bjt amplifier input impedance
It's fairly simple (now that you've posted a schematic): -


Both 100 kΩ bias resistors are considered to be in parallel for input impedance because the 10 volt DC voltage is regarded as a short to AC signals
The emitter load is in parallel with the 5 kΩ emitter resistor and capacitors are regarded as shorts to AC when the circuit is operating mid-band AC.
The effect of the combined emitter resistance (5 kΩ||3 kΩ) is seen at the base as an impedance of 5 kΩ||3 kΩ multiplied by β. You probably know this but \$h_{FE}\$ = β.

So you have 50 kΩ in parallel with 187.5 kΩ = 39.5 kΩ. That's a fairly decent calculation for the AC input impedance mid-band. Trying to calculate it to any greater depth is missing a certain point about single transistor circuits - calculations are at best an approximation due to such large variations in β and its temperature dependency.
If you want to know what it is at low frequencies you need to take into account the capacitors shown in the circuit.