Training Complex Numbers Poles, Zeros & the Complex Plane
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Poles, Zeros & the Complex Plane

30 min Complex Numbers
Poles and zeros are the critical features of complex-valued functions and transfer functions. A zero of f(z) is a point where f(z) = 0; a pole is a point where f(z) → ∞. For a rational function P(z)/Q(z), zeros come from the roots of P and poles from the roots of Q. In control theory and signal processing, the locations of poles and zeros in the complex plane determine system stability (poles in the left half-plane), frequency response, and filter characteristics. The pole-zero plot is one of the most important visualization tools in engineering.

Poles, Zeros & the Complex Plane

In advanced mathematics and engineering, poles and zeros describe where a rational function blows up or vanishes. They are the roots of the denominator and numerator, respectively, and they live in the complex $s$-plane.

Rational Function in $s$

A rational function (also called a transfer function in engineering) has the form:

$$H(s) = \frac{N(s)}{D(s)} = \frac{b_m s^m + b_{m-1}s^{m-1} + \cdots + b_0}{a_n s^n + a_{n-1}s^{n-1} + \cdots + a_0}$$

where $s$ is a complex variable.

Zeros

The zeros of $H(s)$ are the values of $s$ where the numerator $N(s) = 0$. At these points, $H(s) = 0$. Marked with ○ on the $s$-plane.

Poles

The poles of $H(s)$ are the values of $s$ where the denominator $D(s) = 0$. At these points, $H(s) \to \infty$. Marked with × on the $s$-plane.

Finding Poles and Zeros — Method
  1. Write $H(s)$ as a ratio of two polynomials $N(s)/D(s)$.
  2. Zeros: Set $N(s) = 0$ and solve. Use factoring, the quadratic formula, or synthetic division.
  3. Poles: Set $D(s) = 0$ and solve the same way.
  4. Check for cancellations — if a factor appears in both $N(s)$ and $D(s)$, it cancels (removable singularity).
  5. Plot zeros as ○ and poles as × on the complex plane.
Why Poles and Zeros Matter
  • Stability: A system is stable if all poles have negative real parts (left half of the $s$-plane).
  • Frequency response: Near a zero, the output is attenuated. Near a pole, it is amplified.
  • System behavior: Complex poles produce oscillations. Real poles produce exponential decay/growth.
  • Order: The number of poles determines the system order and the number of energy-storage elements.
The $s$-Plane

The complex $s$-plane has the real axis (horizontal, $\sigma$) and the imaginary axis (vertical, $j\omega$).

  • Left half-plane (LHP): $\text{Re}(s) < 0$ — stable poles (decaying signals)
  • Right half-plane (RHP): $\text{Re}(s) > 0$ — unstable poles (growing signals)
  • Imaginary axis: $\text{Re}(s) = 0$ — sustained oscillation (marginally stable)
Example 1 — Basic

Find the poles and zeros of $H(s) = \dfrac{s + 2}{s + 5}$.

  1. Zeros: $s + 2 = 0 \implies s = -2$
  2. Poles: $s + 5 = 0 \implies s = -5$
  3. Both are real and in the LHP → stable.
  4. The zero is at $-2$ and the pole is at $-5$.
Example 2 — Quadratic Denominator

Find the poles and zeros of $H(s) = \dfrac{2s}{s^2 + 4s + 13}$.

  1. Zero: $2s = 0 \implies s = 0$
  2. Poles: $s^2 + 4s + 13 = 0$
  3. Quadratic formula:
  4. $s = \dfrac{-4 \pm \sqrt{16 - 52}}{2} = \dfrac{-4 \pm \sqrt{-36}}{2} = \dfrac{-4 \pm 6i}{2}$
  5. $$s = -2 + 3i \quad \text{and} \quad s = -2 - 3i$$
  6. Complex conjugate poles in the LHP → stable with oscillation.
Example 3 — The Full-Page Problem

Find all poles and zeros, check for cancellations, determine stability, and sketch the pole-zero plot for:

