Feedback loop stability: conjugate roots and a complex locus
An engineering student is checking the stability of a feedback loop. The characteristic equation, with real coefficients, is $p(z) = z^3 - 5z^2 + 11z - 15 = 0$, and the system is stable only if every root has negative real part. The student substitutes $z = 1 + 2i$, finds the arithmetic messy, and asks for help. A lecturer confirms that $1 + 2i$ really is a root and remarks that 'the other complex root comes free'. The student must also draw a Nyquist-style diagram on which the critical boundary is the set of points $z$ in the complex plane satisfying $|z - 3| = |z - (1 + 2i)|$, and explain what that boundary is. Finally the lecturer wants a one-line verdict on whether the loop is stable, justified from the roots rather than from any picture.
1. Verify by direct substitution that $z = 1 + 2i$ satisfies $p(z) = 0$, and explain why $z = 1 - 2i$ must also be a root without doing a second full substitution.
Read the worked answer
Powers. $(1+2i)^2 = 1 + 4i + 4i^2 = -3 + 4i$. Then
$$(1+2i)^3 = (1+2i)(-3+4i) = -3 + 4i - 6i + 8i^2 = -11 - 2i .$$
Substitute.
$$p(1+2i) = (-11 - 2i) - 5(-3 + 4i) + 11(1 + 2i) - 15$$ $$= -11 - 2i + 15 - 20i + 11 + 22i - 15 .$$
Real parts: $-11 + 15 + 11 - 15 = 0$. Imaginary parts: $-2 - 20 + 22 = 0$. So $p(1+2i) = 0 + 0i$.
Why the conjugate is free. All coefficients of $p$ are real. If $p(w) = 0$ then taking complex conjugates of $a_nw^n + \cdots + a_0 = 0$ gives $a_n\bar{w}^n + \cdots + a_0 = 0$, because each $a_k$ is unchanged by conjugation. Hence $\overline{1+2i} = 1 - 2i$ is also a root - no computation needed, and this is a structural fact about real-coefficient polynomials, not a coincidence of this example.
2. Factorise $p(z)$ completely and list all three roots. Give the modulus and argument (in radians) of each complex root.
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Build the quadratic factor. The conjugate pair gives
$$\big(z - (1+2i)\big)\big(z - (1-2i)\big) = z^2 - 2z + (1 + 4) = z^2 - 2z + 5 .$$
Find the remaining factor. A cubic with these two roots has $p(z) = (z^2 - 2z + 5)(z - c)$. Matching the constant term, $5 \times (-c) = -15$, so $c = 3$. The linear factor is $z - 3$; expanding confirms it:
$$(z^2 - 2z + 5)(z - 3) = z^3 - 3z^2 - 2z^2 + 6z + 5z - 15 = z^3 - 5z^2 + 11z - 15 .$$
Roots. $z = 3$, $z = 1 + 2i$, $z = 1 - 2i$ (sanity check: sum $= 3 + 1 + 1 = 5$, matching the $-5z^2$ coefficient).
Polar form. For $1 + 2i$: $|z| = \sqrt{1^2 + 2^2} = \sqrt{5} = 2.24$, $\arg z = \arctan 2 = 1.11$ rad. For $1 - 2i$: $|z| = \sqrt{5} = 2.24$, $\arg z = -1.11$ rad (or $5.18$ rad). The two complex roots are mirror images in the real axis, as conjugate roots always are, and the real root $3$ sits further from the origin at argument $0$.
3. Find the Cartesian equation of the locus $|z - 3| = |z - (1 + 2i)|$ and describe it geometrically. Then give the stability verdict with justification from the roots.
Read the worked answer
Put $z = x + iy$.
$$|z - 3|^2 = (x - 3)^2 + y^2, \qquad |z - (1+2i)|^2 = (x - 1)^2 + (y - 2)^2 .$$
Setting them equal and expanding:
$$x^2 - 6x + 9 + y^2 = x^2 - 2x + 1 + y^2 - 4y + 4$$ $$-6x + 9 = -2x + 5 - 4y \ \Rightarrow\ 4y = 4x - 4 \ \Rightarrow\ y = x - 1 .$$
Geometric description. The locus of points equidistant from two fixed points is their perpendicular bisector. The fixed points are $(3, 0)$ and $(1, 2)$; their midpoint is $(2, 1)$ and the segment joining them has gradient $-1$, so the bisector through $(2,1)$ has gradient $1$ - consistent with $y = x - 1$, a straight line.
Stability verdict. All three roots ($3$, $1 \pm 2i$) have real part $+1$ or $+3$, so none lies in the left half-plane; in fact every root has positive real part, and the cubic has no root with negative real part. The loop is unstable. The locus is a boundary in the $z$-plane for the Nyquist construction, but the verdict follows from the roots themselves: a locus or plot is only a picture of the algebra.