Plotting impedance of an inductor and capacitor in series

The capacitor and inductor are in series, which means that the equivalent impedance of these two elements is

$$Z_{LC}=\frac{1-\omega^2LC}{j\omega L}$$

So for the three cases I got the following results

$$\omega=0\ \ \ \ Z_{LC}\ =\infty$$

$$\omega=\infty\ \ \ \ Z_{LC}\ =\infty$$

$$\omega=\frac{1}{\sqrt{LC}}\ \ \ \ Z_{LC}\ =0$$

If I were to plot these values I am not sure how the graph would look like.

In case the inductor and capcitor are in parallel,this is how the graph would look like. Link to the question and graph

• Your equation is not impedance but admittance.
– Bart
Feb 11 '20 at 0:56
• be1995, @Bart is right. That's not impedance. You can easily see this by re-stating your equation as $\frac1{s\,L}+s\,C$ and noting that this is the sum of the inverse of each separate impedance -- namely, the sum of the admittances to get the total admittance in the parallel case. This is why "X J" below tells you that the denominator is wrong if you are talking about series impedance. It's hard to get the right plot if you start with the wrong expression.
– jonk
Feb 11 '20 at 2:15