Applications — solving polynomial equations with complex roots: Worked examples

Three things we lean on constantly

Before the matrix, let's nail down the symbols and one polar habit that show up in almost every cell, so nothing is used before it's defined.

We'll use the symbol $Δ$ (Greek capital "delta") for the discriminant everywhere below.

The scenario matrix

Now the whole battlefield. A polynomial-with-complex-roots problem always lands in one of these cells:

Cell	Case class	What makes it special	Example
A	Quadratic, $Δ > 0$	two real roots — no $i$ appears	Ex 1
B	Quadratic, $Δ = 0$	one repeated real root (degenerate)	Ex 2
C	Quadratic, $Δ < 0$	conjugate pair off the real line	Ex 3
D	Cubic (odd degree), one complex root given	forced lone real root	Ex 4
E	Quartic, two conjugate pairs	pair-and-divide twice	Ex 5
F	$z^{n} = c$ (here $n = 4$ ): roots evenly spaced on a circle	equal $\frac{360°}{n}$ spacing	Ex 6
G	$z^{n} = c$ with $c$ complex (not on axis)	starting angle $θ \neq = 0$	Ex 7
H	$z^{n} = c$ with $c$ on the negative real axis ( $θ = π$ )	intermediate case, roots tilt	Ex 8
K	Degenerate root of unity ( $n = 1$ , or repeated) / limiting behaviour	what happens as cases collapse	Ex 9
I	Word problem (real-world quantity)	translate → solve → interpret	Ex 10
J	Exam twist: complex coefficients	Conjugate Root Theorem fails	Ex 11

The three most confusable cells are A, B, C — they are the same equation shape $a z^{2} + b z + c = 0$ (always with $a \neq = 0$ ), but the sign of the ==discriminant $Δ = b^{2} - 4 a c$ == decides everything. The next figure shows all three at once.

Figure — Applications — solving polynomial equations with complex roots

Cell A — two real roots ( $Δ > 0$ )

Cell B — repeated real root ( $Δ = 0$ , degenerate)

Cell C — conjugate pair off the axis ( $Δ < 0$ )

Cell D — cubic with one given complex root (forced real root)

Worked example

$z^{3} - 3 z^{2} + 9 z + 13 = 0$ , given $z = 2 + 3 i$ is a root Forecast: odd degree with real coefficients — the mnemonic says complex roots pair up, so an odd degree must leave one lonely real root. Can you guess it before dividing?

Here $P (z) = z^{3} - 3 z^{2} + 9 z + 13$ (recall from the $z$ definition: $P (z)$ names the polynomial, and $P (α) = 0$ says $α$ is a root).

Pair it. Real coefficients ⇒ $\overline{2 + 3 i} = 2 - 3 i$ is also a root. Why this step? Conjugate Root Theorem: for real coefficients, $\overline{P (α)} = P (\overset{α}{ˉ})$ , so if $P (α) = 0$ then $P (\overset{α}{ˉ}) = 0$ too.
Build the real quadratic. Sum $= (2 + 3 i) + (2 - 3 i) = 4$ , product $= 2^{2} + 3^{2} = 13$ : $(z - (2 + 3 i)) (z - (2 - 3 i)) = z^{2} - 4 z + 13.$ Why? A quadratic with roots summing to $S$ and multiplying to $P$ is $z^{2} - S z + P$ — and here it is real, so we can divide by it.
Divide. $z^{3} - 3 z^{2} + 9 z + 13 \div (z^{2} - 4 z + 13)$ . Leading term $z$ ; constant $13 \cdot (?) = 13 \Rightarrow ? = + 1$ : $z^{3} - 3 z^{2} + 9 z + 13 = (z^{2} - 4 z + 13) (z + 1) .$ Why? Degree 3 ÷ degree 2 = degree 1, so the last factor is $z + c$ ; matching the constant $13 c = 13$ gives $c = 1$ .
Last root: $z + 1 = 0 \Rightarrow z = - 1$ .

