Why is imidazole considered an electron deficient heterocycle?
As indicated in the comment below, a rule of thumb for these particular compounds, is that aromatic nitrogens with double bonds (e.g., not pyrrole) are "electron deficient."

The term is a bit ill-defined. It does not mean the aromatic system (which usually still has 4n+2 electrons) actually has a different electron count.

More broadly, there can be several ways of declaring "electron rich" or "electron deficient," for organic compounds, but in the context of a "push-pull" conjugated or aromatic system (in the paper you link), these refer to rings or functional groups that are electron donors or electron acceptors.

So:


Electron deficient - tends to gain negative charge as an electron accepting group or ring
Electron rich - tends to gain positive charge as a donor group or ring


The charge transfer need not be a full electron - it is almost always much less (e.g., a small fraction).

The second part of your question is how you can determine this. 

Conjugated systems are typically imagined as a particle-in-a-box.. that the electrons are highly delocalized. The "electron donor" is a region with higher potential energy and the electron acceptor is a region with lower potential energy. Thus, the electrons have slightly greater probability on the electron acceptor.

Therefore, you determine "electron rich" or "electron deficient" in comparison with the rest of the conjugated system.

Typically, electron deficient rings have more negative HOMO and LUMO energies - making them more difficult to oxidize (i.e., they are "deficient" in electrons and it's hard to give up another) and easier to reduce (i.e., adding an electron makes them less "deficient").

Electron rich rings have more positive HOMO and LUMO energies - making them easier to oxidize and give up an electron.

Again, this is a rule of thumb and it's almost always partial charge transfer.