 Solution

Half exponents in an odd-order root of unity group mean multiplication by the inverse of two modulo its order. This uniquely defines ${\zeta }_n^{a/2}$ inside $K_n$, and the definitions are compatible as $n$ varies. Put $d_a(z)=z^{-a/2}(1-z^a)$ on these roots. The symmetric cyclotomic unit is $c_n(a,b)=d_a({\zeta }_n)/d_b({\zeta }_n)$. Complex conjugation replaces each $d_a$ by $-d_a$, so the ratio is real. It is a global unit: choosing positive representatives for $a,b$ modulo $p^{n+1}$ gives geometric-series expressions for $(1-{\zeta }_n^a)/(1-{\zeta }_n)$ and its inverse, and similarly for $b$. The remaining factor is a root of unity.

The conjugates of ${\zeta }_n$ over $K_{n-1}$ are $\xi {\zeta }_n$ with $\xi \in {\mu }_p$. Because $p\nmid a$, multiplication by $a$ permutes ${\mu }_p$, andAlso ${\prod }_{\xi \in {\mu }_p}\xi =1$ for odd $p$, so $N_{n,n-1}({\zeta }_n)={\zeta }_{n-1}$. Applying the compatible half power proves $N_{n,n-1}(d_a({\zeta }_n))=d_a({\zeta }_{n-1})$. Dividing the analogous identities for $a,b$ proves

Embed the fields into the p-adic cyclotomic tower. Its interpolating Coleman power series isHere $(b-a)/2\in {\mathbb{Z}}_p$, and binomial expansion defines the half power. Both numerator and denominator vanish to order one at $T=0$, with coefficients $-a$ and $-b$, so cancellation gives an integral unit with $f_{a,b}(0)=a/b$. This applies also to negative integers $a,b$. Evaluation at ${\zeta }_n-1$ gives $c_n(a,b)$; the same root-product calculation shows $\mathcal{N}{f}_{a,b}=f_{a,b}$ for the Coleman norm operator.

For clarity, the normalization of the higher logarithmic derivative isIn the formal coordinate $T=e^z-1$, $D=d/dz$ andOnly the derivative of the logarithm is needed, so no choice of its constant term is involved. Differentiating gives the regular-at-zero expressionUsing the Bernoulli numbers defined by $z/(e^z-1)={\sum }_{r\ge 0}{B}_r{z}^r/r!$, with $B_1=-1/2$, givesThe two poles cancel. Taking $k-1$ further derivatives and evaluating at zero therefore yields all the requested values:In particular ${\delta }_1=0$, ${\delta }_2=(a^2-b^2)/12$, and ${\delta }_4=-(a^4-b^4)/120$. The odd case includes $k=1$; substituting $B_1$ into the even-index formula would incorrectly lose the symmetric normalization. All the displayed values lie in ${\mathbb{Z}}_p$, directly from the integral Coleman power series definition, even when the rational Bernoulli expression has an apparent p-adic denominator. For even $k\ge 2$ the same answer is $-(a^k-b^k)\zeta (1-k)$.