The resistor Z_g is there to ensure that the offset voltage at the positive and negative terminals of the opamp is as close to identical as possible (hence not affecting the circuit, as it appears as common-mode offset).

By making the input bias current at each terminal (which should be approximately the same) flow through identical impedances, you should end up with the same voltage offset at each node. Hence Z_g is calculated to equal the same impedance as would be seen "looking out of" the inverting input:

\$Z\_g = Z\_i || Z\_f = {{Z\_i\times Z\_f} \over {Z\_i + Z_f}} \$

As Brian Drummond points out, with the TL081 having a JFET input stage and an input bias current that you would need very expensive equipment to measure, practically speaking it is likely to be redundant in this particular schematic.

The resistor Z_g is there to ensure that the offset voltage at the positive and negative terminals of the opamp is as close to identical as possible (hence not affecting the circuit, as it appears as common-mode offset).

By making the input bias current at each terminal (which should be approximately the same) flow through identical impedances, you should end up with the same voltage offset at each node. Hence Z_g is calculated to equal the same impedance as would be seen "looking out of" the inverting input:

\$Z\_g = Z\_i || Z\_f = {{Z\_i\times Z\_f} \over {Z\_i + Z_f}} \$

As Brian Drummond points out, with the TL081 having a JFET input stage and an input bias current that you would need very expensive equipment to measure, practically speaking it is likely to be redundant in this particular schematic.

However, it is also likely that since the TL081 is the default opamp selected by CircuitLab when drawing a schematic I imagine that your original schematic may be for a different device, where this technique has a greater effect.