Quasar redshift The quasar 3C 273 has a redshift z=0.158. A question in a textbook: could this be a gravitational redshift instead of cosmological (=resulting from the expansion of space)?
My answer: no. Firstly, the visible light from the quasar is produced by the accreting gas at a distance from the central supermassive black hole, not by its "surface" (which doesn't physically exist). In the regions where the light is produced, the gravity gradient (?) or the curvature of space isn't large enough to produce z=0.158. Such a redshift could only be produced very near the Schwarzschild radius.
Secondly, if the redshift were gravitational, we wouldn't see a narrow Balmer line H$\beta$ with z=0.158 but instead a broad line because some light originates from near the "surface" (large gravity gradient) and some light from further away.
Am I right?
 A: Your case is not quite watertight - it hinges in your assertion that the optical light that is seen comes from some way out from the black hole (SMBH). The thing is that gravitational redshift can be larger than 0.2 and it is also aided by the relativistic transverse Doppler effect in the orbiting material.
Some details:
Gravitational redshift around a black hole of mass $M$ is governed by
$$z = \left( 1 - \frac{2GM}{rc^2}\right)^{-1/2} - 1,$$
where $r$ is the radial coordinate of 
a light source in orbit around the black holes. This formula would apply for any spherically symmetric mass distribution.
The last stable, possible circular orbit around a non-rotating black hole is at $r = 6GM/c^2$, where $M$ is the black hole mass. This means the gravitational redshift factor could be as large as $0.22$. 
On top of this you must consider the
Relativistic Doppler shift.
The relativistic doppler shift for a source moving at a speed $v$ at an angle $\theta$ (in the reference frame of the observer), then the emitted and observed frequencies are related by
$$ f_o = \frac{f_e}{\gamma\left[ 1 + (v/c)\cos\theta\right]},$$ 
where $\gamma = (1 - v^2/c^2)^{-1/2}$. This means that even when $\theta=90^{\circ}$ and the source orbiting the black hole is moving neither towards or away from an observer on Earth, that there is a "transverse doppler redshift" of 
$$z = \gamma - 1$$
So although an observer on Earth would see the frequency of a source in orbit around a black hole go up and down due to the doppler shift (a spectral line would be broadened by $\sim \pm v/c$), there would be a net redshift due to the transverse doppler effect.
Material at the innermost circular orbit would have a speed of half the speed of light and $\gamma = 1.15$. Thus the redshift due to the transverse doppler effect would be $z=0.15$ and almost the same as the gravitational redshift. At larger orbital radii, the gravitational redshift becomes smaller but more dominant.
On top of the net tranverse doppler redshift there will be a doppler broadening as the gas orbits the black hole. The amplitude of this will depend on the inclination of the orbit to the line of sight. At its largest, $\theta = 0$, the redshift/blueshift will be factors of 
$$z = \gamma(1 \pm v/c) -1$$.
Thus for a source in orbit at the innermost stable circular orbit, this would lead to a factor of two broadening of any spectral line.
Thus if the gas orbits close enough to the SMBH at 3C273 you probably could produce the redshift seen. The maximum broadening that you should see is much broader than the observed spectral lines, but this could be accounted for by the disc being at a low inclination and the emission coming from a narrow range of radii.
In fact, as you say, the optical emission you see from the broad line region comes from considerably further out than a few Schwarzschild radii, so the effects above, whilst very important for hot X-ray emitting gas in quasars, are probably not a big deal in the optical. So you need a further step in your answer to prove/argue that the optical emission does not arise that close to the SMBH.
