Mathematics · Book 2 · Grades 10–12

High School Mathematics

High School Mathematics · Grades 10–12

28Complex Numbers

The equation x2=−1x^2 = -1 has no real solution. Enlarging R\R with a single new number i\iu whose square is −1-1 produces the field C\C of complex numbers, in which every polynomial equation has solutions — and whose arithmetic encodes the geometry of the plane: translations, rotations and scalings become additions and multiplications.

28.1 Algebraic form

Definition 28.1 (Complex numbers)

The set of complex numbers is

C={a+ib:a,b∈R},\C = \{a + \iu b : a, b \in \R\},

where i\iu is a symbol subject to the single rule i2=−1\iu^2 = -1; addition and multiplication are performed as for real polynomials in i\iu, reducing i2\iu^2 to −1-1. The reals aa and bb are the real part Re⁡(z)\Rea(z) and imaginary part Im⁡(z)\Ima(z) of z=a+ibz = a + \iu b; this expression is the algebraic form of zz, and it is unique: a+ib=a′+ib′a + \iu b = a' + \iu b' if and only if a=a′a = a' and b=b′b = b'.

Example 28.2

(2+3i)(1−i)=2−2i+3i−3i2=5+i(2 + 3\iu)(1 - \iu) = 2 - 2\iu + 3\iu - 3\iu^2 = 5 + \iu.

Definition 28.3 (Conjugate)

The conjugate of z=a+ibz = a + \iu b is z‾=a−ib\conj{z} = a - \iu b.

Proposition 28.4 (Properties of conjugation)

For all z,w∈Cz, w \in \C:

z+w‾=z‾+w‾,zw‾=z‾ w‾,z‾‾=z,z+z‾=2Re⁡(z),z−z‾=2i Im⁡(z),\conj{z + w} = \conj z + \conj w, \quad \conj{zw} = \conj z\,\conj w, \quad \conj{\conj z} = z, \quad z + \conj z = 2\Rea(z), \quad z - \conj z = 2\iu\,\Ima(z),

and zz‾=a2+b2≥0z\conj z = a^2 + b^2 \geq 0 for z=a+ibz = a + \iu b. Moreover z∈Rz \in \R if and only if z=z‾z = \conj z.

Proof. All are direct computations from the algebraic forms; for instance zz‾=(a+ib)(a−ib)=a2−(ib)2=a2+b2z\conj z = (a + \iu b)(a - \iu b) = a^2 - (\iu b)^2 = a^2 + b^2. ∎

Method 28.5 (Dividing complex numbers)

To write a quotient in algebraic form, multiply numerator and denominator by the conjugate of the denominator:

12+i=2−i(2+i)(2−i)=2−i5=25−15i.\frac{1}{2 + \iu} = \frac{2 - \iu}{(2+\iu)(2-\iu)} = \frac{2 - \iu}{5} = \frac25 - \frac15\iu .

Every nonzero complex number therefore has an inverse: C\C is a field.

28.2 The complex plane, modulus and argument

Definition 28.6 (Affix, modulus)

In the plane with orthonormal coordinates, the point M(a,b)M(a, b) is the image of z=a+ibz = a + \iu b, and zz is the affix of MM. The modulus of zz is the distance from MM to the origin:

∣z∣=a2+b2=zz‾.\abs{z} = \sqrt{a^2 + b^2} = \sqrt{z\conj z}.
The point M(a,b) of affix z: the modulus z is the length OM, and the conjugate z is the reflection of z in the real axis.
The point M(a,b)M(a,b) of affix zz: the modulus ∣z∣\abs z is the length OMOM, and the conjugate z‾\conj z is the reflection of zz in the real axis.

Proposition 28.7 (Properties of the modulus)

For z,w∈Cz, w \in \C:

∣zw∣=∣z∣ ∣w∣,∣zw∣=∣z∣∣w∣ (w≠0),∣z‾∣=∣z∣,∣z+w∣≤∣z∣+∣w∣ (triangle inequality).\abs{zw} = \abs z\,\abs w, \qquad \abs{\tfrac zw} = \tfrac{\abs z}{\abs w}\ (w \neq 0), \qquad \abs{\conj z} = \abs z, \qquad \abs{z + w} \leq \abs{z} + \abs{w} \ \text{(triangle inequality)}.

