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Mathematics AA · HL · Unit 1: Number and Algebra

Unit 1Topic 1.14Key Concepts51 min read

Conjugate roots and De Moivre’s theorem

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Complex conjugates and conjugate pairs

For z=a+biz=a+bi, the complex conjugate is z‾=a−bi\overline z=a-bi. Conjugation reflects the point representing z across the real axis while preserving its modulus.

The product of a number and its conjugate is real: zz‾=a2+b2=∣z∣2z\overline z=a^2+b^2=|z|^2. Their sum is 2a2a, and their difference is 2bi2bi.

In polar form, if z=rcis⁡θz=r\operatorname{cis}\theta, then z‾=rcis⁡(−θ)\overline z=r\operatorname{cis}(-\theta). The modulus stays r and the argument changes sign.

These algebraic and geometric facts explain why conjugate pairs combine to produce polynomial factors with real coefficients.

Key terms

Complex conjugate
The number a − bi obtained by reversing the sign of the imaginary part of a + bi.
Conjugate pair
The two numbers a + bi and a − bi.
Reflection
The geometric transformation across the real axis induced by conjugation.

Worked examples

Cartesian conjugate

For z=−3+4iz=-3+4i, find z‾\overline z and zz‾z\overline z.

  1. 1Reverse the imaginary sign to obtain the conjugate.
  2. 2Use zz‾=(−3)2+42z\overline z=(-3)^2+4^2.

Answer: z‾=−3−4i\overline z=-3-4i and zz‾=25z\overline z=25.

Polar conjugate

Write the conjugate of 5cis⁡2π75\operatorname{cis}\frac{2\pi}{7} in polar form.

  1. 1Keep the modulus 5.
  2. 2Negate the argument.

Answer: 5cis⁡(−2π7)5\operatorname{cis}(-\frac{2\pi}{7}).

Recover a conjugate pair

A conjugate pair has sum 6 and product 13. Find the pair.

  1. 1Write the numbers as a±bia\pm bi.
  2. 2Their sum gives 2a=62a=6, so a=3a=3.
  3. 3Their product gives a2+b2=13a^2+b^2=13, so b2=4b^2=4.

Answer: The pair is 3+2i3+2i and (3-2i).

Rules and formulas

Complex conjugate

a+bi‾=a−bi\overline{a+bi}=a-bi

Conjugate product

zz‾=∣z∣2=a2+b2z\overline z=|z|^2=a^2+b^2

Conjugate in polar form

rcis⁡θ‾=rcis⁡(−θ)\overline{r\operatorname{cis}\theta}=r\operatorname{cis}(-\theta)

The conjugate-root theorem

If a polynomial (p(x)p(x)) has real coefficients and a+bia+bi is a non-real root, then (a-bi) is also a root. This is the conjugate-root theorem.

The reason is that conjugation commutes with addition and multiplication. For real coefficients, p(z‾)=p(z)‾p(\overline z)=\overline{p(z)}. If p(z)=0p(z)=0, then p(z‾)=0‾=0p(\overline z)=\overline0=0.

Non-real roots therefore occur in pairs when the coefficients are real. A real polynomial of odd degree must still have at least one real root because the remaining non-real roots pair off.

The theorem does not apply automatically to polynomials with non-real coefficients.

Key terms

Conjugate-root theorem
Non-real roots of a real-coefficient polynomial occur in conjugate pairs.
Real-coefficient polynomial
A polynomial all of whose coefficients are real numbers.
Root
A value zz satisfying p(z)=0p(z)=0.

Worked examples

Find a missing conjugate root

A polynomial with real coefficients has root (2-5i). State another root.

  1. 1Reverse the sign of the imaginary part.

Answer: 2+5i2+5i is also a root.

Use degree information

A real cubic has roots −1-1 and 3+i3+i. Find the third root.

  1. 1The non-real root requires its conjugate.
  2. 2A cubic has three roots counting multiplicity.

Answer: The third root is (3-i).

Check when the theorem does not apply

Does a polynomial with coefficient 2+i2+i have to contain conjugate root pairs?

  1. 1Inspect the coefficient condition.
  2. 2At least one coefficient is non-real.

Answer: No. Conjugate pairing is guaranteed only for real coefficients.

Rules and formulas

Conjugation of a real polynomial

p(z‾)=p(z)‾p(\overline z)=\overline{p(z)}

Conjugate-root implication

p(z)=0  ⟹  p(z‾)=0p(z)=0\implies p(\overline z)=0

Building real polynomial factors from conjugate roots

The conjugate roots a±bia\pm bi produce the real quadratic factor (x−a−bi)(x−a+bi)=(x−a)2+b2(x-a-bi)(x-a+bi)=(x-a)^2+b^2.

Expanded, this factor is x2−2ax+(a2+b2)x^2-2ax+(a^2+b^2). Its discriminant is −4b2<0-4b^2<0 when b≠0b\ne0, confirming that its roots are non-real.

To construct a real polynomial from known roots, include every required conjugate, write one factor per root, combine conjugate factors first, then expand only if requested.

Multiplicity is preserved: if a+bia+bi is a repeated root of multiplicity m in a real polynomial, (a-bi) has the same multiplicity.

Key terms

Linear factor
The factor x − α corresponding to root α.
Real quadratic factor
The factor (x − a)² + b2b^{2} generated by roots a ± bi.
Multiplicity
The number of times a root and its factor are repeated.

Worked examples

Quadratic from a conjugate pair

Find the monic quadratic with roots 3±2i3\pm2i.

