Calculate the length of the curve $c:[0,\pi]\rightarrow \mathbb{H}^2$,$\displaystyle c(t)=\frac{\sin t+ia\cos t}{\cos t-ia\sin t}$ for $a>1$ in hyperbolic metric.
I know that hyberbolic length=$\displaystyle\int_0^\pi\frac{|c'(t)|}{Im c(t)}dt$ but here that calculation for $|c'(t)|$ is very long and it does not seem to be simplifying to a short formula. Is there an easy way of calculating this, rather than directly differentiating it?
My original instinct was totally wrong. This is a parametric equation for a circle of radius $\frac12(a-\frac 1a)$ centered at $\frac12(a+\frac1a)i$. Having found that by divine inspiration, you can reparametrize in the usual way you do a circle and then the integral is not bad: If you have $w=(R\cos\theta)+i(c+R\sin\theta)$, $0\le\theta\le 2\pi$, then you get a hyperbolic length of $\dfrac{2\pi R}{\sqrt{c^2-R^2}}$. (This integral is a standard one with the residue theorem.) By the way, this Euclidean circle is also a hyperbolic circle — with center at $i$.
Here's the interesting algebra. If we consider $z=\cos t+ia\sin t$, then the upper semi-ellipse given by $0\le t\le\pi$ maps bijectively to the unit circle by the map $w=\dfrac z{\bar z} = \left(\dfrac z{|z|}\right)^2$. [So there's implicitly double angle magic going on in this problem.] But $$\frac{\sin t+ia\cos t}{\cos t-ia\sin t} - i\frac{a+\frac1a}2 = i\frac{a-\frac1a}2\frac{\cos t+ia\sin t}{\cos t-ia\sin t},$$ from which we see that $$\left|\frac{\sin t+ia\cos t}{\cos t-ia\sin t} - i\frac{a+\frac1a}2\right| = \frac{a-\frac1a}2,$$ showing that the image is contained in the promised circle. On the other hand, because of the observation above, we get precisely the entire circle.
In hindsight, if we go back to the original parametrization, if we multiply and divide by the conjugate of the denominator, we do in fact see double angles leaping out at us: \begin{align*} \frac{\sin t+ia\cos t}{\cos t-ia\sin t} &= \frac{(\sin t+ia\cos t)(\cos t+ia\sin t)}{\cos^2 t+a^2\sin^2 t} = \frac{(1-a^2)\sin t\cos t + ia}{\cos^2 t+a^2\sin^2 t}\\ &= \frac{\frac{1-a^2}2\sin 2t + ia}{\frac{1-a^2}2\cos 2t+\frac{a^2+1}2}= \frac{i-\frac{a-\frac 1a}2\sin 2t}{\frac{a+\frac 1a}2-\frac{a-\frac 1a}2\cos 2t}. \end{align*} Hmm ... This looks suspicious. Setting $2t=u$, we have $w=\dfrac{i-R\sin u}{c-R\cos u}$, with $c=\frac{a+\frac 1a}2$, $R=\frac{a-\frac 1a}2$, and now the usual $0\le u\le 2\pi$. I'll leave it to you to pursue this approach.