Verify: roots $2 + 3 i, 2 - 3 i, - 1$ . Vieta sum $= 4 + (- 1) = 3 = - (- 3) /1$ ✓; product $= 13 \cdot (- 1) = - 13 = (- 1)^{3} \cdot 13/1$ ✓. The lone real root $- 1$ appeared exactly as forecast.

Cell E — quartic, two conjugate pairs

Worked example

$z^{4} - 4 z^{3} + 14 z^{2} - 36 z + 45 = 0$ , given $z = 3 i$ is a root Forecast: degree 4 with two complex pairs means no real roots at all — the whole solution set floats off the axis. Guess the second pair's real part.

Pair the given root. $\overline{3 i} = - 3 i$ ⇒ factor $(z - 3 i) (z + 3 i) = z^{2} + 9$ . Why? Pure imaginary root $q i$ pairs with $- q i$ ; their product is $z^{2} + q^{2}$ — real, so it divides out.
Divide, showing the multiplication. We claim the other factor is $z^{2} + b z + d$ and pin down $b, d$ by multiplying out: $(z^{2} + 9) (z^{2} + b z + d) = z^{4} + b z^{3} + (d + 9) z^{2} + 9 b z + 9 d .$ Match this term-by-term with $z^{4} - 4 z^{3} + 14 z^{2} - 36 z + 45$ :
- $z^{3}$ : $b = - 4$ .
- constant: $9 d = 45 \Rightarrow d = 5$ .
- check $z^{2}$ : $d + 9 = 5 + 9 = 14$ ✓; check $z$ : $9 b = 9 (- 4) = - 36$ ✓.
So the division is exact and $(z^{2} + 9) (z^{2} - 4 z + 5) = z^{4} - 4 z^{3} + 14 z^{2} - 36 z + 45$ . Why this layout? Instead of hand-waving "the middle terms force it," we expanded the product and matched all five coefficients, so nothing is assumed.
Solve the leftover quadratic $z^{2} - 4 z + 5 = 0$ : $Δ = 16 - 20 = - 4$ , so $z = \frac{4 \pm 2 i}{2} = 2 \pm i$ . Why? $- 4 = 2 i$ — the second conjugate pair.
All roots: $3 i, - 3 i, 2 + i, 2 - i$ .

Verify: Vieta sum $= 0 + 0 + 2 + 2 = 4 = - (- 4) /1$ ✓; product $= (9) (5) = 45 = (- 1)^{4} \cdot 45/1$ ✓. Two clean conjugate pairs — Cell E.

Cell F — $z^{n} = c$ with $n = 4$ : roots evenly spaced on a circle

Worked example Solve

$z^{4} = 16$ using De Moivre's Theorem Forecast: four roots, evenly spaced $\frac{360°}{4} = 90°$ apart (that is $\frac{2 π}{4} = \frac{π}{2}$ radians) on a circle. Look at figure s02 and guess the radius. (Note: here the four roots land on the axes — the boundaries between quadrants — because the starting angle is $0$ .)

Write $16$ in polar form: modulus $r = 16$ , angle $θ = 0$ , so $16 = 16 (cos 0 + i sin 0)$ . Why this step? De Moivre needs the target in $r (cos θ + i sin θ)$ form to read off angle and size.
Radius of every root: $r^{1/4} = 1 6^{1/4} = 2$ . Why? If $z^{4}$ has modulus 16 then $z$ has modulus $416 = 2$ — the fourth-root of the size.
Angles: $\frac{0 + 2 π k}{4} = \frac{π k}{2}$ for $k = 0, 1, 2, 3$ : that is $0, \frac{π}{2}, π, \frac{3 π}{2}$ radians (i.e. $0°, 90°, 180°, 270°$ ). Why the $+ 2 π k$ ? Angle $0$ and $0 + 2 π$ (a full turn) point the same way, but dividing by 4 spreads them into 4 distinct directions, $90° = \frac{π}{2}$ apart.
Convert: $z_{0} = 2, z_{1} = 2 i, z_{2} = - 2, z_{3} = - 2 i$ — evenly spaced, one on each axis direction.

