Resonance of Parallel RLC circuit
Impedance is \$\dfrac{R + j\omega L}{R + j\omega L +\frac{1}{j\omega C}}\cdot\dfrac{1}{j\omega C}\$ = \$\dfrac{R + j\omega L}{j\omega RC + j^2\omega^2 LC +1}\$ = \$\dfrac{R + j\omega L}{j\omega RC - \omega^2 LC +1}\$
If you were to multiply both the numerator and the denominator by the denominator's complex conjugate you'd get a complex equation on the top and a non-complex (real) equation on the bottom. Resonance is when the imaginary part of the new numerator equals zero so, working only on the new numerator we get: -
New Numerator = \$(R+j\omega L)\cdot (-j\omega RC - \omega^2 LC +1)\$
The imaginary part boils down to \$j(\omega L - \omega^3 L^2 C - \omega C R^2)\$
Equating this to zero we get, \$\omega L - \omega^3 L^2 C - \omega C R^2 = 0\$
Therefore, \$\omega^2 L^2 C = L - CR^2\$ and \$\omega^2 = \dfrac{1}{LC} - \dfrac{R^2}{L^2}\$

EDIT section that describes other &quot;resonances&quot;
It's also interesting to note that there is another resonance at play. Firstly, the &quot;resonance&quot; above is defined by the impedance looking into the LCR network being purely resistive - this is slightly different to the &quot;natural&quot; resonant frequency (also purely resistive) when R is ignored.
In this situation \$\omega = \sqrt{\dfrac{1}{LC}}\$
However, the real peak of the response (as would be seen on a spectrum analyser and not purely resistive) is found by equating the denominator to zero and solving for s where s = j\$\omega\$.
\$s^2LC +sRC +1=0\$ or
\$s^2 + s\frac{R}{L} + \frac{1}{LC} = 0\$
Using the general solution to a quadratic, we get: -
\$s= \dfrac{-\frac{R}{L} +/- \sqrt{\frac{R^2}{L^2} - \frac{4}{LC}}}{2}\$
If we use a trick of reversing the signs inside the square root part and bring the square root of -1 (j) outside we get: -
\$s = \dfrac{-\frac{R}{L} +/- j\sqrt{\frac{4}{LC} - \frac{R^2}{L^2}}}{2}\$
Now, divide thru by two and we get: -
\$s = -\frac{R}{2L} +/- j\sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}\$
As we should know the &quot;jw&quot; part is what we see in a bode plot so the peak magnitude of the response (not when the impedance is purely resistive) is when: -
\$\omega = \sqrt{\dfrac{1}{LC} - \dfrac{R^2}{4L^2}}\$
If you drew a pole zero diagram, the imaginary parts of the poles would be at
+/-\$\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{L^2}}\$ i.e. slightly different to +/-\$\sqrt{\dfrac{1}{LC} - \dfrac{R^2}{4L^2}}\$