How to draw Root locus for two gains In the book I'm reading, one of the exercises asks to draw the root locus for this closed-loop  system
$$
\frac{Y(s)}{U(s)} = \frac{1}{s^2 + K_2 s + K_1}
$$
I never encounter such a problem for two gains. I usually do it for a single gain where I can change the gain and observe how the poles of the closed-loop system move in the s-plane. Any suggestions for handling this?
 A: Because we want to calculate our transfer function in the complex plane modulus and phases are sufficient to trace it.
Let $$H : s \to \dfrac{1}{s^2+K_1s+K_2}$$
Setting $s=j \omega$
$$ H(jw)=\dfrac{1}{(K_2-\omega^2)+j\omega K_1}$$
which is by multiplying by conjugate quantity
$$ H(jw)=\dfrac{(K_2-\omega^2)-j\omega K_1}{(K_2-\omega^2)^2+(\omega K_1)^2}$$
Now for gain:
$$ |H(j\omega)|=\dfrac{1}{\sqrt{(K_2-\omega^2)^2+(\omega K_1)^2}}\\H(0)=1 \\ |H(j\omega)_{\omega \to \infty}|=0$$
Now for phase:
$$ Arg(H(j\omega))= \arctan\left(\dfrac{-\omega K_1}{K_2-\omega^2}\right) +\mathbb{1}_{\omega \geq \sqrt{K2}}(\omega)\pi$$
So here to gain many time it is fundamental to take time to study the phase of $H$.
Variations are as follow (phases must be continous)
$$ \text{On} \ [0,\sqrt{K2}] \ , \text{decreasing}\\
    \text{At} \ \sqrt{K_2} \  \text{we get a phase of} \ -\pi/2 \mod \pi\\\text{Then decreasing to } -\pi \mod \pi $$
So you will get this form :

The modulus at intersection with imaginary axis will be $\frac{1}{K_1\sqrt{K_2}}$.
So the diagramm consists in fixing $K_1$ and trace your diagramm in function and making it for different value of $K_1$ as it is made on the picture ( don't take in account the letter $z$)
