Series-parallel RLC resistor voltage
Let's take this slowly:

Ok, we know the that the complex impedance of a capacitor is

\$Z_C = \frac1{j\omega C}\$

and the of an inductor

\$Z_L = {j\omega L}\$,

with \$\omega=2\pi f\$.

Inserting values:
\begin{align}
Z_C &amp;= \frac1{j2\pi f C}\\
Z_L &amp;= {j2\pi f L}\\
Z_{L||C} &amp;= \frac1{\frac1{Z_C}+\frac1{Z_L}}\\
&amp;= \frac1{j2\pi f C+\frac1{j2\pi f L}}&amp;\text{extending elegantly yields}\\
&amp;= \frac{j2\pi f C-\frac1{j2\pi f L}}{\left({j2\pi f C+\frac1{j2\pi f L}}\right)\left({j2\pi f C-\frac1{j2\pi f L}}\right)}\\
&amp;= \frac{j2\pi f C-\frac1{j2\pi f L}}{\left({j2\pi f C}\right)^2-\left({\frac1{j2\pi f L}}\right)^2}\\
&amp;= \frac{j2\pi f C+\frac j{2\pi f L}}{-4\pi^2f^2C^2 + \frac1{4\pi^2f^2L^2} } \\
&amp;= \frac{j\left(2\pi f C+\frac 1{2\pi f L}\right)}{-4\pi^2f^2C^2 + \frac1{4\pi^2f^2L^2} }\\
&amp;= j\frac{2\pi f C+\frac 1{2\pi f L}}{-4\pi^2f^2C^2 + \frac1{4\pi^2f^2L^2} }
\end{align}

As you can see, the complex value of that sub-circuit is purely imaginary!

Now, do the usual voltage divider calculation for the voltage across R1, and you will find the voltage drop as a function of frequency \$f\$.