Moreover ∣zB−zA∣\abs{z_B - z_A} is the distance between the points of affixes zAz_A and zBz_B.

Proof. ∣zw∣2=zw zw‾=zz‾ ww‾=∣z∣2∣w∣2\abs{zw}^2 = zw\,\conj{zw} = z\conj z\, w \conj w = \abs z^2 \abs w^2 gives the first; the second and third are similar. For the triangle inequality:

∣z+w∣2=(z+w)(z‾+w‾)=∣z∣2+∣w∣2+2Re⁡(zw‾)≤∣z∣2+∣w∣2+2∣zw‾∣=(∣z∣+∣w∣)2,\abs{z+w}^2 = (z+w)(\conj z + \conj w) = \abs z^2 + \abs w^2 + 2\Rea(z \conj w) \leq \abs z^2 + \abs w^2 + 2\abs{z\conj w} = \bigl(\abs z + \abs w\bigr)^2,

using Re⁡(u)≤∣u∣\Rea(u) \leq \abs u. The distance statement is the Pythagorean formula applied to the coordinates of zB−zAz_B - z_A. ∎

Definition 28.8 (Argument, trigonometric and exponential forms)

Let z≠0z \neq 0. An argument of zz, written arg⁡(z)\Arg(z), is any measure θ\theta of the angle from the positive real axis to the ray OMOM; it is defined up to 2kπ2k\pi. Writing r=∣z∣r = \abs z,

z=r(cos⁡θ+isin⁡θ)(trigonometric form).z = r(\cos\theta + \iu\sin\theta) \qquad\text{(trigonometric form)}.

Introducing the notation

eiθ=cos⁡θ+isin⁡θ,\eu^{\iu\theta} = \cos\theta + \iu\sin\theta,

this becomes the exponential form z=r eiθz = r\,\eu^{\iu\theta}.

Exponential form: z is determined by its distance r to the origin and the angle from the positive real axis.
Exponential form: zz is determined by its distance rr to the origin and the angle θ\theta from the positive real axis.

Theorem 28.9

For all θ,θ′∈R\theta, \theta' \in \R:

eiθ eiθ′=ei(θ+θ′),eiθ‾=e−iθ,∣eiθ∣=1.\eu^{\iu\theta}\,\eu^{\iu\theta'} = \eu^{\iu(\theta + \theta')}, \qquad \conj{\eu^{\iu\theta}} = \eu^{-\iu\theta}, \qquad \abs{\eu^{\iu\theta}} = 1 .

Consequently, for z=reiθz = r\eu^{\iu\theta} and z′=r′eiθ′z' = r'\eu^{\iu\theta'} nonzero:

zz′=rr′ ei(θ+θ′),zz′=rr′ ei(θ−θ′):zz' = rr'\,\eu^{\iu(\theta+\theta')}, \qquad \frac{z}{z'} = \frac{r}{r'}\,\eu^{\iu(\theta - \theta')} :

moduli multiply, arguments add.

Proof. The first identity is the pair of addition formulas (Proposition 24.3):

(cos⁡θ+isin⁡θ)(cos⁡θ′+isin⁡θ′)=(cos⁡θcos⁡θ′−sin⁡θsin⁡θ′)+i(sin⁡θcos⁡θ′+cos⁡θsin⁡θ′).(\cos\theta + \iu\sin\theta)(\cos\theta' + \iu\sin\theta') = (\cos\theta\cos\theta' - \sin\theta\sin\theta') + \iu(\sin\theta\cos\theta' + \cos\theta\sin\theta').

The rest follows by direct computation. ∎

Corollary 28.10 (De Moivre’s formula)

For θ∈R\theta \in \R and n∈Zn \in \Z: (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ\bigl(\cos\theta + \iu\sin\theta\bigr)^n = \cos n\theta + \iu \sin n\theta.

Proof. Induction on n≥0n \geq 0 using Theorem 28.9; extend to n<0n < 0 by taking inverses. ∎

Method 28.11 (Switching between forms)

From algebraic to exponential: compute r=∣z∣r = \abs z, then find θ\theta with cos⁡θ=ar\cos\theta = \frac ar, sin⁡θ=br\sin\theta = \frac br (locate the correct quadrant before invoking an inverse cosine). From exponential to algebraic: expand rcos⁡θ+i rsin⁡θr\cos\theta + \iu\, r\sin\theta. Use the algebraic form for sums, the exponential form for products, powers and quotients.

