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Equations and Polynomials



Key Terms
Polynomial Equations with Complex Solutions
The Fundamental Theorem of Algebra
Factoring Polynomials over C\mathbb{C}
Vieta's Formulas
Polynomials with Real Coefficients
Quadratic Equations with Complex Coefficients
Solving zn=wz^n = w
Solving zn=zˉz^n = \bar{z}
Equations Involving Conjugates
Applications
Summary: Equation Types and Their Solution Methods
Complex Equations FAQ



Where Every Polynomial Finds Its Roots

The real numbers leave many polynomial equations unsolved — x2+1=0x^2 + 1 = 0 has no real answer, and neither does x4+4=0x^4 + 4 = 0. Complex numbers fill this gap completely. The Fundamental Theorem of Algebra guarantees that every polynomial equation has solutions in C\mathbb{C}, and the number of solutions matches the polynomial's degree exactly. This completeness makes the complex field the natural setting for polynomial theory.

Key Terms

Roots of Unity— solutions to zn=1z^n = 1
Complex Conjugate— roots of real-coefficient polynomials come in conjugate pairs
Exponential Form— used to express nnth roots
Modulus— all roots of unity have modulus 11

See All Complex Numbers Definitions →


Polynomial Equations with Complex Solutions

Real numbers solve many polynomial equations but fail systematically on others. The equation x2−4=0x^2 - 4 = 0 yields x=±2x = \pm 2, both real. The equation x2+1=0x^2 + 1 = 0 yields nothing — no real number squared equals −1-1. This gap plagued mathematics for centuries.

Complex numbers close the gap entirely. The equation x2+1=0x^2 + 1 = 0 has two solutions in C\mathbb{C}: x=ix = i and x=−ix = -i. Each satisfies i2=−1i^2 = -1 and (−i)2=−1(-i)^2 = -1. The imaginary unit exists precisely to solve this equation.

Higher-degree equations with no real solutions also yield to complex numbers. Consider x4+4=0x^4 + 4 = 0, equivalent to x4=−4x^4 = -4. No real fourth power is negative, yet four complex solutions exist. Using De Moivre's theorem, we write −4=4cis(180°)-4 = 4\text{cis}(180°) and extract fourth roots: zk=2cis(45°+90°k)z_k = \sqrt{2}\text{cis}(45° + 90°k) for k=0,1,2,3k = 0, 1, 2, 3. The solutions are 1+i1 + i, −1+i-1 + i, −1−i-1 - i, and 1−i1 - i.

Even equations mixing real and complex solutions find all their roots in C\mathbb{C}. The equation x3+x=0x^3 + x = 0 factors as x(x2+1)=0x(x^2 + 1) = 0, giving x=0x = 0 (real) and x=±ix = \pm i (complex). Three roots total, matching the cubic degree.

The pattern holds universally: every polynomial equation of degree nn has exactly nn roots when we work in C\mathbb{C} and count roots according to their multiplicity.

The Fundamental Theorem of Algebra

The cornerstone of polynomial theory makes a sweeping guarantee: every non-constant polynomial with complex coefficients has at least one complex root. This is the Fundamental Theorem of Algebra, first rigorously proved by Carl Friedrich Gauss in 1799.

The statement seems modest — just one root guaranteed. But its consequences are profound. If polynomial p(z)p(z) of degree n≥1n \geq 1 has a root z1z_1, then p(z)=(z−z1)q(z)p(z) = (z - z_1)q(z) where q(z)q(z) has degree n−1n - 1. If n−1≥1n - 1 \geq 1, the theorem applies again: q(z)q(z) has a root z2z_2, so q(z)=(z−z2)r(z)q(z) = (z - z_2)r(z). Continue until only a constant remains.

The conclusion: every polynomial of degree nn factors completely as:

Fundamental Theorem of Algebra
p(z)=an(z−z1)(z−z2)⋯(z−zn)p(z) = a_n(z - z_1)(z - z_2) \cdots (z - z_n)
Learn more about this formula: Fundamental Theorem of Algebra →


where ana_n is the leading coefficient and z1,z2,…,znz_1, z_2, \ldots, z_n are the nn roots (possibly with repetitions). A polynomial of degree nn has exactly nn roots in C\mathbb{C}, counted with multiplicity.

