Transimpedance/Transconductance Amp: Low/High Rin & Rout, App Usage

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peripatein
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Why do transimpedance and transconductance amplifiers have low Rin and Rout and high Rin and Rout, respectively? Also, I am having difficulties finding applications which require the use of current amplifiers.
I'd appreciate it if someone would kindly provide feedback on these three questions.
 
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Transimpedance/ transresistance gain is express as this
Ar = Vout/Iin
As you can see the input signal for transimpedance amplifier is current. And this is why we want Rin = 0 ohm. Because only if Rin = 0 ohm our amplifier has no influence on input current. The input of our amplifier should behave like an ideal ammeter.
We have the same situation for the output signal. The output of a transimpedance amplifier should behave like an ideal voltage. And we can achieve that only if Rout = 0 ohm
Because for Rout larger than 0 ohm. The output voltage will drop if we connect the load. Vout = V* Rload/(Rout + Rload).
So only if Rout = 0, Vout = V.
https://www.physicsforums.com/threads/output-gain-of-2-stage-amp.696241/#post-4410720

As for transconductance amplifiers again the we have the same story.
The gain is Ag = Iout/Vin,
In this case the input signal is a voltage (input of our amplifier should behave like an ideal voltmeter), so only if Rin = ∞ amplifier do not have any effects on the input voltage. Amplifier do not load the input source.
But output signal is a current, so again only if Rout = ∞ all output current will flow through the Rload (current divider).
Iout = I * Rout/(Rour + Rload)

As for the current amplifier. What about BJT common emitter amplifier?
 
Thank you very much! :-)