Reading figure s02: the violet circle has radius $2$ (labelled). The four coloured arrows point from the origin to the four roots; each is $90°$ from its neighbours. The horizontal navy line is the real axis, the vertical navy line the imaginary axis, so you can read off $z_{0} = 2$ (right), $z_{1} = 2 i$ (up), $z_{2} = - 2$ (left), $z_{3} = - 2 i$ (down). Because $θ = 0$ , the roots land on the axes — precisely the quadrant boundaries, not inside the quadrants.

Verify: each satisfies $z^{4} = 16$ : $2^{4} = 16$ , $(2 i)^{4} = 16 i^{4} = 16$ , $(- 2)^{4} = 16$ , $(- 2 i)^{4} = 16$ ✓. Sum $= 2 + 2 i - 2 - 2 i = 0$ ✓ (no $z^{3}$ term). Cell F: the full evenly-spaced circle.

Cell G — $z^{n} = c$ with $c$ complex (nonzero start angle)

Worked example Solve

$z^{3} = 8 i$ Forecast: the target $8 i$ points straight up ( $90° = \frac{π}{2}$ ), not along the real axis. So the three cube-roots are tilted. Guess whether any root is purely real.

Polar form of $8 i$ : modulus $= 8$ , angle $θ = \frac{π}{2}$ (straight up, i.e. $90°$ ), so $8 i = 8 (cos \frac{π}{2} + i sin \frac{π}{2})$ . Why? $8 i$ has no real part and positive imaginary part → it sits on the positive imaginary axis.
Radius of roots: $8^{1/3} = 2$ .
Angles: $\frac{π /2 + 2 π k}{3}$ for $k = 0, 1, 2$ : $k = 0 : \frac{π}{6} = 30°, k = 1 : \frac{π /2 + 2 π}{3} = \frac{5 π}{6} = 150°, k = 2 : \frac{π /2 + 4 π}{3} = \frac{3 π}{2} = 270°.$ Why? Same spread-by- $120°$ ( $= \frac{2 π}{3}$ ) as any cube root, but the starting angle is $30°$ (one-third of $90°$ ) because $θ \neq = 0$ .
Convert (radius 2): $z_{0} = 2 (cos 30° + i sin 30°) = 3 + i, z_{1} = 2 (cos 150° + i sin 150°) = - 3 + i, z_{2} = 2 (cos 270° + i sin 270°) = - 2 i .$

Verify: $z_{2} = - 2 i$ : $(- 2 i)^{3} = - 8 i^{3} = - 8 (- i) = 8 i$ ✓. Sum $= (3 + i) + (- 3 + i) + (- 2 i) = 0$ ✓ (no $z^{2}$ term). None is purely real — the tilt from $θ = 90°$ pushed them off the axis, as forecast.

Cell H — $z^{n} = c$ with $c$ on the negative real axis ( $θ = π$ )

Worked example Solve

$z^{3} = - 8$ Forecast: the target $- 8$ is a negative real number, so it points left along the real axis — angle $θ = π = 180°$ , the natural halfway case between "straight right" ( $θ = 0$ , Cell F) and "general complex" (Cell G). Guess: does $z = - 2$ work, and are there hidden complex roots?