This completeness property means C\mathbb{C} is algebraically closed. Every polynomial equation solvable anywhere is solvable in C\mathbb{C}. No further extension of the number system is needed to find polynomial roots — complex numbers are the final destination.

The theorem's proof lies beyond elementary methods, requiring tools from analysis or topology. But its implications pervade all of algebra: we can always factor polynomials, always find roots, always reduce polynomial equations to products of linear factors over C\mathbb{C}.

Factoring Polynomials over C\mathbb{C}

The Fundamental Theorem guarantees that every polynomial splits into linear factors over C\mathbb{C}. The factorization reveals the polynomial's roots explicitly and connects algebraic structure to geometric information in the complex plane.

For a polynomial p(z)=anzn+an−1zn−1+⋯+a1z+a0p(z) = a_nz^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0, the complete factorization takes the form:

p(z)=an(z−z1)(z−z2)⋯(z−zn)p(z) = a_n(z - z_1)(z - z_2)\cdots(z - z_n)


The roots z1,z2,…,znz_1, z_2, \ldots, z_n may repeat. A root appearing mm times is said to have multiplicity mm. The polynomial p(z)=(z−2)3(z+i)(z−i)p(z) = (z - 2)^3(z + i)(z - i) has degree 55, with root 22 of multiplicity 33 and simple roots ±i\pm i.

The factorization of z3−1z^3 - 1 illustrates the interplay with roots of unity. The roots satisfy z3=1z^3 = 1, giving z0=1z_0 = 1, z1=cis(120°)=−12+32iz_1 = \text{cis}(120°) = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, and z2=cis(240°)=−12−32iz_2 = \text{cis}(240°) = -\frac{1}{2} - \frac{\sqrt{3}}{2}i. Writing ω=cis(120°)\omega = \text{cis}(120°):

z3−1=(z−1)(z−ω)(z−ω2)z^3 - 1 = (z - 1)(z - \omega)(z - \omega^2)


Multiplying the last two factors produces a real quadratic: (z−ω)(z−ω2)=z2+z+1(z - \omega)(z - \omega^2) = z^2 + z + 1. Thus z3−1=(z−1)(z2+z+1)z^3 - 1 = (z - 1)(z^2 + z + 1), the familiar factorization over R\mathbb{R}. Over C\mathbb{C}, we factor further into linear terms.

Every polynomial admits such complete factorization. Real polynomials may resist factoring over R\mathbb{R} into linear terms, but over C\mathbb{C}, every polynomial yields completely.

Vieta's Formulas

A remarkable correspondence links a polynomial's coefficients to symmetric combinations of its roots. These relationships, known as Vieta's formulas after François Viète, allow information to flow between the algebraic expression of a polynomial and the location of its roots.

For a monic quadratic z2+bz+c=0z^2 + bz + c = 0 with roots z1z_1 and z2z_2:

Vieta's Quadratic
z1+z2=−ba,z1⋅z2=caz_1 + z_2 = -\frac{b}{a}, \qquad z_1 \cdot z_2 = \frac{c}{a}
Learn more about this formula: Vieta's Quadratic →


The sum of roots equals the negative of the linear coefficient. The product of roots equals the constant term. Given roots 33 and −5-5, the polynomial is z2−(−2)z+(−15)=z2+2z−15z^2 - (-2)z + (-15) = z^2 + 2z - 15.

For a monic cubic z3+bz2+cz+d=0z^3 + bz^2 + cz + d = 0 with roots z1z_1, z2z_2, z3z_3:

z1+z2+z3=−bz_1 + z_2 + z_3 = -b

z1z2+z1z3+z2z3=cz_1z_2 + z_1z_3 + z_2z_3 = c

z1z2z3=−dz_1z_2z_3 = -d


The pattern continues for higher degrees. The kk-th symmetric sum of roots (all products of kk roots) equals (−1)k(-1)^k times the coefficient of zn−kz^{n-k} in a monic polynomial of degree nn. Formally:

