What is the total circuit impedance of a parallel LC (6.8mH and 0.1uF) in series with a 1.2kOhm resistor, at 2.5kHz?
This answer is not for the purists: -


  why do I need to use j?


There is a good reason and that is because, in an inductor the instantaneous ratio of voltage and current (impedance) is just NOT a constant value like it is in a resistor. This means you have to use a frig-factor (purists will hate me calling it that of course). The frig factor is "j" but first, remind yourself about inductor and capacitor current and voltage waveforms when sine waves are used: -



And for a resistor it is: -



To be able to express the ratio of voltage and current (in a capacitor) as a ratio, you multiply the current by "j" and, in doing so, you have correctly shifted the current by 90 degrees. This is really important; multiplying a sinewave by j shifts it 90 degrees. Multiplying something by \$j^2\$ is the same as shifting it by 180 degrees which happens to be the same as multiplying a number by -1. OK so far?

So, the current in a capacitor (compared to it's voltage) is multiplied by j to signify that it leads the voltage by 90 degrees. For an inductor, it follows that the current is "marked" with -j to signify it lags the voltage by 90 degrees.

Incidentally, if \$j^2\$ = -1 then it has to follow that \$j^3\$ = -j and if you did some algebra you would find that j = \$\sqrt{-1}\$. You might have heard this mentioned somewhere? 

It also quite simply follows that \$ -j = \dfrac{1}{j}\$ (I'll be using this below)...

So, back the the impedance problem. The parallel combination of C and L have to be treated as proper complex numbers to do the math justice and the impedance is of course product divided by sum: -

Z = \$\dfrac{\frac{1}{jwC}\times\frac{wL}{-j}}{\frac{1}{jwC}+\frac{wL}{-j}}\$ = \$\dfrac{jwL}{1+ j^2w^2LC}\$ = \$\dfrac{jwL}{1- w^2LC}\$

Of note here is the denominator; when \$w^2LC = 1\$ the impedance is infinite. 

Or, put another way, resonance occurs when \$w = \dfrac{1}{\sqrt{LC}}\$ or \$F_{RES} = \dfrac{1}{2\pi\sqrt{LC}}\$.

Anyway that's the impedance of the parallel combo of L and C and, if you put a resistor in series the impedance becomes: -

Impedance = \$ R + \dfrac{jwL}{1- w^2LC}\$ 

I suspect this isn't going to help much because you haven't grasped the concept of "j" but keep plugging at it and "ask"!