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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
Formula Summary
Common Pitfalls
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 non-constant 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 plenty of polynomial equations.
x2−4=0x^2 - 4 = 0 has the solutions x=2x = 2 and x=−2x = -2. On the graph of y=x2−4y = x^2 - 4 they are the two places where the curve crosses the xx-axis.
x² + 1 = 0 has no real solution at all.
Squaring a real number never gives a negative result, so x2+1x^2 + 1 is always at least 11. Its graph sits entirely above the xx-axis and never crosses it.
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.
y = x^2 - 4 crosses the x-axis at -2 and 2; y = x^2 + 1 never reaches itxy−3−2−1123−6−4−2246810y = x² − 4y = x² + 1never reaches 0x² − 4 = 0 at x = ±2; x² + 1 stays at 1 or above
The definition
Working in C\mathbb{C}, a root (or solution) of a polynomial equation p(x)=0p(x) = 0 is any complex number zz with p(z)=0p(z) = 0 — real or not.
So x2+1=0x^2 + 1 = 0 has two roots, ii and −i-i, just as negative numbers were once admitted so that x+5=0x + 5 = 0 would have one.
A complex root is a point of the complex plane where the polynomial is zero.
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.
The roots of x^2 + 1: i and -i on the unit circleReIm1−1i90°−i270°i and −i are conjugatesx² + 1 = (x − i)(x + i): two roots, off the real axis
The roots of x2+1x^2 + 1: ii and −i-i, above and below the real axis.
Nothing was lost: every real root is still a root.
Real numbers are complex numbers with imaginary part 00, so x2−4=0x^2 - 4 = 0 still has exactly the roots 22 and −2-2. Some equations now have real and non-real roots side by side.
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.
What is lost is the picture: complex roots do not show up on the ordinary graph.
The graph of y=x2+1y = x^2 + 1 is drawn over real xx only, so a curve that misses the xx-axis means "no real roots", not "no roots". Non-real roots have to be found by algebra, or drawn in the complex plane instead.
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 next section states this as the Fundamental Theorem of Algebra; the page then covers factoring over C\mathbb{C}, Vieta's formulas, real coefficients, and solving equations such as zn=wz^n = w.

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 proved in essence by Carl Friedrich Gauss in 1799; a gap in his argument was closed only in 1920.

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.
The argument above, one root at a time: each use of the theorem splits off one linear factor.
DownloadPNG2×PDF
Derivep(z)=an(z−z1)(z−z2)⋯(z−zn)p(z) = a_n(z - z_1)(z - z_2) \cdots (z - z_n)
1
p(z)=(z−z1)q(z)p(z) = (z - z_1)q(z)
one rootA polynomial p(z)p(z) of degree n≥1n \geq 1 has a root z1z_1; q(z)q(z) has degree n−1n - 1
2
q(z)=(z−z2)r(z)q(z) = (z - z_2)r(z)
key stepIf n−1≥1n - 1 \geq 1, the theorem applies again: q(z)q(z) has a root z2z_2
3
p(z)=an(z−z1)(z−z2)⋯(z−zn)p(z) = a_n(z - z_1)(z - z_2) \cdots (z - z_n)
repeatContinue until only a constant remains: the leading coefficient ana_n
nn roots in C\mathbb{C}
A polynomial of degree nn has exactly nn roots in C\mathbb{C}, counted with multiplicity.
From one root to a full factorizationLearn Math Class
The theorem promises only one root, but applied again to each quotient it delivers all nn of them.
Fact: Gauss stated it with real factors only
Gauss's 1799 doctoral thesis, written at 22, states the theorem without complex numbers: every real polynomial splits into real factors of degree one or two. Pairing each non-real root with its conjugate gives exactly those quadratic factors.

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}.
One guaranteed root, used again and again, splits every polynomial of degree nn into exactly nn linear factors over C\mathbb{C}.

