We have the closed loop transfer function:
$$T(s)=\frac{L(s)}{1+L(s)}$$
So as far as I understand we check along the $j\omega$-axis on the Bode plot whether $L(s)=-1$, cause that's when $T(s)$ has pole on the axis and that leads to instability.
My question is, why don't we need to check the entire right half plane whether a gain/phase modification leads to $L(s)=-1$ there? It doesn't make sense from my standpoint, cause we don't want any poles on the right half plane.
The test known as Nyquist Stability Criterion is that -1 is not encircled clockwise when the open loop frequency response of $L(s)$ is plotted on a complex plane. This applies to open loop systems that are stable (no poles on the right half plane) as the number of encirclements of -1 is equal to the number of poles in the right half plane for the closed loop system. For systems that have open loop poles in the right half plane or RHP, the number of encirclements will be reduced by the number of open loop poles on the RHP so we need to account for that in those cases.
This is a more robust test than the similar Bode plot test that the phase must not exceed -180 degrees (-1) when the gain crosses 0 dB and is derived from Cauchy’s Argument Principle which I have further detailed here:
Nyquist's Stability Criterion and Cauchy's Argument Principle
