Can they do it via light sails as is proposed here?

Can they be one-time and as short as 5 milliseconds just because of flickering due to rotation of the transmitter considering the fact, that the transmitter must aim to the target star for years? Let's do some math.

FRB 121102 is believed to be co-located in a dwarf galaxy about 3*10^9 light-years from Earth. There are two more FRBs believed to be from approx. 1Gpc = 3.26*10^9 light-years and 6*10^9 light-years.

But laser and maser beams undergo dispersion. The lower limit for dispersion is: disp_angle=lambda/(pi*omega), where lambda is wavelength of the beam and omega is diameter of the emitter

So let's rewrite the equation to use frequency instead of wavelength and put numbers into it.

disp_angle=c/(pi*omega*f)=3*10^8/(pi*3*10^7*10^9)=10^-8/pi where c is speed of light (approx.), f is 1GHz, which is approx. frequency of FRBs and omega is 30000 kilometers - slightly more, than twice the Earth diameter.

We're 3*10^9 light-years away, which is 2.284*10^16 meters, so the beam will diverge at least 2*2.284*10^16*tg(10^-8/pi)=1.45*10^8 meters near our solar system.

How fast the beam will move due to rotation of the galaxy? Sun in the Milky Way does one rotation in approx 240 millions of years, how much it will move in just 5 milliseconds?

2*pi/(240*10^6*365*86400*200)=4.15*10^-18 rads

How big the distance is in 3*10^9 light years?

4.15*10^-18*2.84*10^16=.117 meters. That's not much.

How fast the beam will move due to moving of the target star? The star with the highest angular velocity on our sky is the Barnard's star. It has angular velocity sqrt(798.71^2+10337.77^2)=10368 mas/yr. Wikipedia has extra value for movement in right ascension and in declination, therefore the Pythagorean theorem, mas is mili arc second, so it's slightly over 10 arc seconds per year.

10.368*(2*pi/(3600*360))/(365*86400*200)=7.9695*10^-15 rads per 5 milliseconds. 360*3600 is how many arc seconds are in the full circle and 365*86400*200 is how many 5ms pieces of time fits to one year (approximatelly).

8*10^-15*2.84*10^16=227.2 meters

So the beam travels on the earth 227.2 meters per 5 miliseconds. That looks reasonable right?

No, it isn't, because it's 1.45*10^8 meters big, so it will last about 1.45*10^8/227.2/200=3191 seconds on one of our recievers.

- How many real expensive antennas you'll need to reasonable travel inside the solar system? Hunreds of them? Thousands? That seems unlikely for me (and it isn't what was proposed in the original paper).

- How fast the target star must fly to let us observe what we're observing? More than 2900 times speed of light if it's the same far from the source as Barnard's star from us. It must be 1.45 million times closer to not exceed orbital velocity in the galaxy (using number for our Galaxy, but this will vary 10 times, 100 times maybe, but not 3000 or 1.5 million times). That's 2.17 light minutes - less than half of the smallest distance from Earth to Mars. Which is not interstellar travel.

- But in extreme gravitational fields objects are moving faster, right? Yes, if some extraterrestrial civilization lives near a supermassive black hole and they're travelling to another star orbiting it, it may be possible. But gravitational disturbances form other closely packed stars will make their solar system highly unstable.

- What about pulsed power? Our radars does that too. But continuous sources delivers more power to target in the long term and they does it cheaper. And there is no need to achive high peak power levels using light sail. Power times time is what counts. Their repeat rate should be something like once an hour to look like one-time phenomenas for us. That will decrease delivered power at least 150000 times which doesn't make sense for this usage.