Factoring Polynomials over C\mathbb{C}

The Fundamental Theorem guarantees that every non-constant 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:

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 →


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°):

Factoring Polynomials over C\mathbb{C}
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 non-constant polynomial admits such complete factorization. Real polynomials may resist factoring over R\mathbb{R} into linear terms, but over C\mathbb{C}, every non-constant polynomial factors completely.
The roots of z^3 - 1: 1, omega and omega squared on the unit circleReIm−1i−i10°ω = −1/2 + (√3/2)i120°ω² = −1/2 − (√3/2)i240°ω and ω² are conjugates: together they give the real factor z² + z + 1z³ − 1 = (z − 1)(z − ω)(z − ω²) = (z − 1)(z² + z + 1)
Three linear factors, three roots

Each linear factor of z³ − 1 is one root in the plane: 1 on the real axis, and ω, ω² at 120° and 240° on the unit circle. Over the reals only the factor z − 1 splits off; the other two roots are a conjugate pair, and multiplying their factors gives (z − ω)(z − ω²) = z² + z + 1, which has real coefficients but no real root.

Over ℂ every factor is linear; over ℝ a conjugate pair stays joined as one quadratic factor.

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 quadratic az2+bz+c=0az^2 + bz + c = 0 (a≠0a \neq 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 →


For a monic quadratic (a=1a = 1), 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:

Vieta's Formulas
z1+z2+z3=−bz_1 + z_2 + z_3 = -b
Vieta's Formulas
z1z2+z1z3+z2z3=cz_1z_2 + z_1z_3 + z_2z_3 = c
Vieta's Formulas
z1z2z3=−dz_1z_2z_3 = -d

Pitfall: Dropping the alternating sign
Taking the sum of the roots of z3−z2+z−1z^3 - z^2 + z - 1 as b=−1b = -1 and their product as d=−1d = -1 leaves out the signs. Its roots are 11, ii, −i-i: the sum is 1=−b1 = -b and the product is 1⋅i⋅(−i)=1=−d1 \cdot i \cdot (-i) = 1 = -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:

Polynomials with Real Coefficients
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})

Pitfall: Pairing roots without real coefficients
Taking ii as a root of z2+2iz−1z^2 + 2iz - 1 because −i-i is one ignores the coefficient 2i2i, which is not real. Substituting gives i2+2i⋅i−1=−1−2−1=−4i^2 + 2i \cdot i - 1 = -1 - 2 - 1 = -4; the polynomial is (z+i)2(z + i)^2, and −i-i is its only root.

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.
The conjugation chain above, split at each equals sign, then the step from p(z0)=0p(z_0) = 0 to p(z0ˉ)=0p(\bar{z_0}) = 0.
DownloadPNG2×PDF
Provep(z0)=0  ⇒  p(z0‾)=0p(z_0) = 0 \;\Rightarrow\; p(\overline{z_0}) = 0 for real coefficients
1
p(z)‾=anzn+⋯+a0‾\overline{p(z)} = \overline{a_nz^n + \cdots + a_0}
write outLet 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
2
=anzn‾+⋯+a0= a_n\overline{z^n} + \cdots + a_0
key stepUses ak‾=ak\overline{a_k} = a_k, since each coefficient is real
3
=an(zˉ)n+⋯+a0= a_n(\bar{z})^n + \cdots + a_0
powersThe property zk‾=(zˉ)k\overline{z^k} = (\bar{z})^k
4
=p(zˉ)= p(\bar{z})
collectThe same polynomial, evaluated at zˉ\bar{z}
5
p(z0)‾=0ˉ=0\overline{p(z_0)} = \bar{0} = 0
rootNow suppose p(z0)=0p(z_0) = 0
p(z0ˉ)=0p(\bar{z_0}) = 0
The conjugate is also a root.
Proving the conjugate root theoremLearn Math Class
Real coefficients are what make it work: since ak‾=ak\overline{a_k} = a_k, conjugating the value of pp only conjugates its input.
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.

COROLLARY
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:

Quadratic Equations with Complex Coefficients
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:

Quadratic Equations with Complex Coefficients
Δ=1−4(1−i)=1−4+4i=−3+4i\Delta = 1 - 4(1 - i) = 1 - 4 + 4i = -3 + 4i

Pitfall: Splitting the square root over a sum
Writing −3+4i\sqrt{-3 + 4i} as −3+4i=2+(2+3)i\sqrt{-3} + \sqrt{4i} = \sqrt{2} + (\sqrt{2} + \sqrt{3})i gives a number whose square is not −3+4i-3 + 4i. The square roots are ±(1+2i)\pm(1 + 2i), since (1+2i)2=1+4i−4=−3+4i(1 + 2i)^2 = 1 + 4i - 4 = -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 nonzero 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)

Pitfall: Stopping at one root
Using only k=0k = 0 for z4=−16z^4 = -16 gives z=2(1+i)z = \sqrt{2}(1 + i), one solution out of four. The values k=1,2,3k = 1, 2, 3 supply the other three: 2(−1+i)\sqrt{2}(-1 + i), 2(−1−i)\sqrt{2}(-1 - i) and 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. Consecutive roots sit a quarter turn apart, because the angles π+2πk4\frac{\pi + 2\pi k}{4} grow by π2\frac{\pi}{2} with each step in kk, and the modulus 2⋅2=2\sqrt{2} \cdot \sqrt{2} = 2 is the same for all of them.

