BJT Circuit DC Analysis
Assuming:


\$Q_1\$ and \$Q_2\$ are forward-biased
\$V_s\$ is DC
the transistors are physically identical (temperature, doping concentrations, \$\beta\$, etc.)


Then, we determine that (from KVL on \$Q_2\$ base-emitter junction):
\$0 - V_{BE} = V_E \$  
\$V_E = -V_{BE}\$

Then, we determine that (from KVL on \$Q_1\$ base-emitter junction):
\$V_s - I_{B1}(10000) - V_{BE} = V_E = -V_{BE}\$  
\$V_s - I_{B1}(10000) = 0\$  
\$V_s - \frac{I_{E1}(10000)}{\beta + 1} = 0\$  
\$I_{E1} = \frac{V_s(\beta + 1)}{10000}\$  

Using KCL at the emitter node:
\$I_{E1} + I_{E2} = 0.001\$  
\$I_{E2} = 0.001 - I_{E1} = 0.001 - \frac{V_s(\beta + 1)}{10000}\$  

Conclusion:
\$g_{m1} = \frac{I_{E1}}{\alpha V_T} = \frac{V_s(\beta + 1)}{10000\alpha V_T}\$
\$g_{m2} = \frac{I_{E2}}{\alpha V_T} = \frac{0.001 - \frac{V_s(\beta + 1)}{10000}}{\alpha V_T}\$