There are better gyrator circuits so let me give you the downsides of this circuit first.
The idea behind this gyrator circuit is that at the emitter there is a voltage that connects back to the input via R1. If the emitter voltage is phase shifted to the input, it will take a current (via R1) from the input that appears to be reactive i.e. it looks like an inductor's current.
However the input is also feeding the network (C1 etc.) which does the phase shifting and so this "capacitive circuit" is in parallel with the "intentional" inductive current via R1. This makes it a band-pass filter but, it can look like an inductor across a range of frequencies.
Better gyrators use an op-amp or another transistor to buffer C1, Anyway, the analysis: -
At point B (base) the AC voltage relative there to the input voltage is: -
\$\dfrac{R_2}{R_2+\dfrac{1}{sC_1}}\$ and this voltage is also at the emitter (the emitter voltage is fractionally less in AC terms but this can be largely ignored). The emitter also acts as a reasonably good ideal voltage source so we don't have to worry about its output impedance of a few ohms.
The current into R1 is the voltage across it divided by R1 (I = V/R): -
Current = \$\dfrac{V_{IN}}{R_1}(1-\dfrac{sC_1 R_2}{sC_1 R_2+1})\$
The impedance, Z into R1 is \$V_{IN}\$ divided by current: -
Z = \$\dfrac{R_1}{1-\dfrac{sC_1 R_2}{sC_1 R_2+1}}\$ = \$\dfrac{R_1+sC_1 R_1 R_2}{1+sC_1 R_2-sC_1 R_2} = R_1+sC_1 R_1 R_2\$
In other words the impedance looking into R1 is an inductance of C1*R1*R2 in series with a resistor of R1 ohms. Remember there is current through the capacitor but this can be ignored if R2 is a lot bigger than R1 and the gain of the transistor is high.
The major disadvantage of a gyrator is that it needs DC power and a DC bias to set the circuit up correctly. The advantage of an inductor of course is that this is not needed.
Another disadvantage of a gyrator is that they are all (from what I've seen of the simple transistor/op-amp type) all ground referenced at one end of the inductor i.e. if you wanted an inductor that was ungrounded on both pins then it becomes a lot more complex.
Many inductors are used in power applications (such as switch mode power supplies). In simple words, gyrators are useless in these applications.
The main area that I see a lot of gyrators used is in audio graphic equalizers where a low power and low to medium Q factor is needed. Also frequencies are moderately low i.e. limited to no-more than 20kHz.
Best Answer
The way I approach this is to initially assume that the combined impedance of C and R is very high. Then I calculate the output voltage of the op-amp: -
$$V_{IN}\cdot\dfrac{sCR}{1+sCR}$$
Then I calculate the voltage across \$R_L\$: -
$$V_{IN}\cdot (1 - \dfrac{sCR}{1+sCR})$$ $$=\dfrac{V_{IN}}{1+sCR}$$
And, because: - $$I_{IN} = \dfrac{V_{IN}}{R_L+sCRR_L}$$
$$Z_{IN}=\dfrac{V_{IN}}{I_{IN}}=R_L +sCRR_L$$
In other words an inductance of \$CRR_L\$ in series with resistance \$R_L\$.