  1. 1Write (x−(3+2i))(x−(3−2i))(x-(3+2i))(x-(3-2i)).
  2. 2Use (x−3)2+4(x-3)^2+4.
  3. 3Expand.

Answer: x2−6x+13x^2-6x+13.

Quartic from two non-real roots

Find the monic real polynomial whose roots include 2+i2+i and −1+2i-1+2i, with no additional roots.

  1. 1Include (2-i) and (-1-2i).
  2. 2Form ((x−2)2+1)((x+1)2+4)((x-2)^2+1)((x+1)^2+4).
  3. 3Expand the real quadratics.

Answer: x4−2x3+2x2−10x+25x^4-2x^3+2x^2-10x+25.

Repeated conjugate roots

A real polynomial has 1+3i1+3i as a double root. What factor must divide it?

  1. 1The conjugate (1-3i) is also a double root.
  2. 2Combine both repeated factors.

Answer: ((x−1)2+9)2\left((x-1)^2+9\right)^2 must divide the polynomial.

Rules and formulas

Conjugate factor pair

(x−(a+bi))(x−(a−bi))=(x−a)2+b2(x-(a+bi))(x-(a-bi))=(x-a)^2+b^2

Expanded real quadratic

x2−2ax+(a2+b2)x^2-2ax+(a^2+b^2)

Repeated conjugate factor

((x−a)2+b2)m\left((x-a)^2+b^2\right)^m

De Moivre's theorem

De Moivre's theorem states that for an integer n, raising a polar complex number to the nth power raises its modulus to n and multiplies its argument by n.

For z=rcis⁡θz=r\operatorname{cis}\theta, zn=rncis⁡(nθ)z^n=r^n\operatorname{cis}(n\theta). In full trigonometric form, (cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta).

For positive integers, the result follows by repeated polar multiplication. For negative integers it also holds when z≠0z\ne0, using reciprocals.

The theorem turns a long Cartesian expansion into two simple operations, but the original number must first be converted to polar form accurately.

Key terms

De Moivre's theorem
The integer-power rule for a complex number in polar form.
Integer exponent
A positive, zero, or negative whole-number power.
Argument multiplication
The rule that the direction of zⁿ is n times an argument of z, modulo 2π.

Worked examples

Unit-modulus power

Simplify (cis⁡π7)5(\operatorname{cis}\frac{\pi}{7})^5.

  1. 1Keep modulus 1.
  2. 2Multiply the argument by 5.

Answer: cis⁡5π7\operatorname{cis}\frac{5\pi}{7}.

Power with non-unit modulus

Calculate (2cis⁡2π3)4(2\operatorname{cis}\frac{2\pi}{3})^4.

  1. 1Raise the modulus: 24=162^4=16.
  2. 2Multiply the argument: 4⋅2π3=8π34\cdot\frac{2\pi}{3}=\frac{8\pi}{3}.
  3. 3Subtract 2π2\pi if a simpler coterminal angle is desired.

Answer: 16cis⁡2π3=−8+83i16\operatorname{cis}\frac{2\pi}{3}=-8+8\sqrt3 i.

Negative integer power

Simplify (3cis⁡π6)−2(3\operatorname{cis}\frac{\pi}{6})^{-2}.

  1. 1Raise the modulus: 3−2=193^{-2}=\frac19.
  2. 2Multiply the argument by −2-2: −π3-\frac{\pi}{3}.

Answer: 19cis⁡(−π3)\frac19\operatorname{cis}(-\frac{\pi}{3}).

Rules and formulas

De Moivre's theorem

(rcis⁡θ)n=rncis⁡(nθ)(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)
nn
integer exponent

Unit-modulus form

(cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)

Exponential equivalent

(reiθ)n=rneinθ(re^{i\theta})^n=r^ne^{in\theta}

Finding powers of complex numbers

To calculate a large power, convert the base to polar form, apply De Moivre's theorem, normalize the angle if useful, and convert back to Cartesian form only if requested.

Exact standard angles make the final conversion exact. Keep radical moduli such as 2\sqrt2 symbolic before raising them to a power.

A reasonableness check is immediate: ∣zn∣=∣z∣n|z^n|=|z|^n. If the computed Cartesian result has a different modulus, revisit the arithmetic.

When the multiplied argument completes whole turns, the power may become real even if the original number is non-real.

Key terms

Power workflow
Convert, apply De Moivre, normalize, then convert only if required.
Coterminal normalization
Adding or subtracting 2π without changing a direction.

Worked examples

A high power of 1 + i

Calculate (1+i)8(1+i)^8.

  1. 1Write 1+i=2cis⁡π41+i=\sqrt2\operatorname{cis}\frac{\pi}{4}.
  2. 2Apply De Moivre: (2)8cis⁡(2π)(\sqrt2)^8\operatorname{cis}(2\pi).
  3. 3Use (2)8=16(\sqrt2)^8=16 and cis⁡(2π)=1\operatorname{cis}(2\pi)=1.

Answer: (1+i)8=16(1+i)^8=16.

A fifth power

Calculate (−3+i)5(-\sqrt3+i)^5 in Cartesian form.

  1. 1Write −3+i=2cis⁡5π6-\sqrt3+i=2\operatorname{cis}\frac{5\pi}{6}.
  2. 2Raise to obtain 32cis⁡25π632\operatorname{cis}\frac{25\pi}{6}.
  3. 3Normalize to 32cis⁡π632\operatorname{cis}\frac{\pi}{6} and convert.

