Given this frequency response:
I need to find the maximum sensitivity for closed system for automatic regulation $ M_\text{s} $. I now that $$ M_\text{s} = \max|S(j\omega)| = \max\left|\frac{1}{1+G(wj)C(wj)}\right|$$ however $G(j\omega)$ and $C(j\omega)$ are not given.
The teacher has given us the following hint: $$[1+G(j\omega)C(j\omega)]^2 = -0.402^2 + (-0.506)^2$$ Although I get some idea on how to proceed with solving the task I don't get from where $$[1+G(j\omega)C(j\omega)]^2 = -0.402^2 + (-0.506)^2$$ comes.
Consider the two points $s_\text{critical}=-1$ (can be interpreted as the vector from the origin to the critical point) and the point on the Nyquist curve $G(j\omega)C(j\omega)$ (can be interpreted as the vector from the origin to the Nyquist curve). I refer to them as points because they have real and imaginary components. We can also think of these points as vectors from the origin.
The vector $s_0$ connecting the critical point and the Nyquist curve can be calculated from
$$ s_\text{critical} + s_0 - G(j\omega)C(j\omega)=0$$
This equation can be interpreted as "go from the origin to the critical point" ($s_\text{critical}$) then "go from the critical point to the Nyquist curve" ($s_0$) then "go from the Nyquist curve to the origin" ($-G(j\omega)C(j\omega)$). If we add these vectors we will come back to the origin $0$. Note, that $|s_0|$ is just the distance between the critical point and the Nyquist curve.
We can rewrite the previous equation as: $$\implies s_0 = -s_\text{critical} +G(j\omega)C(j\omega)$$ $$\implies s_0 = -(-1) +G(j\omega)C(j\omega)$$ $$\implies s_0 = 1+G(j\omega)C(j\omega).$$
Hence, $s_0$ is the inverse of the sensitivity function. Instead of maximizing the magnitude of the sensitivity function $\max |S(j\omega)|$ we can as well minimize the inverse of the magnitude of the sensitivity function. This is equivalent to minimizing $|s_0|$. But $|s_0|$ is just the distance between the critical point $s_\text{critical}=-1$ and the Nyquist curve $G(j\omega)C(j\omega)$. So we need to find a point at the Nyquist curve $G(j\omega)C(j\omega)$ such that the distance to the critical point $s_\text{critical}=-1$ is minimized.
As we only have the graphical representation of the Nyquist curve we need to draw circles centered at the critical point $s_\text{critical}$ and increas the radius of the circle until the circle is a tangent (graphically) to the Nyquist curve. The point in which the circle touches the Nyquist curve is what we are looking for. Let us call this point $s_\text{min}=\sigma_\text{min}+j\omega_\text{min}$. Then $$M_\text{s}=\max|S(j\omega)|=\max\left|\dfrac{1}{1+G(j\omega)C(j\omega)} \right|$$ $$=\max \left|\dfrac{1}{s_0}\right|=\min|s_0|$$ $$=|\sigma_\text{min}+j\omega_\text{min}|=\sqrt{\sigma_\text{min}^2+\omega_\text{min}^2}.$$