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 (w≠0w \neq 0), which has exactly nn roots, this equation has n+2n + 2 solutions for n≥2n \geq 2.

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:

Solving zn=zˉz^n = \bar{z}
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}.

Pitfall: Dividing out the modulus
Dividing r2=rr^2 = r by rr in z2=zˉz^2 = \bar{z} leaves only r=1r = 1 and three solutions on the unit circle. Factoring instead, r(r−1)=0r(r - 1) = 0 keeps r=0r = 0 as well, so z=0z = 0 is a fourth solution.

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 and n≥2n \geq 2). 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.
The four solutions of z^2 = conj zReIm−1i−i10°−1/2 + (√3/2)i120°−1/2 − (√3/2)i240°0r = 1: three solutions on the unit circler = 0: the solution z = 0, off the circlea root and its conjugate: mirror images across the real axisz² = z̄ has 2 + 2 = 4 solutions: z = 0, and three with |z| = 1at 0°, 120°, 240°, where 3θ is a multiple of 360°
The n + 2 solutions for n = 2

Matching moduli in z² = z̄ gives r² = r, so r = 0 or r = 1. The case r = 0 is the single point z = 0, drawn filled. The case r = 1 puts the rest on the unit circle, where matching arguments gives 3θ = 360°k: three solutions at 0°, 120° and 240°. That makes 1 + 3 = 4 = n + 2, two more than the n roots of zⁿ = w for w ≠ 0.

The circle solutions are the cube roots of unity: for r = 1, z̄ = 1/z, so z² = z̄ becomes z³ = 1.

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:
Equations Involving Conjugates
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.

Fact: Three roots share every border point
Run Newton's method on z3=1z^3 = 1 and colour each starting point by the root it reaches. Cayley settled the case of two roots in 1879, where a straight line splits the plane; with three roots the regions meet in a fractal, and every border point touches all three.

Numerical Root Finding: Software finds the roots of a polynomial pp by iteration rather than by formula. Newton's method starts from a guess z0z_0 and repeats zk+1=zk−p(zk)p′(zk)z_{k+1} = z_k - \frac{p(z_k)}{p'(z_k)}; close to a simple root, each step roughly doubles the number of correct digits. The rule works unchanged with complex numbers, so a complex starting point can lead to a complex root.

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 (w≠0w \neq 0) 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 non-constant 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 for w ≠ 0, evenly spaced on a circle (regular n-gon)
Power-equals-conjugate zⁿ = z̄ write z = reⁱᵒ; match moduli and arguments separately n + 2 for 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
The table as a checklist. The special forms are checked first because each has a shortcut; a polynomial that matches none of them falls through to the general case.
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every route at once
equations in ℂ

Which type of equation?

Locate the form; the method and the number of solutions come with it.
1
Quadratic: az² + bz + c = 0?any complex coefficients
yes
Quadratic formula
z = (−b ± √Δ)/(2a); for a complex Δ, find √Δ in trigonometric or exponential form→ Complex coefficients
2 solutions (1 repeated if Δ = 0)
no
2
A pure power: zⁿ = w?
yes
De Moivre's root formula
write w in trigonometric or exponential form, then take the n roots→ Solving zⁿ = w
exactly n solutions for w ≠ 0, the corners of a regular n-gon
no
3
A power equal to the conjugate: zⁿ = z̄?
yes
Match moduli and arguments
write z = re^(iθ) and compare the two sides part by part→ Solving zⁿ = z̄
n + 2 solutions for n ≥ 2: the origin and n + 1 on the unit circle
no
4
Built from z and z̄?z + z̄ = k, z · z̄ = k, …
yes
Translate to a locus
rewrite as conditions on Re(z), Im(z) or |z|; intersect loci for systems→ Equations with conjugates
a line, a circle or an axis, or finitely many points when intersected
no
Factor over ℂ
factor the polynomial, or use numeric methods→ Factoring over ℂ
exactly n solutions, counted with multiplicity
Which type of complex equation?·/complex-numbers/equations-polynomialsLearn Math Class
Every route ends with a count as well as a method. For a polynomial of degree nn the count is fixed in advance by the Fundamental Theorem of Algebra: exactly nn solutions. Among the power equations, zn=zˉz^n = \bar{z} is the one whose count is not the power: it has n+2n + 2 solutions for n≥2n \geq 2.

