Do resistors consume reactive power?
By KCL, some of the current i flows through the resistor and some through the inductor. You can use the current divider rule to find the current through the resistor. Let's call that \$i_R\$.

Whatever \$i_R\$ is, the voltage across the resistor will be \$(435\ \Omega)i_R\$.
If we represent \$i_R\$ as \$I\angle\theta\$, then the voltage will be \$RI\angle \theta\$.

So the phase difference between the resistors current and voltage will be 0. 


  it came to my notice that if I find the power across this load and then find the current that passes through the resistor of interest and finally multiply current by voltage the resulted power would be a complex number.


This isn't clear. How did you use the power into the load to find the current through the resistor without first separating the resistive and inductive parts? You would have to assume that the inductor is responsible for all reactive power and the resistor is responsible for all real power. Then of course you will have the resistor power being purely real.