$$H(s) = \frac{s^3 + 5s^2 + 8s + 4}{s^4 + 6s^3 + 13s^2 + 12s + 4}$$

  1. Step 1: Factor the numerator.
  2. Try $s = -1$: $(-1)^3 + 5(1) + 8(-1) + 4 = -1 + 5 - 8 + 4 = 0$ ✓
  3. Divide by $(s+1)$: $s^3+5s^2+8s+4 = (s+1)(s^2+4s+4) = (s+1)(s+2)^2$
  4. Step 2: Factor the denominator.
  5. Try $s = -1$: $1 - 6 + 13 - 12 + 4 = 0$ ✓
  6. Divide by $(s+1)$: $s^4+6s^3+13s^2+12s+4 = (s+1)(s^3+5s^2+8s+4)$
  7. But we just factored $s^3+5s^2+8s+4 = (s+1)(s+2)^2$, so:
  8. $$D(s) = (s+1)^2(s+2)^2$$
  9. Step 3: Write the factored form.
  10. $$H(s) = \frac{(s+1)(s+2)^2}{(s+1)^2(s+2)^2}$$
  11. Step 4: Cancel common factors.
  12. Cancel $(s+2)^2$ and one $(s+1)$:
  13. $$H(s) = \frac{1}{s+1}$$
  14. Step 5: Identify remaining poles and zeros.
  15. Zeros: None (numerator is constant)
  16. Poles: $s = -1$ (simple pole)
  17. Cancelled: $s = -2$ (double, removable), $s = -1$ (one copy cancelled)
  18. Step 6: Stability.
  19. The only pole is at $s = -1$, which has $\text{Re}(s) = -1 < 0$ → stable.
  20. Step 7: Interpretation.
  21. Despite starting as a 4th-order system, pole-zero cancellation reduces it to a simple first-order system with time constant $\tau = 1$ second.
Example 4 — Unstable System

Is $H(s) = \dfrac{s - 1}{s^2 - 4}$ stable?

  1. Zeros: $s = 1$
  2. Poles: $s^2 - 4 = (s-2)(s+2) = 0 \implies s = 2, \; s = -2$
  3. Pole at $s = 2$ is in the RHP ($\text{Re}(s) = 2 > 0$) → unstable.
Interactive Explorer: Pole-Zero Plotter & Stability Analyzer

✨ Use Sliders mode to drag and watch poles move in real-time, or Presets for specific transfer functions.

Practice Problems

1. Find the poles and zeros of $H(s) = \dfrac{s + 3}{s + 7}$.
2. Find the poles of $H(s) = \dfrac{1}{s^2 + 6s + 9}$. What kind of pole is this?
3. Find the poles and zeros of $H(s) = \dfrac{s^2 - 1}{s^2 + 2s + 5}$. Are the poles complex?
4. Is $H(s) = \dfrac{3}{s^2 + s - 6}$ stable? Factor the denominator and find both poles.
5. $H(s) = \dfrac{s + 4}{s^2 + 4s}$. Find a cancellation and simplify. What pole remains?
6. Find the poles of $H(s) = \dfrac{1}{s^2 + 4}$. Where do they lie on the $s$-plane?
7. A system has poles at $s = -1 \pm j2$. Find $|s|$ (distance from origin).
8. Find all poles and zeros of $H(s) = \dfrac{(s+1)(s-3)}{(s+2)(s+1)(s+5)}$. Simplify first.
9. If a pole is at $s = -3 + 4j$, what must the conjugate pole be? (Real coefficients.)
10. Factor $s^2 + 2s + 10$ using the quadratic formula. Are the roots in the LHP?
11. $H(s) = \dfrac{s^2 + 4s + 4}{s^3 + 4s^2 + 4s}$. Factor everything. Find poles and zeros after cancellation.
12. A system has zero at $s = 0$ and poles at $s = -5, -10$. Write $H(s)$ with a DC gain of $1$.
Show Answer Key

1. Zero: $s = -3$. Pole: $s = -7$.

2. $s^2+6s+9 = (s+3)^2$. Double pole at $s = -3$ (repeated root).

3. Zeros: $s = \pm 1$. Poles: $s = \frac{-2 \pm \sqrt{4-20}}{2} = -1 \pm 2i$. Yes, complex conjugate pair in LHP → stable.

4. $s^2+s-6 = (s+3)(s-2)$. Poles at $s = -3$ and $s = 2$. Pole in RHP → unstable.

5. $s^2+4s = s(s+4)$. Cancel $(s+4)$: $H(s) = 1/s$. Pole at $s = 0$ (integrator, marginally stable).

6. $s = \pm 2j$ (purely imaginary — on the imaginary axis). Marginally stable (sustained oscillation).

7. $|s| = \sqrt{1+4} = \sqrt{5} \approx 2.236$

8. Cancel $(s+1)$: $H(s) = \frac{s-3}{(s+2)(s+5)}$. Zero: $s = 3$. Poles: $s = -2, -5$.

9. $s = -3 - 4j$

10. $s = \frac{-2 \pm \sqrt{4-40}}{2} = -1 \pm 3i$. Real part $= -1 < 0$ → yes, LHP.

11. $N = (s+2)^2$, $D = s(s+2)^2$. Cancel $(s+2)^2$: $H(s) = 1/s$. Zero: none. Pole: $s = 0$. Cancelled: double at $s = -2$.

12. $H(s) = Ks/((s+5)(s+10))$. DC gain: $H(0) = 0$, so DC gain of $1$ isn't achievable. With steady-state gain at $\omega = 1$: typically $K = 50$ gives $H(s) = 50s/((s+5)(s+10))$.

Complex Roots & Applications Placement Test Practice — Complex Numbers