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 $4(2-K)>(3+K)^2$ 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) further confirms that $-4 \pm i$ is not a closed loop pole.

[![Root Locus][1]][1]

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][2])


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 should result in the break out close to -1.4 as shown in the root locus above. Combining that with the other characteristics of the root locus as summarized on Dr. Cheever's page can be used to prove the solution does not exist (as demonstrated by the root locus itself).

  [1]: https://i.sstatic.net/SEwSw.png
  [2]: http://lpsa.swarthmore.edu/Root_Locus/Example2/Example2.html