Polar form of $- 8$ : modulus $= 8$ , angle $θ = π$ (pointing left, i.e. $180°$ ), so $- 8 = 8 (cos π + i sin π)$ . Why? A negative real number sits on the negative real axis, which is the direction $180° = π$ from the positive axis.
Radius of roots: $8^{1/3} = 2$ .
Angles: $\frac{π + 2 π k}{3}$ for $k = 0, 1, 2$ : $k = 0 : \frac{π}{3} = 60°, k = 1 : \frac{π + 2 π}{3} = π = 180°, k = 2 : \frac{π + 4 π}{3} = \frac{5 π}{3} = 300°.$ Why? Start angle $60°$ (one-third of $180°$ ), then step by $120° = \frac{2 π}{3}$ . The middle one lands exactly at $180°$ → the real root.
Convert (radius 2): $z_{0} = 2 (cos 60° + i sin 60°) = 1 + 3 i, z_{1} = 2 (cos 180° + i sin 180°) = - 2, z_{2} = 2 (cos 300° + i sin 300°) = 1 - 3 i .$

Verify: $z_{1} = - 2$ : $(- 2)^{3} = - 8$ ✓. $z_{0} = 1 + 3 i$ : $(1 + 3 i)^{3} = - 8$ ✓ (real part checks below). Sum $= (1 + 3 i) + (- 2) + (1 - 3 i) = 0$ ✓ (no $z^{2}$ term). One real root $- 2$ plus a conjugate pair — Cell H, the intermediate $θ = π$ case.

Cell K — degenerate / limiting behaviour of $z^{n} = c$

Worked example What are the solutions of

$z^{1} = 5$ and of $z^{3} = 0$ ? Forecast: the "spread on a circle" idea should still work — but what happens when $n = 1$ (nothing to spread) or when $c = 0$ (radius zero)?

$z^{1} = 5$ : here $n = 1$ . The De Moivre formula gives exactly one angle $\frac{θ + 2 π \cdot 0}{1} = θ$ and radius $5^{1/1} = 5$ . Why? With $n = 1$ there is only $k = 0$ ; " $n$ equally-spaced points" degenerates to a single point. So $z = 5$ .
$z^{3} = 0$ : modulus of $0$ is $0$ , so every root has radius $0^{1/3} = 0$ . Why? All three roots collapse to the same point, the origin. So $z = 0$ with multiplicity 3, i.e. $z^{3} = 0 \Rightarrow (z - 0)^{3} = 0$ .
Limiting picture: as $c \to 0$ in $z^{3} = c$ , the three points on the circle shrink toward the centre and merge — the degenerate case of roots of unity.

Verify: $z = 5$ : $5^{1} = 5$ ✓. $z = 0$ : $0^{3} = 0$ ✓ and multiplicity 3 matches degree 3. Degenerate cells covered.

Cell I — word problem (real-world quantity)

Worked example A series RLC circuit's natural frequencies solve

$L s^{2} + R s + \frac{1}{C} = 0$ . With $L = 1, R = 2, \frac{1}{C} = 5$ , find the (complex) frequencies $s$ and say whether the circuit oscillates. Note on notation: engineers call the unknown $s$ instead of $z$ , but it plays the exact same role (see the $z$ definition) — a possibly-complex root of a quadratic. Everything from Cell C applies unchanged; only the letter differs.

Forecast: in engineering, a complex root means the system oscillates while decaying; a real root means it just decays. Which will we get? Check the discriminant sign first.

Equation: $s^{2} + 2 s + 5 = 0$ , so $a = 1, b = 2, c = 5$ (and $a \neq = 0$ , so it is a genuine quadratic). Why? Same quadratic shape — we're reusing Cell C machinery in a physical dress, with $s$ standing in for $z$ .
$Δ = 2^{2} - 4 (1) (5) = 4 - 20 = - 16 < 0$ . Why? Negative $Δ$ → conjugate pair → oscillatory behaviour (the imaginary part is the ringing frequency).
$s = \frac{- 2 \pm - 16}{2} = \frac{- 2 \pm 4 i}{2} = - 1 \pm 2 i$ . Why? $- 16 = 4 i$ ; the real part $- 1$ is the decay rate, the $\pm 2 i$ is the oscillation frequency (rad/s).
Interpret: real part $- 1 < 0$ ⇒ amplitude decays; nonzero imaginary part $\pm 2$ ⇒ it rings at $2$ rad/s while decaying. A damped oscillator.