Vieta's General
∑iri=−an−1an,∏iri=(−1)na0an\sum_{i} r_i = -\frac{a_{n-1}}{a_n}, \qquad \prod_{i} r_i = (-1)^n \frac{a_0}{a_n}
Learn more about this formula: Vieta's General →


These formulas prove useful in both directions. Given a polynomial, we can compute sums and products of roots without finding the roots explicitly. Given information about roots, we can reconstruct polynomial coefficients. Problems asking for the sum of squares of roots or similar expressions often yield to Vieta's formulas combined with algebraic identities like (z1+z2)2=z12+2z1z2+z22(z_1 + z_2)^2 = z_1^2 + 2z_1z_2 + z_2^2.
Degree Monic polynomial Sum of roots Product of roots Middle symmetric sums
Quadratic z² + bz + c z₁ + z₂ = −b z₁ · z₂ = c —
Cubic z³ + bz² + cz + d z₁ + z₂ + z₃ = −b z₁z₂z₃ = −d z₁z₂ + z₁z₃ + z₂z₃ = c
Quartic z⁴ + bz³ + cz² + dz + e Σz_i = −b z₁z₂z₃z₄ = e Σ z_i z_j = c (pairs); Σ z_i z_j z_k = −d (triples)
General n zⁿ + aₙ₋₁zⁿ⁻¹ + … + a₀ −aₙ₋₁ (−1)ⁿ · a₀ k-th symmetric sum = (−1)ᵏ · aₙ₋ₖ

Polynomials with Real Coefficients

Polynomials whose coefficients are all real numbers exhibit special structure: their non-real roots always come in conjugate pairs. If z0z_0 is a root, so is z0ˉ\bar{z_0}. Formally:

Conjugate Root Theorem
p(z0)=0  ⇒  p(z0‾)=0(real coefficients)p(z_0) = 0 \;\Rightarrow\; p(\overline{z_0}) = 0 \quad \text{(real coefficients)}
Learn more about this formula: Conjugate Root Theorem →


The proof exploits how conjugation interacts with polynomial evaluation. Let p(z)=anzn+an−1zn−1+⋯+a0p(z) = a_nz^n + a_{n-1}z^{n-1} + \cdots + a_0 with all aka_k real. For any complex zz:

p(z)‾=anzn+⋯+a0‾=anzn‾+⋯+a0=an(zˉ)n+⋯+a0=p(zˉ)\overline{p(z)} = \overline{a_nz^n + \cdots + a_0} = a_n\overline{z^n} + \cdots + a_0 = a_n(\bar{z})^n + \cdots + a_0 = p(\bar{z})


The key step uses ak‾=ak\overline{a_k} = a_k since each coefficient is real, and the property zk‾=(zˉ)k\overline{z^k} = (\bar{z})^k.

Now suppose p(z0)=0p(z_0) = 0. Then p(z0)‾=0ˉ=0\overline{p(z_0)} = \bar{0} = 0, so p(z0ˉ)=0p(\bar{z_0}) = 0. The conjugate is also a root.

Consequences abound. A real quadratic with complex roots has the form (z−z0)(z−z0ˉ)(z - z_0)(z - \bar{z_0}) with conjugate pair z0,z0ˉz_0, \bar{z_0}. This product expands to z2−2Re(z0)z+∣z0∣2z^2 - 2\text{Re}(z_0)z + |z_0|^2, a real quadratic as expected.

A real polynomial of odd degree must have at least one real root. Complex roots pair off, consuming an even number of the nn roots. With nn odd, at least one root remains unpaired — and an unpaired root of a real polynomial must be real (otherwise its conjugate would also be a root).

This explains why every real cubic crosses the real axis: it must have either three real roots or one real root plus a conjugate pair.
Property Statement Consequence
Conjugate pairing if z₀ is a non-real root, so is z̄₀ non-real roots always come in pairs
Real quadratic factor (z − z₀)(z − z̄₀) = z² − 2 Re(z₀) z + |z₀|² real polynomials factor over ℝ into linear and irreducible-quadratic factors only
Odd degree → real root if n is odd, at least one root is real every real cubic, quintic, etc. crosses the real axis
Non-real root count is even non-real roots, counted with multiplicity, total an even number (n − # real roots) is even

Quadratic Equations with Complex Coefficients

The quadratic formula extends unchanged into the complex domain. For the equation az2+bz+c=0az^2 + bz + c = 0 with complex coefficients aa, bb, cc (and a≠0a \neq 0):

Quadratic Formula in Complex
z=−b±b2−4ac2az = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Learn more about this formula: Quadratic Formula in Complex →


The formula remains valid, but a new challenge emerges: the discriminant Δ=b2−4ac\Delta = b^2 - 4ac may itself be complex, requiring us to compute the square root of a complex number.

