Complex length of geodesic added in hyperbolic Dehn surgery

Question

Suppose $M$ is a cusped finite-volume hyperbolic $3$-manifold, say with a single cusp for simplicity. Following [NZ, Section 4] we can parametrize deformations of the hyperbolic structure with a complex paramter $u$, chosen so that $u = 0$ gives the complete hyperbolic structure on $M$. We can identify $u$ with the derivative of the holonomy of a meridian acting on the cusp torus. There is a second parameter $v$ associated with the longitude. A point $u$ is compatible with $p/q$ Dehn surgery on $M$ if $pu + qv = 2\pi i$.

I want to understand the complex length $$ \operatorname{L}_{\mathbb C} (\gamma) = \operatorname{length}(\gamma) + i \operatorname{torsion} \gamma $$ of the geodesic core $\gamma$ of the torus added in the surgery. If we choose $r,s \in \mathbb{Z}$ with $ps - qr = 1$, then $(p,q)$ and $(r,s)$ are a basis for the homology of the torus near the cusp, and $(p,q)$ is the curve sent to zero by the surgery, so $\gamma$ corresponds to $(r,s)$ up to orientation and we conclude that $$ \operatorname{L}_{\mathbb C} (\gamma) = \pm(ru + sv). $$

[NZ] determine that the right sign is $-(ru + sv)$. Furthermore they observe that $$ \operatorname{length}(\gamma) = -\Re(ru + sv) = -\frac{1}{2\pi} \Im(u \overline v) = \frac{1}{2\pi} \Re(i u \overline v) $$ In particular, the last expression $\Re(i u \overline v)/2\pi$ makes sense on all of generalized Dehn surgery space, not just at the points $u$ with $pu + qv = 2\pi i$ for coprime integers $p,q$.

The obvious conjecture is that the complex length is also a function $$ \operatorname{L}_{\mathbb C} (\gamma) = \frac{1}{2\pi} i u \overline v \tag{1} $$ that makes sense on all of Dehn surgery space. However, I don't see an obvious way to prove this from the method of [NZ]. My question is: Is formula (1) (or something closely related) true? How could one prove it?

[NZ] W. D. Neumann and D. Zagier, Volumes of hyperbolic three-manifolds. Topology 24, 307--332 DOI 10.1016/0040-9383(85)90004-7

Answer 1

After more thought, I realize this can't possibly be true, by an easy extension of the methods in [NZ]. We can re-write our two equations $pu + qv = 2\pi i$ and $\operatorname{L}_{\mathbb C} (\gamma) = -(ru + sv)$ as the matrix equation $$ \begin{pmatrix} p & q \\ r & s \end{pmatrix} \begin{pmatrix} \Re u & \Im u \\ \Re v & \Im v \end{pmatrix} = \begin{pmatrix} 0 & 2 \pi \\ - \operatorname{length}(g) & - \operatorname{torsion}(g) \end{pmatrix}. $$ Taking the determinant is what gives $\operatorname{length}(\gamma) = \frac{1}{2\pi} \Re(i u \overline v)$. If we subtract $\left( \begin{smallmatrix} 2\pi & - 2\pi \\ 0 & 0 \end{smallmatrix}\right)$ from both sides and take the determinant again we get $$ (\Re(u) + 2 \pi s)( \Im (v) + 2 \pi r) - (\Re(v) - 2 \pi r) (\Im(u) + 2 \pi s) = -2\pi \operatorname{torsion}(g). $$ which is not the imaginary part of $i u \overline v$.