Answer: 163+16i16\sqrt3+16i.

Power and modulus check

Find ∣(3−4i)6∣|(3-4i)^6|.

  1. 1First find ∣3−4i∣=5|3-4i|=5.
  2. 2Use ∣zn∣=∣z∣n|z^n|=|z|^n.

Answer: ∣(3−4i)6∣=56=15625|(3-4i)^6|=5^6=15625.

Rules and formulas

Power modulus

∣zn∣=∣z∣n|z^n|=|z|^n

Power argument

arg⁡(zn)=narg⁡z(mod2π)\arg(z^n)=n\arg z\pmod{2\pi}

Using De Moivre to derive multiple-angle identities

De Moivre's theorem can generate trigonometric identities by expanding (cos⁡θ+isin⁡θ)n(\cos\theta+i\sin\theta)^n algebraically and comparing real and imaginary parts with cos⁡(nθ)+isin⁡(nθ)\cos(n\theta)+i\sin(n\theta).

The real parts must be equal, and the imaginary parts must be equal. This produces exact formulas for cos⁡(nθ)\cos(n\theta) and sin⁡(nθ)\sin(n\theta).

For n=3n = 3, expansion gives the triple-angle identities. Equivalent forms can be obtained using sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta or cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta.

Show the comparison step explicitly; a final identity without the complex expansion may omit the central reasoning.

Key terms

Multiple-angle identity
A trigonometric identity connecting nθ with powers of sin θ and cos θ.
Compare components
Equating the real parts and imaginary parts of equal complex numbers.

Worked examples

Derive cos 3θ

Use De Moivre to express cos⁡3θ\cos3\theta in terms of cos⁡θ\cos\theta.

  1. 1Expand (cos⁡θ+isin⁡θ)3(\cos\theta+i\sin\theta)^3.
  2. 2The real part is cos⁡3θ−3cos⁡θsin⁡2θ\cos^3\theta-3\cos\theta\sin^2\theta.
  3. 3Replace sin⁡2θ\sin^2\theta by 1−cos⁡2θ1-\cos^2\theta.

Answer: cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta.

Derive sin 3θ

Use De Moivre to express sin⁡3θ\sin3\theta in terms of sin⁡θ\sin\theta.

  1. 1The imaginary part of the cubic expansion is 3cos⁡2θsin⁡θ−sin⁡3θ3\cos^2\theta\sin\theta-\sin^3\theta.
  2. 2Replace cos⁡2θ\cos^2\theta by 1−sin⁡2θ1-\sin^2\theta.

Answer: sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta=3\sin\theta-4\sin^3\theta.

Evaluate an exact cosine

Use cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta with θ=π9\theta=\frac{\pi}{9} to relate cos⁡π3\cos\frac{\pi}{3} and cos⁡π9\cos\frac{\pi}{9}.

  1. 1Substitute 3θ=π33\theta=\frac{\pi}{3}.
  2. 2Use cos⁡π3=12\cos\frac{\pi}{3}=\frac12.

Answer: 4cos⁡3π9−3cos⁡π9=124\cos^3\frac{\pi}{9}-3\cos\frac{\pi}{9}=\frac12.

Rules and formulas

Triple-angle cosine

cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta

Triple-angle sine

sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta=3\sin\theta-4\sin^3\theta

Finding nth roots of a complex number

To solve zn=wz^n=w, write w=Rcis⁡θw=R\operatorname{cis}\theta with R>0R>0, and write z=ρcis⁡ϕz=\rho\operatorname{cis}\phi. De Moivre gives zn=ρncis⁡(nϕ)z^n=\rho^n\operatorname{cis}(n\phi).

Equating moduli gives ρ=R1n\rho=R^{\frac{1}{n}}. Equating directions requires nϕ=θ+2kπn\phi=\theta+2k\pi, so ϕ=θ+2kπn\phi=\frac{\theta+2k\pi}{n}.

Exactly n distinct roots are obtained by using k=0,1,…,n−1k=0,1,\ldots,n-1. Other integer values of k repeat these roots.

The roots all have the same modulus R1nR^{\frac{1}{n}} and successive arguments differ by 2πn\frac{2\pi}{n}.

Key terms

Nth root
A complex number z satisfying zⁿ = w.
Root index k
The integer selecting one of the n distinct angular solutions.
Angular spacing
The constant difference 2π/n between consecutive root arguments.

Worked examples

Derive cube-root angles

If z3=Rcis⁡θz^3=R\operatorname{cis}\theta, state the three root arguments. Assume R>0R>0.

  1. 1Use ϕk=θ+2kπ3\phi_k=\frac{\theta+2k\pi}{3}.
  2. 2Substitute k=0,1,2k=0,1,2.

Answer: θ3\frac{\theta}{3}, θ+2π3\frac{\theta+2\pi}{3}, and θ+4π3\frac{\theta+4\pi}{3}.

Why only n values of k?

Explain why k=nk = n repeats the k=0k = 0 root.

  1. 1Compare the arguments for k=nk = n and k=0k = 0.
  2. 2Their difference is 2nπn=2π\frac{2n\pi}{n}=2\pi.

Answer: They are coterminal and represent the same root.

Root modulus

Find the modulus of every fifth root of a complex number of modulus 32.

  1. 1Use ρ=R1n\rho=R^{\frac{1}{n}}.
  2. 2Calculate 321532^{\frac{1}{5}}.

Answer: Every root has modulus 2.

