Determining Impedances of an Op-Amp Circuit
The op-amps themselves don't have zero output impedance but when configured with negative feedback they do.


  Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals. 


That's OK.


  However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case. 


I find that the 0 V input case is less helpful than, say a 1 V input case as it is too easy to balance out circuits with 0s everywhere. Let's use 1 V. And lets add some external Rout to make the output obviously non-ideal.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Rout is the output impedance of the op-amp.

With Rout in circuit, ask yourself what does the output of OA1 have to do to get 1 V at the inverting input? Answer: it has to get the voltage at OUT = \$ \frac {Z_3}{Z_2+Z_3} V_{IN} \$. That means that the output of the op-amp actually has to go some distance beyond Vout
to compensate for the voltage drop across the output resistance.


  It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance. 


Again, it's not the op-amp itself but rather the feedback that gives the circuit this nature.

Does that help?