Computing rlc circuit in parallel
Resistors and capacitors don't add algebraically. If your resistor is 1\$\Omega\$ and the magnitude of the capacitor's impedance is 1\$\Omega\$ their series circuit won't give you 2\$\Omega\$, but 1.4\$\Omega\$. That's because you add vectors which are at 90° with each other, which is represented by multiplying with \$j\$. The 1.4 is \$\sqrt{2}\$, which you get if you vectorially add two perpendicular vectors with modulus 1. So you'll have this factor \$j\$ in your equations.
Also, the impedance of capacitors and inductors is frequency dependent, it's not just \$C\$ or \$L\$. In the equation below this frequency dependence shows as the factor \$\omega\$.


  \$ Z = \dfrac{1}{\dfrac{1}{R} + \dfrac{1}{Z_L} + \dfrac{1}{Z_C}} = \dfrac{1}{\dfrac{1}{R} + \dfrac{1}{j \cdot \omega \cdot L} + j \cdot \omega \cdot C} = \dfrac{1}{\dfrac{1}{R} + j \left(\omega \cdot C - \dfrac{1}{\omega \cdot L} \right)} \$  


Note the negative term between the brackets. This means that the imaginary part can become zero, namely when  


  \$ \omega = \sqrt{\dfrac{1}{L C}} \$  


In that case the parallel circuit becomes purely resistive with  


  \$ Z = R \$