Understanding of two stage DC coupled complementary amplifier
First I must say that it is easier to design such a circuit then to find the DC operating point.
But let us try



simulate this circuit &ndash; Schematic created using CircuitLab
You can use any network analysis technique you are familiar with.
I will use KVL.
First KVL loop looks like this:

So the first equation will look like this:
$$(I_{C1} - I_{B2})R_{C1} =(I_{E2} + I_{B1})R_{E2} + I_{E2}R_{E3} + V_{BE2}  \:\:\: (1) $$
The second KVL loop will look like this:

And the KVL equation around this loop:
$$V_{CC} = (I_{E2} + I_{B1})R_{E2}+I_{B1}R_F+V_{BE1}+I_{E1}R_{E1} \:\:\: (2) $$
And from &quot;transistor action&quot; we know that:
\$I_E = I_B(\beta +1)\$ ; \$I_B =\frac{I_C}{\beta} \$ and \$I_B =\frac{I_E}{\beta+1} \$
Thus
$$(I_{C1} - \frac{I_{E2}}{\beta_2 +1})R_{C1} = V_{BE2}+ I_{E2}R_{E3} + (I_{E2} + \frac{I_{C1}}{\beta_1})R_{E2} \:\:\: (1) $$
$$V_{CC} = (I_{E2} +\frac{I_{C1}}{\beta_1} )R_{E2}+V_{BE1}+\frac{I_{C1}}{\beta_1 }\left(R_F+(\beta_1+1)R_{E1}\right) \:\:\: (2) $$
And for \$\beta_1 = 67 ;\: \beta_2=130\$ and \$V_{BE1} = V_{BE2} = 0.6V\$
I've got:
\$I_{C1} = 1.343mA\$
\$I_{E2} = 2.53mA\$
$$I_{C1} \approx  \frac{(R_{C1} + \beta (R_{E2} + R_{E3}))V_{CC} - (R_{C1} + \beta R_{RE3}) V_{BE}}{\frac{R_{C1} (R_{E2} + \beta(R_{E1} + \beta R_{E2}) + R_F)}{\beta} + (R_{E2} + R_{E3})R_F + \beta R_{E1}(R_{E2} + R_{E3}) +  R_{E2}R_{E3}}$$
And the voltage at \$Q_1 \$ collector is:
$$V_{C1} = V_{CC} - I_{C1}R_{C1} = 12V - 1.343mA \times 4.7k\Omega = 12V - 6.3V = 5.7V$$
and the voltage at \$Q_2\$ collector is:
$$V_{C2} = I_{C2}R_{C2} = I_{E2}\frac{\beta_2}{\beta_2+1} \times R_{C2} = 2.51mA \times 1.6k\Omega = 4V $$
For the AC (small-signal) analysis the equivalent circuit will look like this:



simulate this circuit
And the mid-frequency voltage gain is equal to:
$$\frac{V_{OUT}}{V_{IN}} \approx \frac{R_{C1}||(\beta +1)(R_{E3} + r_{e2})}{R_{E1} + r_{e1}} \times \frac{R_{C2}||R_L}{R_{E3} + r_{e2}}$$
If \$C_1\$ and \$C_2\$ have a higher capacity so that in the &quot;mid-frequency&quot; we can treat them as short circuit the gain will increase to:
$$\frac{V_{OUT}}{V_{IN}} \approx \frac{R_{C1}||(\beta +1)r_{e2}}{r_{e1}} \times \frac{R_{C2}||R_L}{r_{e2}}$$
Where
\$r_{e1} = \frac{V_T}{I_{E1}} \approx \frac{26mV}{I_{C1}}\$
\$r_{e2} = \frac{V_T}{I_{E2}} \approx \frac{26mV}{I_{C2}}\$
The End