Example 28.12

z=1+iz = 1 + \iu: ∣z∣=2\abs z = \sqrt2 and cos⁡θ=sin⁡θ=12\cos\theta = \sin\theta = \frac{1}{\sqrt2}, so θ=π4\theta = \frac\pi4 and z=2 eiπ/4z = \sqrt2\,\eu^{\iu\pi/4}. Hence z20=210 e5iπ=−1024z^{20} = 2^{10}\,\eu^{5\iu\pi} = -1024.

28.3 Second-degree equations and roots of unity

Theorem 28.13 (Quadratic equations with real coefficients)

Let a,b,c∈Ra, b, c \in \R, a≠0a \neq 0, and Δ=b2−4ac\Delta = b^2 - 4ac. The equation az2+bz+c=0az^2 + bz + c = 0 has solutions in C\C:

Δ>0: z=−b±Δ2a (two real);Δ=0: z=−b2a (double);Δ<0: z=−b±i−Δ2a (two conjugate).\begin{aligned} &\Delta > 0:\ z = \frac{-b \pm \sqrt{\Delta}}{2a} \ (\text{two real}); \qquad \Delta = 0:\ z = \frac{-b}{2a} \ (\text{double}); \\ &\Delta < 0:\ z = \frac{-b \pm \iu\sqrt{-\Delta}}{2a} \ (\text{two conjugate}). \end{aligned}

Proof. Complete the square: az2+bz+c=a((z+b2a)2−Δ4a2)az^2 + bz + c = a\left(\left(z + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2}\right). If Δ<0\Delta < 0, then Δ4a2=(i−Δ2a)2\frac{\Delta}{4a^2} = \left(\iu\frac{\sqrt{-\Delta}}{2a}\right)^2, and the difference of squares factors as usual. ∎

Definition 28.14 (Roots of unity)

For n≥1n \geq 1, the nn-th roots of unity are the solutions of zn=1z^n = 1.

Theorem 28.15

The equation zn=1z^n = 1 has exactly nn solutions:

ωk=e2ikπ/n,k=0,1,…,n−1.\omega_k = \eu^{2\iu k\pi/n}, \qquad k = 0, 1, \dots, n-1 .

Their images are the vertices of a regular nn-gon inscribed in the unit circle, and for n≥2n \geq 2 their sum is 00.

Proof. Each ωk\omega_k satisfies ωkn=e2ikπ=1\omega_k^n = \eu^{2\iu k\pi} = 1, and they are pairwise distinct since their arguments lie in [0, 2π)\intco{0}{2\pi}. Conversely, if zn=1z^n = 1 then ∣z∣n=1\abs z^n = 1, so ∣z∣=1\abs z = 1 (positive real) and z=eiθz = \eu^{\iu\theta} with nθ≡0(mod2π)n\theta \equiv 0 \pmod{2\pi}, i.e. θ=2kπn\theta = \frac{2k\pi}{n}; reducing kk modulo nn lands among the ωk\omega_k. The vertices are equally spaced by angle 2πn\frac{2\pi}{n}: a regular nn-gon. Finally, with ω=ω1\omega = \omega_1, ω≠1\omega \neq 1 and the geometric sum gives

∑k=0n−1ωk=∑k=0n−1ωk=ωn−1ω−1=0.\sum_{k=0}^{n-1}\omega_k = \sum_{k=0}^{n-1}\omega^k = \frac{\omega^n - 1}{\omega - 1} = 0 . \qedhere

∎

The fifth roots of unity, = 2 π/5: the vertices of a regular pentagon inscribed in the unit circle ().
The fifth roots of unity, ω=e2iπ/5\omega = \eu^{2\iu\pi/5}: the vertices of a regular pentagon inscribed in the unit circle (Exercise 28.10).

Example 28.16

The cube roots of unity are 11, j=e2iπ/3=−12+i32j = \eu^{2\iu\pi/3} = -\frac12 + \iu\frac{\sqrt3}{2} and j‾=j2\conj j = j^2, satisfying 1+j+j2=01 + j + j^2 = 0.