Consider z2+2iz−1=0z^2 + 2iz - 1 = 0. Here a=1a = 1, b=2ib = 2i, c=−1c = -1. The discriminant is:

Δ=(2i)2−4(1)(−1)=−4+4=0\Delta = (2i)^2 - 4(1)(-1) = -4 + 4 = 0


With discriminant zero, one repeated root exists: z=−2i2=−iz = \frac{-2i}{2} = -i.

For a case with complex discriminant, take z2+z+(1−i)=0z^2 + z + (1 - i) = 0. The discriminant is:

Δ=1−4(1−i)=1−4+4i=−3+4i\Delta = 1 - 4(1 - i) = 1 - 4 + 4i = -3 + 4i


Finding −3+4i\sqrt{-3 + 4i} requires the technique from De Moivre's theorem. Convert −3+4i-3 + 4i to trigonometric form: modulus 9+16=5\sqrt{9 + 16} = 5, argument θ=arctan⁡(4/(−3))\theta = \arctan(4/(-3)) adjusted to the second quadrant. The square roots have modulus 5\sqrt{5} and arguments θ/2\theta/2 and θ/2+180°\theta/2 + 180°.

The process is lengthier than real quadratics but entirely systematic. Every quadratic with complex coefficients has two roots in C\mathbb{C} (counting multiplicity), accessible through the same formula that handles real coefficients.

Solving zn=wz^n = w

Equations of the form zn=wz^n = w, where ww is a given complex number, have exactly nn solutions. The technique applies De Moivre's theorem systematically.

The procedure:

1. Express ww in trigonometric form: w=Rcisϕw = R\text{cis}\phi
2. Apply the root formula: zk=R1/ncis(ϕ+360°kn)z_k = R^{1/n}\text{cis}\left(\frac{\phi + 360°k}{n}\right) for k=0,1,…,n−1k = 0, 1, \ldots, n-1
3. Convert to algebraic form if needed

Example: Solve z4=−16z^4 = -16.

First write −16-16 in trigonometric form. The modulus is 1616, and the argument is 180°180° (negative real axis). So −16=16cis(180°)-16 = 16\text{cis}(180°).

The fourth roots have modulus 161/4=216^{1/4} = 2 and arguments 180°+360°k4=45°+90°k\frac{180° + 360°k}{4} = 45° + 90°k:

z0=2cis(45°)=2⋅22(1+i)=2(1+i)z_0 = 2\text{cis}(45°) = 2 \cdot \frac{\sqrt{2}}{2}(1 + i) = \sqrt{2}(1 + i)

z1=2cis(135°)=2⋅22(−1+i)=2(−1+i)z_1 = 2\text{cis}(135°) = 2 \cdot \frac{\sqrt{2}}{2}(-1 + i) = \sqrt{2}(-1 + i)

z2=2cis(225°)=2⋅22(−1−i)=2(−1−i)z_2 = 2\text{cis}(225°) = 2 \cdot \frac{\sqrt{2}}{2}(-1 - i) = \sqrt{2}(-1 - i)

z3=2cis(315°)=2⋅22(1−i)=2(1−i)z_3 = 2\text{cis}(315°) = 2 \cdot \frac{\sqrt{2}}{2}(1 - i) = \sqrt{2}(1 - i)

Four roots forming a square on a circle of radius 22, rotated 45°45° from the axes. Each satisfies (zk)4=−16(z_k)^4 = -16.

Solving zn=zˉz^n = \bar{z}

The equation zn=zˉz^n = \bar{z} combines exponentiation with conjugation, producing a distinctive solution set. Unlike zn=wz^n = w which has exactly nn roots, this equation has n+2n + 2 solutions.