Rules and formulas

Nth-root formula

zk=R1ncis⁡(θ+2kπn),k=0,1,…,n−1z_k=R^{\frac{1}{n}}\operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right),\qquad k=0,1,\ldots,n-1

Root modulus

∣zk∣=R1n|z_k|=R^{\frac{1}{n}}

Root angular spacing

Δθ=2πn\Delta\theta=\frac{2\pi}{n}

Calculating and checking complex roots

Use the nth-root formula systematically: express the target in polar form, find the root modulus, create the n arguments, and convert roots to Cartesian form only if required.

A complete answer should list all distinct roots. Order them by increasing argument or by increasing k so omissions are easy to detect.

Check one root by raising it to the nth power. Check the whole set geometrically: equal modulus and equal angular spacing are required.

When the target is a negative real number, use argument π\pi under the principal convention before generating all roots.

Key terms

Complete root set
All n distinct values satisfying zⁿ = w.
Root verification
Raising a proposed root to the nth power and checking that it returns w.

Worked examples

Cube roots of 8

Find all solutions of z3=8z^3=8.

  1. 1Write 8=8cis⁡08=8\operatorname{cis}0.
  2. 2The root modulus is 2.
  3. 3Use arguments 0,2π3,4π30,\frac{2\pi}{3},\frac{4\pi}{3}.

Answer: z=2,−1+3i,−1−3iz=2,-1+\sqrt3 i,-1-\sqrt3 i.

Fourth roots of −16

Find all solutions of z4=−16z^4=-16 in polar form.

  1. 1Write −16=16cis⁡π-16=16\operatorname{cis}\pi.
  2. 2The root modulus is 2.
  3. 3Use π+2kπ4\frac{\pi+2k\pi}{4} for k=0,1,2,3k=0,1,2,3.

Answer: 2cis⁡π42\operatorname{cis}\frac{\pi}{4}, 2cis⁡3π42\operatorname{cis}\frac{3\pi}{4}, 2cis⁡5π42\operatorname{cis}\frac{5\pi}{4}, 2cis⁡7π42\operatorname{cis}\frac{7\pi}{4}.

Cube roots of 8i

Solve z3=8iz^3=8i in Cartesian form.

  1. 1Write 8i=8cis⁡π28i=8\operatorname{cis}\frac{\pi}{2}.
  2. 2The root modulus is 2.
  3. 3The root arguments are π6,5π6,3π2\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}.
  4. 4Convert each to Cartesian form.

Answer: z=3+i,−3+i,−2iz=\sqrt3+i,-\sqrt3+i,-2i.

Rules and formulas

Root angles

ϕk=θ+2kπn\phi_k=\frac{\theta+2k\pi}{n}

Root check

(zk)n=w(z_k)^n=w

Geometric arrangement of nth roots

The n roots of a non-zero complex number lie on the circle centered at the origin with radius R1nR^{\frac{1}{n}}. Their arguments are equally spaced by 2πn\frac{2\pi}{n}.

Joining consecutive roots forms a regular n-gon. Rotating one root by 2πn\frac{2\pi}{n} reaches the next root.

Changing the argument θ of the target rotates the entire regular polygon by θn\frac{\theta}{n}, while changing the modulus R changes only its radius to R1nR^{\frac{1}{n}}.

The sum of all n roots is zero for n>1n>1, because the symmetric position vectors cancel. This also follows from the coefficient of zn−1z^{n-1} in zn−wz^n-w.

Key terms

Regular n-gon
A polygon with n equal sides and n equal angles formed by equally spaced roots.
Rotational symmetry
Invariance of the root set under rotation by 2π/n.
Vector cancellation
The symmetric root position vectors add to zero.

Worked examples

Geometry of fourth roots

Describe the Argand arrangement of the fourth roots of −16-16.

  1. 1Every root has modulus 2.
  2. 2Successive arguments differ by 2π4=π2\frac{2\pi}{4}=\frac{\pi}{2}.

Answer: The roots are the vertices of a square centered at the origin on the circle ∣z∣=2|z|=2.

Geometry of fifth roots

A complex number has modulus 243. Describe the arrangement of its fifth roots.

  1. 1Each root has modulus 24315=3243^{\frac{1}{5}}=3.
  2. 2The angular spacing is 2π5\frac{2\pi}{5}.

Answer: They form a regular pentagon centered at the origin on ∣z∣=3|z|=3.

Sum of a root set

Find the sum of all solutions of z6=64z^6=64.

  1. 1The six roots form a regular hexagon centered at the origin.
  2. 2Opposite vectors cancel.

Answer: The sum is 0.

Rules and formulas

Root circle

∣z∣=R1n|z|=R^{\frac{1}{n}}

Consecutive-root rotation

zk+1=zkcis⁡2πnz_{k+1}=z_k\operatorname{cis}\frac{2\pi}{n}

Root sum

∑k=0n−1zk=0,n>1\sum\limits_{k=0}^{n-1}z_k=0,\qquad n>1

Roots of unity

The nth roots of unity solve zn=1z^n=1. Since 1=cis⁡(2mπ)1=\operatorname{cis}(2m\pi), the distinct roots are ωk=cis⁡2kπn\omega_k=\operatorname{cis}\frac{2k\pi}{n} for k=0,1,…,n−1k=0,1,\ldots,n-1.

All roots of unity have modulus 1 and form a regular n-gon on the unit circle. The root 11 always occurs at angle 0.

Six equally spaced sixth roots of unity forming a regular hexagon on an Argand diagram.The sixth roots lie on the unit circle with angular separation π3\frac{\pi}{3}, forming a regular hexagon.