28.4 Complex numbers and plane geometry

Proposition 28.17 (Geometric dictionary)

Let A,B,C,DA, B, C, D be points of affixes a,b,c,da, b, c, d (a≠ba \neq b, c≠dc \neq d).

  1. The translation by the vector of affix tt is z↦z+tz \mapsto z + t.
  2. The rotation of center ω\omega (affix) and angle θ\theta is z↦ω+eiθ(z−ω)z \mapsto \omega + \eu^{\iu\theta}(z - \omega).
  3. The scaling (homothety) of center ω\omega and ratio k∈R∗k \in \R^* is z↦ω+k(z−ω)z \mapsto \omega + k(z - \omega).
  4. d−cb−a\dfrac{d - c}{b - a} has modulus CDAB\dfrac{CD}{AB} and argument the angle between the vectors AB→\vect{AB} and CD→\vect{CD}. In particular, AB⊥CDAB \perp CD if and only if d−cb−a\frac{d-c}{b-a} is purely imaginary, and A,B,CA, B, C are collinear (C≠AC \ne A, C≠BC\ne B) if and only if c−ab−a∈R\frac{c-a}{b-a} \in \R.

Proof. (1) and (3) restate the coordinate formulas. For (2), the map w↦eiθww \mapsto \eu^{\iu\theta} w multiplies moduli by 11 and adds θ\theta to arguments: it is the rotation of angle θ\theta about the origin; conjugating by the translation sending ω\omega to 00 gives the general center. For (4), moduli and arguments of quotients subtract (Theorem 28.9). ∎

28.5 Exercises

Exercise 28.1 ★

Write in algebraic form: (3−2i)(1+4i)(3 - 2\iu)(1 + 4\iu), 2+i1−i\dfrac{2 + \iu}{1 - \iu}, i2027\iu^{2027}.

Solution

Solution of Exercise 28.1.

(3−2i)(1+4i)=3+12i−2i−8i2=11+10i(3-2\iu)(1+4\iu) = 3 + 12\iu - 2\iu - 8\iu^2 = 11 + 10\iu.

2+i1−i=(2+i)(1+i)(1−i)(1+i)=1+3i2=12+32i\dfrac{2+\iu}{1-\iu} = \dfrac{(2+\iu)(1+\iu)}{(1-\iu)(1+\iu)} = \dfrac{1 + 3\iu}{2} = \dfrac12 + \dfrac32\iu.

i2027\iu^{2027}: since i4=1\iu^4 = 1 and 2027=4×506+32027 = 4 \times 506 + 3, i2027=i3=−i\iu^{2027} = \iu^3 = -\iu.

Exercise 28.2 ★

Solve in C\C: (a) z2−4z+13=0z^2 - 4z + 13 = 0; (b) z2+z+1=0z^2 + z + 1 = 0. Check in each case that the two solutions are conjugates and compute their product.

Solution

Solution of Exercise 28.2.

(a) Δ=16−52=−36\Delta = 16 - 52 = -36: z=4±6i2=2±3iz = \dfrac{4 \pm 6\iu}{2} = 2 \pm 3\iu. Conjugate pair; product =4+9=13=ca= 4 + 9 = 13 = \frac ca. (b) Δ=1−4=−3\Delta = 1 - 4 = -3: z=−1±i32z = \dfrac{-1 \pm \iu\sqrt3}{2} (the cube roots of unity jj and j‾\conj j); product =1+34=1= \frac{1 + 3}{4} = 1.

Exercise 28.3 ★

Put in exponential form: z1=−1+iz_1 = -1 + \iu, z2=3−iz_2 = \sqrt3 - \iu, and compute z1z2\dfrac{z_1}{z_2} in exponential then algebraic form. Deduce the exact value of cos⁡11π12\cos\dfrac{11\pi}{12}.

Solution

Solution of Exercise 28.3.

