One of the comments to your question mentioned performing the analysis using complex mathematics.  Here is an outline:

The impedance of a resistor is R
$$$$
The complex impedance of a capacitor is 
$$\frac{-j}{\omega   \cdot C}$$

The complex impedance of an inductor is 
$$j \cdot \omega  \cdot L$$

The definition of j is:
$$j=\sqrt{-1}$$

Therefore:
$$j\cdot j=-1$$

and:
$$-j\cdot j=1$$

What you need, is to work out the impedance using your understanding of series and parallel connected components, substituting these impedance expressions.  If you call the cap on the left C1 and the cap on the right C2, you will end up with the j terms which are complex numbers.  It'll end up looking like:


$$\frac{\left( \frac{R \cdot \frac{-j}{\omega  \cdot C_2}}{R+\frac{-j}{\omega   \cdot C_2}} +( j \cdot \omega  \cdot L)\right) \cdot \frac{-j}{\omega   \cdot C_1}}{\frac{R \cdot \frac{-j}{\omega  \cdot C_2}}{R+\frac{-j}{\omega   \cdot C_2}} +( j \cdot \omega  \cdot L)+\frac{-j}{\omega  \cdot C_1}}$$

Expand the mathematic equation, multiply by the complex conjugate and take the square root.
The complex expression z
$$\frac{a+b\cdot j}{c - d\cdot j }$$

Has the conjugate z^*
$$\frac{a-b\cdot j}{c + d\cdot j }$$

You flip the polarity of the j expressions from positive to negative or vice versa

Multiplying the two
$$z \cdot z^*=\frac{a+b\cdot j}{c - d\cdot j }\cdot \frac{a-b\cdot j}{c + d\cdot j }=\frac{a^2+b^2}{c^2+d^2}$$
Yeilds a real expression (not complex), but it is the square of the real magnitude, so you square root the answer
$$\sqrt{\frac{a^2+b^2}{c^2+d^2}}$$