# Electronic – Wien Bridge Oscillator: Why does equating the real part to 0 give the gain equation

impedanceresonancewien-bridge

I am learning about Wien bridge oscillators. Following Experiment No. 9: WIEN BRIDGE OSCILLATOR USING OPAMP, they use the following schematic: They then arrive at the following equation half-way down page 3: They then equate the imaginary part to 0 to find the resonant frequency, i.e: This makes sense, because at resonance the current is in phase with the voltage, hence the imaginary part goes to 0. I don't understand the next part where they:

To obtain the condition for gain at the frequency of oscillation, equate the imaginary part to zero.

which expressed as an equation gives you: Why can you just equate the imaginary part to zero (i.e. the real part goes to 0) to find the gain needed at resonance?

EDIT: After @TimWescott's great answer I created a re-written version of the equation which shows the "missing link", i.e. Im() + Re() = 0: Because they're starting with the Barkhausen Criterion; loop gain = 1. The loop gain is $$\left( 1 + \frac{R_3}{R_4} \right)\left(\frac{RC s}{(RC s)^2 + 3RCs + 1} \right)$$.
Everything else comes from setting that to one, and then doing some math. Basically, they set that to one: $$\left( 1 + \frac{R_3}{R_4} \right)\left(\frac{RC s}{(RC s)^2 + 3RCs + 1} \right)$$
It happens, conveniently, that in the circuit as shown, the real part depends on $$\R\$$, $$\C\$$, and the frequency of oscillation, $$\\omega\$$. So setting it to zero finds you $$\\omega\$$ as a function of $$\R\$$ and $$\C\$$.
Again, conveniently, once you know the frequency of oscillation (because you set the real part to zero), $$\R\$$, $$\C\$$, and $$\\omega\$$ drop out of the equation and you can calculate the gain.