∣z1∣=2\abs{z_1} = \sqrt2, argument 3π4\frac{3\pi}{4} (second quadrant): z1=2 e3iπ/4z_1 = \sqrt2\,\eu^{3\iu\pi/4}. ∣z2∣=2\abs{z_2} = 2, argument −π6-\frac\pi6: z2=2 e−iπ/6z_2 = 2\,\eu^{-\iu\pi/6}. Hence

z1z2=22 ei(3π4+π6)=22 e11iπ/12.\frac{z_1}{z_2} = \frac{\sqrt2}{2}\,\eu^{\iu\left(\frac{3\pi}{4} + \frac{\pi}{6}\right)} = \frac{\sqrt2}{2}\,\eu^{11\iu\pi/12}.

Algebraically,

z1z2=−1+i3−i=(−1+i)(3+i)4=−3−1+i(3−1)4.\frac{z_1}{z_2} = \frac{-1+\iu}{\sqrt3 - \iu} = \frac{(-1+\iu)(\sqrt3+\iu)}{4} = \frac{-\sqrt3 - 1 + \iu(\sqrt3 - 1)}{4}.

Identifying real parts: 22cos⁡11π12=−3−14\frac{\sqrt2}{2}\cos\frac{11\pi}{12} = \frac{-\sqrt3-1}{4}, so

cos⁡11π12=−6+24.\cos\frac{11\pi}{12} = -\frac{\sqrt6 + \sqrt2}{4}.

Exercise 28.4 ★★

Describe geometrically the set of points MM of affix zz such that: (a) ∣z−2∣=∣z+i∣\abs{z - 2} = \abs{z + \iu}; (b) ∣z−1−i∣=3\abs{z - 1 - \iu} = 3; (c) z−1z+1\dfrac{z - 1}{z + 1} is purely imaginary.

Solution

Solution of Exercise 28.4.

(a) ∣z−2∣=∣z−(−i)∣\abs{z - 2} = \abs{z - (-\iu)} means MM is equidistant from A(2,0)A(2, 0) and B(0,−1)B(0, -1): the perpendicular bisector of [AB][AB].

(b) Distance 33 from the point (1,1)(1,1): the circle of center 1+i1 + \iu and radius 33.

(c) By Proposition 28.17(4), z−1z+1\frac{z-1}{z+1} purely imaginary means the vectors from A(1,0)A(1,0) to MM and from B(−1,0)B(-1,0) to MM are orthogonal: MM sees the segment [AB][AB] under a right angle. The set is the circle of diameter [AB][AB] (the unit circle) minus the points AA and BB themselves.

Exercise 28.5 ★★

Using De Moivre’s formula and the binomial theorem, express cos⁡3θ\cos 3\theta as a polynomial in cos⁡θ\cos\theta, and sin⁡3θ\sin 3\theta in terms of sin⁡θ\sin\theta.

Solution

Solution of Exercise 28.5.

De Moivre: cos⁡3θ+isin⁡3θ=(cos⁡θ+isin⁡θ)3\cos3\theta + \iu\sin3\theta = (\cos\theta + \iu\sin\theta)^3. Expanding by the binomial theorem with c=cos⁡θc = \cos\theta, s=sin⁡θs = \sin\theta:

(c+is)3=c3+3ic2s−3cs2−is3=(c3−3cs2)+i(3c2s−s3).(c + \iu s)^3 = c^3 + 3\iu c^2 s - 3 c s^2 - \iu s^3 = (c^3 - 3cs^2) + \iu(3c^2 s - s^3).

Identifying parts and using s2=1−c2s^2 = 1 - c^2, c2=1−s2c^2 = 1 - s^2:

cos⁡3θ=4cos⁡3θ−3cos⁡θ,sin⁡3θ=3sin⁡θ−4sin⁡3θ.\cos3\theta = 4\cos^3\theta - 3\cos\theta, \qquad \sin3\theta = 3\sin\theta - 4\sin^3\theta .

Exercise 28.6 ★★

(Linearization.) Using cos⁡θ=eiθ+e−iθ2\cos\theta = \frac{\eu^{\iu\theta} + \eu^{-\iu\theta}}{2}, linearize cos⁡3θ\cos^3\theta (write it as a combination of cos⁡kθ\cos k\theta), and deduce ∫0π/2cos⁡3θ  ⁣dθ\displaystyle\int_0^{\pi/2}\cos^3\theta\,\dd\theta.