Begin by writing z=reiθz = re^{i\theta} in exponential form. Then zn=rneinθz^n = r^ne^{in\theta} and zˉ=re−iθ\bar{z} = re^{-i\theta}. The equation becomes:

rneinθ=re−iθr^ne^{in\theta} = re^{-i\theta}


Two conditions emerge from matching moduli and arguments.

Moduli: rn=rr^n = r, giving rn−r=r(rn−1−1)=0r^n - r = r(r^{n-1} - 1) = 0. Either r=0r = 0 or rn−1=1r^{n-1} = 1.

Arguments: nθ=−θ+360°kn\theta = -\theta + 360°k for integer kk, so (n+1)θ=360°k(n+1)\theta = 360°k, giving θ=360°kn+1\theta = \frac{360°k}{n+1}.

The solution r=0r = 0 yields z=0z = 0.

The solutions with rn−1=1r^{n-1} = 1 require r=1r = 1 (since r≥0r \geq 0). These solutions lie on the unit circle with arguments θ=360°kn+1\theta = \frac{360°k}{n+1} for k=0,1,…,nk = 0, 1, \ldots, n.

Counting: one solution at the origin, plus n+1n + 1 solutions on the unit circle, totaling n+2n + 2 solutions.

Example: z2=zˉz^2 = \bar{z} has 2+2=42 + 2 = 4 solutions.

From r2=rr^2 = r: r=0r = 0 or r=1r = 1.
From 3θ=360°k3\theta = 360°k: θ=0°,120°,240°\theta = 0°, 120°, 240°.

Solutions: z=0z = 0, z=1z = 1, z=cis(120°)=−12+32iz = \text{cis}(120°) = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, z=cis(240°)=−12−32iz = \text{cis}(240°) = -\frac{1}{2} - \frac{\sqrt{3}}{2}i.

Equations Involving Conjugates

Equations relating zz and zˉ\bar{z} often describe geometric loci in the complex plane. The conjugate identities translate these equations into conditions on real and imaginary parts.

Type 1: z+zˉ=kz + \bar{z} = k for real constant kk.

Since z+zˉ=2Re(z)z + \bar{z} = 2\text{Re}(z), the equation becomes 2a=k2a = k, or a=k/2a = k/2. The solution set is the vertical line where the real part equals k/2k/2. Every point on this line satisfies the equation; no point off the line does.

Type 2: z−zˉ=kiz - \bar{z} = ki for real constant kk.

Since z−zˉ=2biz - \bar{z} = 2bi, the equation becomes 2bi=ki2bi = ki, so b=k/2b = k/2. The solution set is the horizontal line where the imaginary part equals k/2k/2.

Type 3: z⋅zˉ=kz \cdot \bar{z} = k for positive real constant kk.

Since z⋅zˉ=∣z∣2z \cdot \bar{z} = |z|^2, the equation becomes ∣z∣2=k|z|^2 = k, or ∣z∣=k|z| = \sqrt{k}. The solution set is the circle centered at the origin with radius k\sqrt{k}.

Combining conditions produces intersections of these loci. The system:
z+zˉ=4andz⋅zˉ=5z + \bar{z} = 4 \quad \text{and} \quad z \cdot \bar{z} = 5


The first equation restricts to the vertical line Re(z)=2\text{Re}(z) = 2. The second restricts to the circle ∣z∣=5|z| = \sqrt{5}. Their intersection: points with real part 22 on the circle of radius 5\sqrt{5}. Writing z=2+biz = 2 + bi, we need 4+b2=54 + b^2 = 5, so b=±1b = \pm 1. Solutions: z=2+iz = 2 + i and z=2−iz = 2 - i.

This geometric approach often simplifies what algebraic manipulation would make tedious.
Equation Algebraic meaning (z = a + bi) Geometric locus
z + z̄ = k  (k real) 2 Re(z) = k → a = k⁄2 vertical line at x = k⁄2
z − z̄ = ki  (k real) 2bi = ki → b = k⁄2 horizontal line at y = k⁄2
z · z̄ = k  (k > 0) |z|² = k → |z| = √k circle centered at origin, radius √k
z̄ = z −b = b → b = 0 the real axis
z̄ = −z a = −a → a = 0 the imaginary axis

Applications

Complex roots of polynomials appear throughout science and engineering, carrying meaning far beyond pure mathematics. The location of roots in the complex plane determines the behavior of physical and computational systems.