Open graph full size (opens in a new tab)

The visual shows the sixth roots of unity: six points on one circle, separated by 2π6=π3\frac{2\pi}{6}=\frac{\pi}{3}, forming a regular hexagon.

A primitive nth root of unity generates every nth root by repeated powers. The root cis⁡2kπn\operatorname{cis}\frac{2k\pi}{n} is primitive exactly when k and n are coprime.

Key terms

Root of unity
A complex number satisfying zⁿ = 1.
Primitive root of unity
An nth root whose powers generate all nth roots of unity.
Coprime
Two integers whose greatest common divisor is 1.
Unit circle
The circle |z| = 1 in the Argand plane.

Worked examples

Fourth roots of unity

Find all fourth roots of unity.

  1. 1Use angles 2kπ4\frac{2k\pi}{4} for k=0,1,2,3k=0,1,2,3.
  2. 2Convert the four standard directions to Cartesian form.

Answer: 1,i,−1,−i1,i,-1,-i.

Cube roots of unity

Let ω=cis⁡2π3\omega=\operatorname{cis}\frac{2\pi}{3}. State the cube roots of unity and a useful sum.

  1. 1The roots are ω0=1\omega_0=1, ω1=ω\omega_1=\omega, and ω2=ω2\omega_2=\omega^2.
  2. 2Their symmetric vector sum is zero.

Answer: The roots are 1,ω,ω21,\omega,\omega^2, and 1+ω+ω2=01+\omega+\omega^2=0.

Identify primitive sixth roots

Which sixth roots cis⁡2kπ6\operatorname{cis}\frac{2k\pi}{6} are primitive?

  1. 1Require gcd⁡(k,6)=1\gcd(k,6)=1.
  2. 2Among k=0,1,2,3,4,5k=0,1,2,3,4,5, only 1 and 5 are coprime to 6.

Answer: The primitive roots are cis⁡π3\operatorname{cis}\frac{\pi}{3} and cis⁡5π3\operatorname{cis}\frac{5\pi}{3}.

Rules and formulas

Nth roots of unity

ωk=cis⁡2kπn,k=0,1,…,n−1\omega_k=\operatorname{cis}\frac{2k\pi}{n},\qquad k=0,1,\ldots,n-1

Root-of-unity equation

zn−1=0z^n-1=0

Primitive-root condition

gcd⁡(k,n)=1\gcd(k,n)=1

Sum of nth roots of unity

∑k=0n−1ωk=0,n>1\sum\limits_{k=0}^{n-1}\omega_k=0,\qquad n>1

Solving polynomial equations using complex roots

Polynomial equations can be solved by combining factorization, conjugate pairing, and nth-root methods. First identify the algebraic structure before choosing a technique.

For zn=wz^n=w, use the nth-root formula directly. For a real polynomial with one known non-real root, add its conjugate and divide by the resulting real quadratic factor.

After reducing the polynomial degree, solve the remaining real or complex factor by familiar methods. Count roots with multiplicity to confirm completeness.

The fundamental theorem of algebra guarantees exactly n complex roots counting multiplicity for a degree-n polynomial.

Key terms

Fundamental theorem of algebra
A degree-n polynomial has n complex roots counting multiplicity.
Polynomial reduction
Dividing by known factors to obtain a lower-degree equation.
Completeness check
Comparing the listed root count with the polynomial degree.

Worked examples

Solve a pure-power equation

Solve z6=64z^6=64.

  1. 1Write 64=64cis⁡064=64\operatorname{cis}0.
  2. 2The root modulus is 6416=264^{\frac{1}{6}}=2.
  3. 3Use arguments 2kπ6=kπ3\frac{2k\pi}{6}=\frac{k\pi}{3} for k=0,…,5k=0,\ldots,5.

Answer: zk=2cis⁡kπ3z_k=2\operatorname{cis}\frac{k\pi}{3}, k=0,1,…,5k=0,1,\ldots,5.

Use a known conjugate pair

A monic real quartic has roots 1+2i1+2i, 3, and −1-1. Find its fourth root and polynomial.

  1. 1The fourth root is (1-2i).
  2. 2The conjugate pair gives (x−1)2+4=x2−2x+5(x-1)^2+4=x^2-2x+5.
  3. 3Multiply by (x−3)(x+1)=x2−2x−3(x-3)(x+1)=x^2-2x-3.

Answer: The polynomial is x4−4x3+6x2−4x−15x^4-4x^3+6x^2-4x-15.

Factor using cube roots of unity

Factor z3−8z^3-8 over the complex numbers.

  1. 1The roots of z3=8z^3=8 are 22, −1+3i-1+\sqrt3 i, and −1−3i-1-\sqrt3 i.
  2. 2Write one linear factor for each root.

Answer: z3−8=(z−2)(z+1−3i)(z+1+3i)z^3-8=(z-2)(z+1-\sqrt3 i)(z+1+\sqrt3 i).

Rules and formulas

Factor theorem

p(α)=0  ⟺  (x−α) is a factor of p(x)p(\alpha)=0\iff(x-\alpha)\text{ is a factor of }p(x)

Root-count rule

deg⁡p=n  ⟹  n complex roots counting multiplicity\deg p=n\implies n\text{ complex roots counting multiplicity}

Mixed De Moivre and root problems

Advanced questions may combine a power, a root condition, and a polynomial requirement. Keep the modulus and argument work separate until the final form is requested.

When an equation can be rearranged to zn=wz^n=w, use the complete root family rather than taking only a principal real-style root.