Solution

Solution of Exercise 28.6.

cos⁡3θ=(eiθ+e−iθ2) ⁣3=e3iθ+3eiθ+3e−iθ+e−3iθ8=cos⁡3θ+3cos⁡θ4.\cos^3\theta = \left(\frac{\eu^{\iu\theta} + \eu^{-\iu\theta}}{2}\right)^{\!3} = \frac{\eu^{3\iu\theta} + 3\eu^{\iu\theta} + 3\eu^{-\iu\theta} + \eu^{-3\iu\theta}}{8} = \frac{\cos3\theta + 3\cos\theta}{4}.

Hence

∫0π/2cos⁡3θ  ⁣dθ=14[sin⁡3θ3+3sin⁡θ]0π/2=14(−13+3)=23.\int_0^{\pi/2}\cos^3\theta\,\dd\theta = \frac14\left[\frac{\sin3\theta}{3} + 3\sin\theta\right]_0^{\pi/2} = \frac14\left(-\frac13 + 3\right) = \frac{2}{3}.

(Consistent with W3=23W_3 = \frac23 in Exercise 25.9, since ∫0π/2cos⁡3=∫0π/2sin⁡3\int_0^{\pi/2}\cos^3 = \int_0^{\pi/2}\sin^3 by the symmetry θ↦π2−θ\theta \mapsto \frac\pi2 - \theta.)

Exercise 28.7 ★★

Let AA and BB have affixes a=1+ia = 1 + \iu and b=3+2ib = 3 + 2\iu. Determine the two points CC making the triangle ABCABC equilateral, as the images of BB under the rotations of center AA and angles ±π3\pm\frac{\pi}{3}.

Solution

Solution of Exercise 28.7.

c=a+e±iπ/3(b−a)c = a + \eu^{\pm\iu\pi/3}(b - a) with b−a=2+ib - a = 2 + \iu and e±iπ/3=12±i32\eu^{\pm\iu\pi/3} = \frac12 \pm \iu\frac{\sqrt3}{2}:

eiπ/3(2+i)=(12+i32)(2+i)=(1−32)+i(12+3),\eu^{\iu\pi/3}(2+\iu) = \left(\tfrac12 + \iu\tfrac{\sqrt3}{2}\right)(2+\iu) = \left(1 - \tfrac{\sqrt3}{2}\right) + \iu\left(\tfrac12 + \sqrt3\right),

so c1=(2−32)+i(32+3)c_1 = \left(2 - \frac{\sqrt3}{2}\right) + \iu\left(\frac32 + \sqrt3\right), and similarly with the angle −π3-\frac\pi3: c2=(2+32)+i(32−3)c_2 = \left(2 + \frac{\sqrt3}{2}\right) + \iu\left(\frac32 - \sqrt3\right).

Exercise 28.8 ★★★

Solve z4=−4z^4 = -4 in C\C and plot the solutions. Factor z4+4z^4 + 4 into two quadratic polynomials with real coefficients.

Solution

Solution of Exercise 28.8.

−4=4 eiπ-4 = 4\,\eu^{\iu\pi}, so the solutions of z4=−4z^4 = -4 are

zk=2 ei(π4+kπ2),k=0,1,2,3,z_k = \sqrt2\,\eu^{\iu\left(\frac\pi4 + \frac{k\pi}{2}\right)}, \quad k = 0, 1, 2, 3,

i.e. 1+i1 + \iu, −1+i-1 + \iu, −1−i-1 - \iu, 1−i1 - \iu: the vertices of a square of circumradius 2\sqrt 2. Grouping conjugate roots,

z4+4=(z−(1+i))(z−(1−i))(z+(1−i))(z+(1+i))=(z2−2z+2)(z2+2z+2).z^4 + 4 = \bigl(z - (1{+}\iu)\bigr)\bigl(z - (1{-}\iu)\bigr) \bigl(z + (1{-}\iu)\bigr)\bigl(z + (1{+}\iu)\bigr) = (z^2 - 2z + 2)(z^2 + 2z + 2).

