How fast would a clock falling into a blackhole tick relative to the reflection of a clock stationed far away? Lets say you have two clocks, you attach one to a mirror and drop in into a black hole, while you keep the other next to you, point it towards the black hole and view it's reflection on the mirror.
Which would tick faster and by how much?
 A: If you drop it from rest at radial coordinate $\rm r=r_0$ the free fall velocity at $\rm r=r_1$ is
$$\rm v=c \ \sqrt{\frac{r_s \ (r_0-r_1)}{r_1 \ (r_0-r_s)}}$$
which in the limit of $\rm r_0=\infty$ reduces to the escape velocity
$$\rm v_{esc}=c \ \sqrt{r_s/r_1}$$
Therefore you get a kinematic doppler of
$$\rm d_k= \sqrt{\frac{1-v/c}{1+v/c}}$$
and a gravitational time dilation of
$$\rm d_g= \sqrt{\frac{1-r_s/r_1}{1-r_s/r_0}}$$
so the observed frequency $\rm f_o$ compared to the initial frequency $\rm f_i$ is
$$\rm f_o = f_i \ d_k \ d_g$$
and for the reflection of your clock in the moving mirror
$$f_{\rm o} = {\rm f_i} \ \rm d_k^2$$
The observer falling with the mirror would see your clock and its reflection with
$${ {{ }_F}_{\rm o}} = {\rm f_i} \rm \ d_k/d_g$$
If you drop it from $\rm r_0=4GM/c^2$ to $\rm r_1=3GM/c^2$ the mirror would gain a local velocity of $\rm c/\surd 3$, and the observed frequency from your position at $\rm r_0$ would be
$$\rm f_{\rm o} = (1-1/\surd 3) \ f_{\rm i} = 0.42265 \ {\rm f_i} $$
and for the reflection of your own clock in the falling mirror
$$f_{\rm o} = \rm (2-\surd 3) \ f_i = 0.26795 \ f_i $$
The falling observer would see his own clock ticking normal, and yours with
$${ {{ }_F}_{\rm o}} = \rm (3-\surd 3)/2 \ f_i = 0.63397 \ f_i$$
If $\rm r_1$ is the Schwarzschild radius $\rm r_s=2GM/c^2$ the redshift observed by the stationary observer at $\rm r_0$ is infinite and ${\rm f_o}=f_{\rm o}=0$.
${ {{ }_F}_{\rm o}}$ however is neither $0$ nor $\infty$ at $\rm r_s$, but you have to use the limes to avoid an undefined $0/0$ or switch to other coordinates, but to keep it short and simple the limit $\rm r_1 \to r_s$ also does the job if we use regular coordinates:
the falling observer sees the stationary clock at his initial position at $\rm r_0$ ticking with its half rate at the moment when he falls through the horizon at $\rm r_s$ (the gravitational and the kinematic time dilation cancel, so in this frame of reference the pure classical doppler remains).
The gravitational time dilation in the reflection of your own clock in the mirror would cancel out, since the frequency of your signal gains as much on its way down as it loses on the way back up, so for that only the kinematic part remains (but squared since it's a reflection).
If the mirror below you were at rest only the gravitational component remains, and you would see your clock's reflection ticking at its original speed, and the lower clock slower:

The local observer below you on the other hand would see his own clock ticking at normal rate, and your reflection faster than normal:

If the mirror is not at rest it depends on whether the doppler due to the velocity or the gravitational time dilation is larger, since the first is relative (you always see the other clock slower) while the latter is absolute (the higher clock ticks faster than the lower one in both systems).
The calculations are assuming that we have a Schwarzschild black hole, for rotating Kerr black holes the whole procedure is a little bit more complicated.