Symmetry can simplify expressions involving roots of unity. For example, 1+ω+ω2=01+\omega+\omega^2=0 for the non-real cube roots of unity.

Use exact polar values first; convert to Cartesian form only for factorization, addition, or a requested final answer.

Key terms

Root family
The indexed set of all n distinct solutions of zⁿ = w.
Symmetry identity
An algebraic relation arising from the regular geometric arrangement of roots.

Worked examples

Solve a shifted pure-power equation

Solve (z−1)3=8i(z-1)^3=8i.

  1. 1Let w=z−1w=z-1.
  2. 2The cube roots of 8i8i are 3+i\sqrt3+i, −3+i-\sqrt3+i, and −2i-2i.
  3. 3Add 1 to each w value.

Answer: z=1+3+i,1−3+i,1−2iz=1+\sqrt3+i,1-\sqrt3+i,1-2i.

Use a cube-root identity

If ω≠1\omega\ne1 and ω3=1\omega^3=1, simplify 1+ω+ω21+\omega+\omega^2.

  1. 1Factor ω3−1=(ω−1)(ω2+ω+1)\omega^3-1=(\omega-1)(\omega^2+\omega+1).
  2. 2Since ω≠1\omega\ne1, divide by ω−1\omega-1.

Answer: 1+ω+ω2=01+\omega+\omega^2=0.

Power followed by a root condition

Find all z satisfying z4=16cis⁡2π3z^4=16\operatorname{cis}\frac{2\pi}{3}.

  1. 1The root modulus is 2.
  2. 2Use arguments 2π3+2kπ4=π6+kπ2\frac{\frac{2\pi}{3}+2k\pi}{4}=\frac{\pi}{6}+\frac{k\pi}{2}.
  3. 3Use k=0,1,2,3k=0,1,2,3.

Answer: zk=2cis⁡(π6+kπ2)z_k=2\operatorname{cis}(\frac{\pi}{6}+\frac{k\pi}{2}), k=0,1,2,3k=0,1,2,3.

Rules and formulas

Shifted power substitution

(z−a)n=wwith u=z−a(z-a)^n=w\quad\text{with }u=z-a

Root-family completeness

k=0,1,…,n−1k=0,1,\ldots,n-1

Calculator/GDC methods and verification

Use complex mode and radian angle mode. A GDC can evaluate powers, convert between Cartesian and polar form, and numerically solve many polynomial equations.

For nth roots, a calculator may display only one principal root. The mathematical formula with k=0,1,…,n−1k=0,1,\ldots,n-1 is still required to generate the complete set.

When a polynomial solver returns decimals, identify exact conjugate structure where possible. Values such as 1.732051.73205 may represent 3\sqrt3, but claim an exact value only when algebra supports it.

Use the GDC to verify each proposed root by evaluating the original expression. Small numerical residues such as 10−1210^{-12} are rounding effects, not additional roots.

For an Argand sketch, plot all roots and check equal radii and angular gaps. The visual check complements the numerical check.

Key terms

Principal root output
A calculator's single default root, which is not the full nth-root set.
Numerical residual
A small non-zero calculator result caused by floating-point rounding.
Exact recognition
Identifying a decimal as an exact radical or standard-angle value using algebraic evidence.

Worked examples

Check a high power

Use a GDC to verify (1+i)8=16(1+i)^8=16.

  1. 1Enter (1+i)8(1+i)^8 in complex mode.
  2. 2Compare with the exact De Moivre result.

Answer: The display confirms 16.

Generate roots systematically

Use a GDC to check the four roots of z4=−16z^4=-16.

  1. 1Generate all four angles with the formula, not a single fourth-root key.
  2. 2Enter each root to the fourth power.
  3. 3Confirm each output is approximately −16-16.

Answer: All four roots pass the check.

Check a polynomial factor

Verify numerically that 3+2i3+2i is a root of x2−6x+13x^2-6x+13.

  1. 1Substitute 3+2i3+2i into the polynomial in complex mode.
  2. 2Evaluate the result.

Answer: The result is 0, within calculator precision.

Rules and formulas

Numerical root check

∣p(zk)∣≈0|p(z_k)|\approx0

Geometric root check

∣zk∣=R1n,arg⁡zk+1−arg⁡zk=2πn|z_k|=R^{\frac{1}{n}},\qquad \arg z_{k+1}-\arg z_k=\frac{2\pi}{n}

Common mistakes and IB HL exam strategy

The most frequent errors are forgetting a conjugate root, applying the conjugate-root theorem without real coefficients, using nrnr instead of rnr^n, dividing only the principal argument by n, or listing fewer than n roots.

For powers, show the polar conversion and both De Moivre operations. For roots, show the target polar form, the common root modulus, the indexed angle formula, and the complete k range.

For polynomial construction, pair conjugates before expanding and count the degree. For an Argand answer, state the radius, angular separation, and regular polygon.

Normalize arguments only when the question requests a principal interval; coterminal root angles are mathematically valid but should be presented consistently.

Key terms

Method communication
Writing enough structure for the examiner to follow and award method marks.
Root-count check
Confirming that exactly n roots, counting multiplicity, have been found.

Worked examples

Diagnose an incomplete root answer

A student solves z4=16z^4=16 and gives only z=2z = 2. What is missing?

  1. 1A fourth-degree equation has four complex roots counting multiplicity.
  2. 2Use angles 2kπ4\frac{2k\pi}{4} for k=0,1,2,3k=0,1,2,3.

