(How to ask a Homework Question): Define poles by using proportional controller
Nice work and your conclusion is correct.

Further confirmation is the poles in the closed loop system are the roots of $$1+K G(s)=0$$

Which results in the polynomial with a numerator

$$s^2+(3+K)s+(2-K) $$

In order to have complex roots, using the quadratic formula $\sqrt{b^2-4ac}$ must be negative.

So we want to test if $4(2-K)&gt;(3+K)^2$ for any K

Since $4(2-K)=(3+K)^2$ has roots at approximately -.101, -9.898:

$$K^2+10K+1=0$$

The range of K where the poles can be complex is therefore from -.101 to -9.898.

Showing all poles for the closed loop system as a function of K (root locus) over the range of K=-10.5 to 0 as plotted below confirms that $-4 \pm i$ is not a closed loop pole.



We could mathematically prove this by using the relationship of "Break-Outs" on the root locus:

(See Dr. Cheever's excellent pages at http://lpsa.swarthmore.edu/Root_Locus/Example2/Example2.html)

As Dr. Cheever summarized, break-out or break-in points occur where N(s)D'(s)-N'(s)D(s)=0 where N(s) and D(s) are the numerator and denominator polynomial of the open loop transfer function, and N'(s) and D'(s) are the differentiation.

$N(s)= s-1$

$N'(s) = 1$

$D(s) = s^2+s3+2$

$D'(s) = 2s+3$

Solving this results in roots at approximately -1.4495 and 3.4495 indicating the break out and break in locations. The root locus will break out at -1.4495 when K=-0.101 and follow a trajectory as shown in the diagram above to the break-in at 3.4495 when K=-9.898. Therefore $-4 \pm i$ is not a closed loop pole.

Note: Root locus was plotted using Octave with the rlocus command as part of the control toolbox

pkg load control %if not already loaded
sys=tf([1 -1],[1 3 2]);
rlocus(sys,.1,-10.5,0);