Formula Summary

This summary collects the formulas for polynomial equations over the complex numbers in four families, shown one at a time with the buttons above the cards. Each card gives a formula with a line on what it says and a link to the section that explains it.

The first family is the complete factorization promised by the fundamental theorem of algebra. The second gives Vieta's formulas, which link the roots to the coefficients for quadratics, cubics and any degree. The third shows how real coefficients pair each root with its conjugate, and the last gives the quadratic formula and the roots of a pure power equation.
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Factoring1 formula
What the fundamental theorem of algebra guarantees.
p(z)=an(z−z1)(z−z2)⋯(z−zn)p(z) = a_n(z - z_1)(z - z_2) \cdots (z - z_n)
Every polynomial of degree nn splits into nn linear factors over C\mathbb{C}, so it has exactly nn roots counted with multiplicity (the fundamental theorem of algebra).
Vieta's formulas5 formulas
Sums and products of the roots read from the coefficients.
z1+z2=−ba,z1⋅z2=ca\displaystyle z_1 + z_2 = -\frac{b}{a}, \qquad z_1 \cdot z_2 = \frac{c}{a}
The sum and the product of the two roots of az2+bz+c=0az^2 + bz + c = 0, as Vieta's formulas shows.
∑iri=−an−1an,∏iri=(−1)na0an\displaystyle \sum_{i} r_i = -\frac{a_{n-1}}{a_n}, \qquad \prod_{i} r_i = (-1)^n \frac{a_0}{a_n}
The same idea for any degree nn: the sum and the product of all the roots, from Vieta's formulas.
z1+z2+z3=−bz_1 + z_2 + z_3 = -b
For a monic cubic z3+bz2+cz+d=0z^3 + bz^2 + cz + d = 0, the sum of the three roots (Vieta's formulas).
z1z2+z1z3+z2z3=cz_1z_2 + z_1z_3 + z_2z_3 = c
The sum of the products of the roots two at a time, for the same cubic (Vieta's formulas).
z1z2z3=−dz_1z_2z_3 = -d
The product of all three roots; the signs alternate from −b-b to cc to −d-d, as Vieta's formulas warns.
Real coefficients2 formulas
Non-real roots come in conjugate pairs.
p(z0)=0  ⇒  p(z0‾)=0p(z_0) = 0 \;\Rightarrow\; p(\overline{z_0}) = 0
When every coefficient is real, the conjugate of a root is a root too, as polynomials with real coefficients proves.
(z−z0)(z−z0ˉ)=z2−2Re(z0)z+∣z0∣2(z - z_0)(z - \bar{z_0}) = z^2 - 2\text{Re}(z_0)z + |z_0|^2
A conjugate pair of roots multiplies out to a real quadratic, from polynomials with real coefficients.
Solving2 formulas
Formulas for two common types of equation.
z=−b±b2−4ac2a\displaystyle z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
The quadratic formula works unchanged with complex coefficients, though the discriminant may itself be complex (quadratic equations with complex coefficients).
zk=R1/ncis(ϕ+360°kn)\displaystyle z_k = R^{1/n}\text{cis}\left(\frac{\phi + 360°k}{n}\right)
The nn solutions of zn=wz^n = w with w=R cis ϕw = R\,\text{cis}\,\phi, one for each k=0,1,…,n−1k = 0, 1, \ldots, n-1, as the root procedure shows.
Polynomial equation formulas·All complex number formulasLearn Math Class
Vieta's formulas give a quick check on any set of roots: add them and multiply them, then compare with the coefficients. For a polynomial with real coefficients, also check that every non-real root appears together with its conjugate.