Answer: The full set is 2,2i,−2,−2i2,2i,-2,-2i.

Diagnose a De Moivre error

A student writes (2cis⁡θ)3=6cis⁡(3θ)(2\operatorname{cis}\theta)^3=6\operatorname{cis}(3\theta). Correct it.

  1. 1Cube the modulus instead of multiplying it by 3.

Answer: (2cis⁡θ)3=8cis⁡(3θ)(2\operatorname{cis}\theta)^3=8\operatorname{cis}(3\theta).

Diagnose an invalid conjugate claim

A student says every polynomial containing root ii also contains root (-i). Is this always true?

  1. 1The conjugate-root theorem needs real coefficients.
  2. 2A complex-coefficient polynomial can have i without −i.

Answer: No. Real coefficients guarantee that a nonreal root occurs with its conjugate, but this is not guaranteed for general complex coefficients. For example, x−ix-i has root ii but not −i-i. Some complex-coefficient polynomials, such as i(x2+1)i(x^2+1), still have both.

Rules and formulas

Power checklist

rcis⁡θ→nrncis⁡(nθ)r\operatorname{cis}\theta\xrightarrow{n}r^n\operatorname{cis}(n\theta)

Root checklist

Rcis⁡θ→nth rootsR1ncis⁡(θ+2kπn)R\operatorname{cis}\theta\xrightarrow{n\text{th roots}}R^{\frac{1}{n}}\operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)

IB HL-style practice with worked solutions

Attempt each problem before reading the steps. The questions progress from conjugate-root fundamentals to powers, complete root families, roots of unity and mixed polynomial reasoning.

Use exact values and radian arguments unless a question explicitly requests a decimal approximation.

Key terms

Exact root set
Every distinct solution expressed without premature decimal rounding.
Structured solution
A solution that states the theorem, formula, index range and verification clearly.

Worked examples

Practice 1: conjugate root

A real polynomial has root (4-3i). State the required partner root and quadratic factor.

  1. 1Use the conjugate-root theorem to obtain 4+3i4+3i.
  2. 2Form (x−4)2+9(x-4)^2+9.

Answer: The partner is 4+3i4+3i, and the factor is x2−8x+25x^2-8x+25.

Practice 2: construct a real quartic

Find the monic real polynomial with roots 2±i2\pm i and −1±2i-1\pm2i.

  1. 1Form (x−2)2+1=x2−4x+5(x-2)^2+1=x^2-4x+5.
  2. 2Form (x+1)2+4=x2+2x+5(x+1)^2+4=x^2+2x+5.
  3. 3Multiply the quadratics.

Answer: x4−2x3+2x2−10x+25x^4-2x^3+2x^2-10x+25.

Practice 3: De Moivre power

Calculate (2cis⁡2π3)4(2\operatorname{cis}\frac{2\pi}{3})^4 in Cartesian form.

  1. 1Raise the modulus to obtain 16.
  2. 2Multiply the angle to obtain 8π3\frac{8\pi}{3}.
  3. 3Normalize to 2π3\frac{2\pi}{3} and convert.

Answer: −8+83i-8+8\sqrt3 i.

Practice 4: high Cartesian power

Evaluate (1+i)8(1+i)^8.

  1. 1Write 1+i=2cis⁡π41+i=\sqrt2\operatorname{cis}\frac{\pi}{4}.
  2. 2Apply De Moivre to get 16cis⁡(2π)16\operatorname{cis}(2\pi).

Answer: 1616.

Practice 5: negative power

Simplify (3cis⁡π6)−2(3\operatorname{cis}\frac{\pi}{6})^{-2}.

  1. 1Use modulus 3−2=193^{-2}=\frac19.
  2. 2Use argument −2⋅π6=−π3-2\cdot\frac{\pi}{6}=-\frac{\pi}{3}.

Answer: 19cis⁡(−π3)\frac19\operatorname{cis}(-\frac{\pi}{3}).

Practice 6: cube roots

Solve z3=8iz^3=8i in Cartesian form.

  1. 1Write 8i=8cis⁡π28i=8\operatorname{cis}\frac{\pi}{2}.
  2. 2Use root modulus 2 and arguments π6,5π6,3π2\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}.
  3. 3Convert all three roots.

Answer: z=3+i,−3+i,−2iz=\sqrt3+i,-\sqrt3+i,-2i.

Practice 7: fourth roots

Find all fourth roots of −16-16 in polar form.

  1. 1Write −16=16cis⁡π-16=16\operatorname{cis}\pi.
  2. 2Use modulus 2 and arguments π+2kπ4\frac{\pi+2k\pi}{4} for k=0,1,2,3k=0,1,2,3.

Answer: 2cis⁡π4,2cis⁡3π4,2cis⁡5π4,2cis⁡7π42\operatorname{cis}\frac{\pi}{4},2\operatorname{cis}\frac{3\pi}{4},2\operatorname{cis}\frac{5\pi}{4},2\operatorname{cis}\frac{7\pi}{4}.

Practice 8: roots of unity

Find the sixth roots of unity and state their geometric arrangement.

  1. 1Use ωk=cis⁡2kπ6\omega_k=\operatorname{cis}\frac{2k\pi}{6}, k=0,…,5k=0,\ldots,5.
  2. 2Their common modulus is 1 and separation is π3\frac{\pi}{3}.

Answer: They are cis⁡kπ3\operatorname{cis}\frac{k\pi}{3}, k=0,…,5k=0,\ldots,5, forming a regular hexagon on the unit circle.