Signal Processing: The transfer function of a linear filter is a ratio of polynomials in a complex variable. The filter's frequency response — which frequencies pass through and which are attenuated — depends directly on where the polynomial roots lie. Roots near the unit circle create sharp peaks or nulls in the response. Filter design amounts to placing roots strategically in the complex plane.

Control Theory: A dynamical system's stability hinges on the roots of its characteristic polynomial. If all roots have negative real parts (lying in the left half-plane), the system is stable — disturbances decay over time. Roots with positive real parts indicate instability — small perturbations grow without bound. Control engineers reshape root locations through feedback to achieve desired stability margins.

Geometry: Regular polygons emerge as roots of unity. The nn-th roots of unity form the vertices of a regular nn-gon inscribed in the unit circle. Solving zn=1z^n = 1 constructs the polygon automatically. More generally, the roots of zn=wz^n = w form a regular nn-gon of radius ∣w∣1/n|w|^{1/n}.

Number Theory: Cyclotomic polynomials — the minimal polynomials whose roots are primitive roots of unity — encode deep arithmetic structure. Their coefficients, degrees, and factorization properties connect to prime numbers, Galois theory, and algebraic integers.

The Fundamental Theorem of Algebra ensures that these applications never encounter polynomials without roots. In C\mathbb{C}, every polynomial equation has solutions, making complex numbers the natural language for polynomial-based analysis across all fields.

Summary: Equation Types and Their Solution Methods

Polynomial and equation problems in C\mathbb{C} fall into a handful of recognizable types, each with its own technique and a characteristic number of solutions. The table below collects them as a single decision reference: given an equation's form, locate the row, then read off the method and the expected solution count.
Equation type Form Method Number of solutions
General polynomial aₙzⁿ + … + a₀ = 0 factor over ℂ; or numeric methods exactly n (with multiplicity)
Quadratic (any coefs) az² + bz + c = 0 quadratic formula z = (−b ± √Δ)⁄(2a); for complex Δ use trig/exponential form to find √Δ 2 (1 repeated if Δ = 0)
Pure n-th power zⁿ = w trig/exponential form + De Moivre's root formula exactly n, evenly spaced on a circle (regular n-gon)
Power-equals-conjugate zⁿ = z̄ write z = reⁱᵒ; match moduli and arguments separately n + 2 (origin + n+1 on unit circle)
Conjugate-locus equations z + z̄ = k, z · z̄ = k, etc. translate to conditions on Re(z), Im(z), |z|; intersect loci for systems infinite (a line, circle, or axis) — or finite if intersected

Complex Equations FAQ

What is the Fundamental Theorem of Algebra?

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It states that every non-constant polynomial with complex coefficients has at least one complex root. Applied repeatedly it means a polynomial of degree n has exactly n roots when multiplicity is counted. This is what completes the number system: no polynomial equation ever demands a further extension beyond the complex numbers.Read more →

Why does every real polynomial of odd degree have a real root?

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Because non-real roots of a real polynomial arrive in conjugate pairs, so they always account for an even count. An odd total therefore leaves at least one root unpaired, and an unpaired root must be its own conjugate, which means it is real. The argument needs no calculus at all.Read more →

What are Vieta's formulas?

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They connect the coefficients of a polynomial to symmetric expressions in its roots: the sum of the roots, the sum of their pairwise products, and so on down to their full product. They let you answer questions about roots without finding them, which is often far quicker than solving the equation.Read more →

How do you solve z to the n equals w?

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Convert w to polar form, take the nth root of its modulus, then divide its argument by n and add a full turn divided by n repeatedly to generate all n solutions. Attempting this in algebraic form means expanding an nth power, which is impractical beyond very small exponents.Read more →

What is root multiplicity?

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The number of times a given root appears as a factor of the polynomial. A double root corresponds to a repeated factor and touches the axis without crossing in the real case. Multiplicity is what makes the count of roots match the degree exactly, rather than falling short when roots coincide.Read more →