Common Pitfalls

Five mistakes account for many of the wrong answers when polynomial equations are solved over C\mathbb{C}. Each is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.
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1Dropping the alternating sign
Vieta's formulas are read off the coefficients without the sign (−1)k(-1)^k. In a monic polynomial the sum of the roots is the negative of the second coefficient, and for odd degree the product of the roots is the negative of the constant term.
Start
For z3−z2+z−1z^3 - z^2 + z - 1,
Wrong turn
Error here
With b=−1b = -1, c=1c = 1, d=−1d = -1, the sum of the roots is taken as b=−1b = -1 and the product as d=−1d = -1.
Right path
z1+z2+z3=−b=1z_1 + z_2 + z_3 = -b = 1, z1z2+z1z3+z2z3=c=1z_1z_2 + z_1z_3 + z_2z_3 = c = 1, z1z2z3=−d=1z_1z_2z_3 = -d = 1. The polynomial factors as (z−1)(z2+1)(z - 1)(z^2 + 1), with roots 11, ii, −i-i: their sum is 11, their pairwise products add to i−i+1=1i - i + 1 = 1, and their product is 1⋅i⋅(−i)=11 \cdot i \cdot (-i) = 1.
Why it happensThe signs alternate, so some of the formulas carry a minus sign and some do not, and the pattern is easy to lose.
How to checkBuild the polynomial back from its roots. (z−z1)(z−z2)(z−z3)(z - z_1)(z - z_2)(z - z_3) expands to z3−(z1+z2+z3)z2+⋯z^3 - (z_1 + z_2 + z_3)z^2 + \cdots, so a root sum of 11 must appear as −1-1 in front of z2z^2.
2Pairing roots without real coefficients
The conjugate root theorem is applied to a polynomial whose coefficients are not all real. The theorem rests on ak‾=ak\overline{a_k} = a_k for every coefficient, and that step fails as soon as one coefficient is non-real.
Wrong turn
Error here
z2+2iz−1=0z^2 + 2iz - 1 = 0 has the root −i-i, so its conjugate ii is taken as the second root.
Right path
Substituting z=iz = i gives i2+2i⋅i−1=−1−2−1=−4≠0i^2 + 2i \cdot i - 1 = -1 - 2 - 1 = -4 \neq 0. The discriminant is (2i)2−4(1)(−1)=0(2i)^2 - 4(1)(-1) = 0, and the polynomial is (z+i)2(z + i)^2: its only root is −i-i, with multiplicity 22.
Why it happensConjugate pairs are so familiar from real quadratics that they are expected everywhere.
How to checkLook at the coefficients before pairing. If any coefficient is non-real, substitute the conjugate and see whether it really gives 00.
3Splitting the square root over a sum
The square root of a complex discriminant a+bia + bi is taken term by term, as a+bi\sqrt{a} + \sqrt{bi}. Square roots do not split over sums, for real or for complex numbers.
Wrong turn
Error here
−3+4i=−3+4i=3i+2(1+i)=2+(2+3)i\sqrt{-3 + 4i} = \sqrt{-3} + \sqrt{4i} = \sqrt{3}i + \sqrt{2}(1 + i) = \sqrt{2} + (\sqrt{2} + \sqrt{3})i. Its square is about −7.90+8.90i-7.90 + 8.90i, not −3+4i-3 + 4i.
Right path
Look for x+yix + yi with (x+yi)2=−3+4i(x + yi)^2 = -3 + 4i, or use the trigonometric form with modulus 55. Both lead to ±(1+2i)\pm(1 + 2i), since (1+2i)2=1+4i+4i2=−3+4i(1 + 2i)^2 = 1 + 4i + 4i^2 = -3 + 4i. Then z=−1±(1+2i)2z = \frac{-1 \pm (1 + 2i)}{2} gives z=iz = i and z=−1−iz = -1 - i.
Why it happensThe same slip, x+y=x+y\sqrt{x + y} = \sqrt{x} + \sqrt{y}, is common with real numbers, and splitting the real and imaginary parts feels natural in algebraic form.
How to checkSquare the result. A square root of Δ\Delta must give Δ\Delta back exactly. The modulus is a quick test too: the root must have modulus ∣Δ∣=5\sqrt{|\Delta|} = \sqrt{5}, and ∣1+2i∣=5|1 + 2i| = \sqrt{5}.
4Stopping at one root
An equation zn=wz^n = w is solved with k=0k = 0 only, giving the principal root and nothing else, as if the equation had one solution.
Wrong turn
Error here
z4=−16z^4 = -16, −16=16cis(180°)-16 = 16\text{cis}(180°), so z=2cis(45°)=2(1+i)z = 2\text{cis}(45°) = \sqrt{2}(1 + i), and the work stops there.
Right path
zk=2cis(45°+90°k)z_k = 2\text{cis}(45° + 90°k) for k=0,1,2,3k = 0, 1, 2, 3 gives 2(1+i)\sqrt{2}(1 + i), 2(−1+i)\sqrt{2}(-1 + i), 2(−1−i)\sqrt{2}(-1 - i) and 2(1−i)\sqrt{2}(1 - i).
Why it happensWith real numbers a root sign such as 83=2\sqrt[3]{8} = 2 names one number, and the habit of writing a single answer carries over.
How to checkCount the solutions. A nonzero ww always gives exactly nn solutions of zn=wz^n = w, spaced 360°/n360°/n apart on one circle; four solutions here, 90°90° apart.
5Dividing out the modulus
In zn=zˉz^n = \bar{z} the modulus equation rn=rr^n = r is divided by rr, which silently assumes r≠0r \neq 0 and throws away the solution z=0z = 0.
Start
For z2=zˉz^2 = \bar{z},
Wrong turn
Error here
r2=rr^2 = r becomes r=1r = 1, and 3θ=360°k3\theta = 360°k gives θ=0°,120°,240°\theta = 0°, 120°, 240°. The answer is three solutions.
Right path
r2−r=r(r−1)=0r^2 - r = r(r - 1) = 0, so r=0r = 0 or r=1r = 1. The case r=0r = 0 gives z=0z = 0, and 02=0=0ˉ0^2 = 0 = \bar{0}. With the three points on the unit circle, there are 4=n+24 = n + 2 solutions.
Why it happensCancelling a common factor is a reflex from solving equations, and rr looks like a quantity that cannot be zero.
How to checkTest z=0z = 0 directly in the original equation whenever a factor of rr or zz is cancelled.
Common pitfalls with equations and polynomials·/complex-numbers/equations-polynomialsLearn Math Class
Each slip skips a condition that the result needs. Vieta's formulas alternate in sign, roots come in conjugate pairs only for real coefficients, a square root does not split over a sum, zn=wz^n = w has nn solutions, and cancelling rr can lose the solution z=0z = 0. Expanding back from the roots, squaring the result, or substituting the doubtful value catches each slip.