Practice 9: primitive roots

Identify the primitive eighth roots of unity.

  1. 1Require gcd⁡(k,8)=1\gcd(k,8)=1.
  2. 2Use k=1,3,5,7k=1,3,5,7.

Answer: cis⁡π4,cis⁡3π4,cis⁡5π4,cis⁡7π4\operatorname{cis}\frac{\pi}{4},\operatorname{cis}\frac{3\pi}{4},\operatorname{cis}\frac{5\pi}{4},\operatorname{cis}\frac{7\pi}{4}.

Practice 10: shifted root equation

Solve (z−1)3=8i(z-1)^3=8i.

  1. 1Let w=z−1w=z-1.
  2. 2Use the three cube roots 3+i,−3+i,−2i\sqrt3+i,-\sqrt3+i,-2i.
  3. 3Add 1 to each root.

Answer: z=1+3+i,1−3+i,1−2iz=1+\sqrt3+i,1-\sqrt3+i,1-2i.

Practice 11: factor over the complex numbers

Factor z3−8z^3-8 completely over C\mathbb C.

  1. 1Find the cube roots of 8: 2,−1+3i,−1−3i2,-1+\sqrt3 i,-1-\sqrt3 i.
  2. 2Write one factor per root.

Answer: (z−2)(z+1−3i)(z+1+3i)(z-2)(z+1-\sqrt3 i)(z+1+\sqrt3 i).

Practice 12: derive a triple-angle identity

Use De Moivre's theorem to derive a formula for cos⁡3θ\cos3\theta in terms of cos⁡θ\cos\theta.

  1. 1Expand (cos⁡θ+isin⁡θ)3(\cos\theta+i\sin\theta)^3.
  2. 2Equate the real part to cos⁡3θ\cos3\theta.
  3. 3Replace sin⁡2θ\sin^2\theta by 1−cos⁡2θ1-\cos^2\theta.

Answer: cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos3\theta=4\cos^3\theta-3\cos\theta.

Rules and formulas

Complete workflow

identify theorem→polar form→power or roots→complete set→verify\text{identify theorem}\to\text{polar form}\to\text{power or roots}\to\text{complete set}\to\text{verify}

Topic summary and cheat sheet

Conjugate pairing connects non-real roots to real polynomial factors. De Moivre's theorem then makes powers and complete root families systematic in polar form.

A strong final check combines algebraic substitution, root count, common modulus, equal angular spacing and the polynomial's coefficient conditions.

Key terms

De Moivre fluency
Choosing and applying conjugate, power, root and geometric ideas accurately.

Worked examples

Final checklist

What should be checked before submitting a De Moivre or roots answer?

  1. 1The theorem's conditions are satisfied.
  2. 2The modulus operation is correct.
  3. 3All n roots are listed with k=0k = 0 to n − 1.
  4. 4Arguments and geometry are consistent.
  5. 5Substitution verifies the result.

Answer: Submit an exact, complete, and clearly indexed answer.

Rules and formulas

Conjugate-root factor

(x−(a+bi))(x−(a−bi))=(x−a)2+b2(x-(a+bi))(x-(a-bi))=(x-a)^2+b^2

De Moivre's theorem

(rcis⁡θ)n=rncis⁡(nθ)(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)

Nth roots

zk=R1ncis⁡(θ+2kπn)z_k=R^{\frac{1}{n}}\operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)

Roots of unity

ωk=cis⁡2kπn\omega_k=\operatorname{cis}\frac{2k\pi}{n}

Topic summary

  • The conjugate of a+bia+bi is (a-bi).
  • Conjugation reflects a complex point in the real axis and preserves modulus.
  • Non-real roots of a polynomial with real coefficients occur in conjugate pairs.
  • The conjugate-root theorem is not guaranteed for non-real polynomial coefficients.
  • Roots a±bia\pm bi generate the real quadratic factor (x−a)2+b2(x-a)^2+b^2.
  • Repeated non-real roots have conjugates with the same multiplicity in a real polynomial.
  • De Moivre's theorem is (rcis⁡θ)n=rncis⁡(nθ)(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta) for integer n.
  • For a power, raise the modulus and multiply the argument.
  • Negative powers require a non-zero complex base.
  • The modulus check for powers is ∣zn∣=∣z∣n|z^n|=|z|^n.
  • De Moivre can derive multiple-angle identities by comparing real and imaginary parts.
  • To solve zn=Rcis⁡θz^n=R\operatorname{cis}\theta, use modulus R1nR^{\frac{1}{n}}.
  • The root arguments are θ+2kπn\frac{\theta+2k\pi}{n}.
  • Use k=0,1,…,n−1k=0,1,\ldots,n-1 to obtain exactly n distinct roots.
  • Consecutive nth roots are separated by 2πn\frac{2\pi}{n}.
  • The nth roots form a regular n-gon centered at the origin.
  • The sum of a complete non-trivial nth-root set is zero.
  • The nth roots of unity are cis⁡2kπn\operatorname{cis}\frac{2k\pi}{n}.
  • Every root of unity lies on the unit circle, and 1 is always included.
  • A root cis⁡2kπn\operatorname{cis}\frac{2k\pi}{n} is primitive when gcd⁡(k,n)=1\gcd(k,n)=1.
  • A degree-n polynomial has n complex roots counting multiplicity.
  • Pair conjugate factors before expanding a real polynomial.
  • A GDC principal-root output does not replace the full indexed root family.
  • Verify results by substitution, root count, modulus and equal angular spacing.