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I know that the offset voltage of an op amp can be multiplied by the voltage gain to reveal the error in an output voltage, but what happens when it's a transimpedance amplifier?
The gain of a transimpedance amplifier is \$ Vout/Iin \$.
The gain is dependent on the input current, not the input voltage. Maybe I'm thinking about this the wrong way?
This is a typical TIA: -
Image from here
Because the input is a current source, the effective impedance of that current source is infinite hence, it can be modelled as having a very high value voltage as a source in series with a very high value resistance. Hence, the gain is minus unity (-1).
This means that any input offset will be seen as the same error voltage at the output.
Of course, if the current source is not ideal then it can be modelled with a shunt resistor in parallel with that source. Now, you can regard the input offset as being amplified by \$1 + \dfrac{R_f}{R_{shunt}}\$ and, that voltage will appear at the output.