Verify: Vieta sum $= - 1 - 1 = - 2 = - b / a$ ✓ (units: $rad/s$ ); product $= (- 1)^{2} + 2^{2} = 5 = c / a$ ✓. The circuit oscillates — Cell I answered with interpretation.

Cell J — exam twist: complex coefficients (theorem fails!)

Worked example Solve

$z^{2} - (3 + i) z + (2 + 2 i) = 0$ . Does the conjugate of any root have to be a root too? Forecast: the coefficients contain $i$ — they are not real. Predict: will the roots come in conjugate pairs?

Note the coefficients $- (3 + i)$ and $2 + 2 i$ are complex, so the Conjugate Root Theorem does not apply. Why this matters? The proof needed $\overline{a_{k}} = a_{k}$ (real coefficients). Here it's false, so roots need not be conjugate twins.
Use the quadratic formula anyway (it works for complex coefficients, and $a = 1 \neq = 0$ ): $Δ = (3 + i)^{2} - 4 (2 + 2 i)$ . Compute $(3 + i)^{2} = 9 + 6 i + i^{2} = 8 + 6 i$ ; then $Δ = 8 + 6 i - 8 - 8 i = - 2 i$ . Why? We still need $Δ$ , but now $Δ$ itself is complex.
Find $- 2 i$ . Seek $x + y i$ with $(x + y i)^{2} = - 2 i$ : then $x^{2} - y^{2} = 0$ and $2 x y = - 2$ . So $x = \pm 1, y = \mp 1$ ; take $- 2 i = 1 - i$ . Why? $x^{2} = y^{2}$ gives $x = \pm y$ ; $2 x y = - 2 < 0$ means $x, y$ have opposite signs → $(1, - 1)$ .
Roots: $z = \frac{( 3 + i ) \pm ( 1 - i )}{2}$ . Plus: $\frac{4 + 0 i}{2} = 2$ . Minus: $\frac{2 + 2 i}{2} = 1 + i$ .

Verify: roots $2$ and $1 + i$ are not conjugates of each other — confirming the theorem fails without real coefficients. Vieta sum $= 2 + (1 + i) = 3 + i = - (- (3 + i))$ ✓; product $= 2 (1 + i) = 2 + 2 i$ ✓. Cell J: the twist exposed.

Recall Self-test: name the cell, then solve

Which cell is $z^{2} + 2 z + 2 = 0$ ? ::: Cell C ( $Δ = 4 - 8 = - 4 < 0$ ), roots $- 1 \pm i$ . Which cell is $z^{2} - 8 z + 16 = 0$ ? ::: Cell B ( $Δ = 0$ ), repeated root $z = 4$ . Which cell is $z^{3} = - 8$ ? ::: Cell H ( $c$ on negative real axis, $θ = π$ ), roots $- 2, 1 \pm 3 i$ . Which cell is $z^{5} = 1$ ? ::: Cell F-type / roots of unity — five points evenly spaced on the unit circle. Why can $z^{2} - (1 + i) z + i = 0$ have non-conjugate roots? ::: Complex coefficients — Conjugate Root Theorem needs real coefficients (Cell J). As $c \to 0$ in $z^{n} = c$ , what happens to the $n$ roots? ::: They shrink to the origin and merge (Cell K, degenerate). What if $a = 0$ in $a z^{2} + b z + c = 0$ ? ::: It is not a quadratic; solve the linear $b z + c = 0 \Rightarrow z = - c / b$ .

Connections

Yeh note Hinglish mein padho →
Complex Numbers — Conjugate and Modulus (definition of $i$ , conjugate pairs, modulus for De Moivre)
De Moivre's Theorem (Cells F, G, H, K)
Roots of Unity (limiting degenerate case)
Vieta's Formulas (every "Verify" uses sum/product)
Fundamental Theorem of Algebra (root counting with multiplicity)
Polynomial Division (Cells D, E)