Complex Equations FAQ

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 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 →

If a polynomial's graph never touches the x-axis, does it have no roots?

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No, only no real roots. The graph of y=p(x)y = p(x) is drawn over real xx, so non-real roots never show on it: y=x2+1y = x^2 + 1 stays above the axis, yet x2+1=0x^2 + 1 = 0 has the roots ii and −i-i. Every polynomial of degree n≥1n \geq 1 has nn roots in C\mathbb{C}, counted with multiplicity; the non-real ones have to be found by algebra or drawn in the complex plane instead.Read more →

How do you find the square root of a complex number such as −3+4i-3 + 4i?

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Look for x+yix + yi with (x+yi)2=−3+4i(x + yi)^2 = -3 + 4i, which means x2−y2=−3x^2 - y^2 = -3 and 2xy=42xy = 4. The solutions are x=1,y=2x = 1, y = 2 and x=−1,y=−2x = -1, y = -2, so the square roots are ±(1+2i)\pm(1 + 2i); indeed (1+2i)2=1+4i−4=−3+4i(1 + 2i)^2 = 1 + 4i - 4 = -3 + 4i. Splitting the root over the sum, as −3+4i\sqrt{-3} + \sqrt{4i}, gives a number whose square is about −7.90+8.90i-7.90 + 8.90i. The trigonometric form, with modulus 55, leads to the same pair.Read more →

Who proved the Fundamental Theorem of Algebra?

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Carl Friedrich Gauss gave the first proof that holds up in essence, in his 1799 doctoral thesis, written at 22; a gap in his argument was closed only in 1920. He stated the theorem without complex numbers: every real polynomial splits into real factors of degree one or two. That is the same statement, since pairing each non-real root with its conjugate produces exactly those quadratic factors.Read more →

What happens when Newton's method is used on z3=1z^3 = 1?

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It reaches one of the three cube roots of unity from almost every starting point, but which one depends on the start in an intricate way. Colouring each starting point by the root it reaches splits the plane into three regions whose borders form a fractal, and every border point touches all three regions. With two roots the picture is simple: Cayley showed in 1879 that a straight line divides the plane between them.Read more →