Exercise 28.9 ★★★

For θ∈(0, 2π)\theta \in \intoo{0}{2\pi} and n∈Nn \in \N, compute

S=1+cos⁡θ+cos⁡2θ+⋯+cos⁡nθS = 1 + \cos\theta + \cos 2\theta + \dots + \cos n\theta

by summing the geometric progression ∑k=0neikθ\sum_{k=0}^n \eu^{\iu k\theta} and taking real parts. (Answer: S=sin⁡(n+1)θ2sin⁡θ2cos⁡nθ2S = \dfrac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}} \cos\frac{n\theta}{2}.)

Solution

Solution of Exercise 28.9.

With q=eiθ≠1q = \eu^{\iu\theta} \neq 1:

∑k=0neikθ=ei(n+1)θ−1eiθ−1=ei(n+1)θ/2eiθ/2⋅ei(n+1)θ/2−e−i(n+1)θ/2eiθ/2−e−iθ/2=einθ/2 sin⁡(n+1)θ2sin⁡θ2,\sum_{k=0}^{n} \eu^{\iu k\theta} = \frac{\eu^{\iu(n+1)\theta} - 1}{\eu^{\iu\theta} - 1} = \frac{\eu^{\iu(n+1)\theta/2}}{\eu^{\iu\theta/2}} \cdot\frac{\eu^{\iu(n+1)\theta/2} - \eu^{-\iu(n+1)\theta/2}}{\eu^{\iu\theta/2} - \eu^{-\iu\theta/2}} = \eu^{\iu n\theta/2}\, \frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}},

using eiα−e−iα=2isin⁡α\eu^{\iu\alpha} - \eu^{-\iu\alpha} = 2\iu\sin\alpha (the half-angle factoring). Taking real parts:

S=sin⁡(n+1)θ2sin⁡θ2 cos⁡nθ2.S = \frac{\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\, \cos\frac{n\theta}{2}.

Exercise 28.10 ★★★

Let ω=e2iπ/5\omega = \eu^{2\iu\pi/5}.

  1. Justify that 1+ω+ω2+ω3+ω4=01 + \omega + \omega^2 + \omega^3 + \omega^4 = 0.
  2. Let u=ω+ω4u = \omega + \omega^4 and v=ω2+ω3v = \omega^2 + \omega^3. Show that u+v=−1u + v = -1 and uv=−1uv = -1, and deduce that u=−1+52u = \frac{-1+\sqrt5}{2}.
  3. Conclude that cos⁡2π5=5−14\cos\dfrac{2\pi}{5} = \dfrac{\sqrt5 - 1}{4}.
Solution

Solution of Exercise 28.10.

1. ω\omega is a 55-th root of unity different from 11, so the sum of all five roots vanishes (Theorem 28.15), which reads 1+ω+ω2+ω3+ω4=01 + \omega + \omega^2 + \omega^3 + \omega^4 = 0.

2. u+v=ω+ω2+ω3+ω4=−1u + v = \omega + \omega^2 + \omega^3 + \omega^4 = -1 by point 1. For the product, using ω5=1\omega^5 = 1:

uv=(ω+ω4)(ω2+ω3)=ω3+ω4+ω6+ω7=ω3+ω4+ω+ω2=−1.uv = (\omega + \omega^4)(\omega^2 + \omega^3) = \omega^3 + \omega^4 + \omega^6 + \omega^7 = \omega^3 + \omega^4 + \omega + \omega^2 = -1 .

So u,vu, v are the roots of X2+X−1=0X^2 + X - 1 = 0: {−1+52,−1−52}\left\{\frac{-1+\sqrt5}{2}, \frac{-1-\sqrt5}{2}\right\}. Now u=ω+ω‾=2cos⁡2π5>0u = \omega + \conj\omega = 2\cos\frac{2\pi}{5} > 0 (since 2π5<π2\frac{2\pi}{5} < \frac{\pi}{2}), so u=−1+52u = \frac{-1 + \sqrt5}{2}.

3. cos⁡2π5=u2=5−14\cos\frac{2\pi}{5} = \frac u2 = \frac{\sqrt5 - 1}{4}.

28.6 Problem: The pentagon in the mirror of unity

Problem 28.1

Weekend problem — the fifth roots of unity compute cos⁡72∘\cos 72^\circ exactly, the golden ratio signs the result, and multiplication turns out to be rotation

Euclid could construct the regular pentagon but the reason why it yields to ruler and compass — while the humble 20∘20^\circ angle does not — stayed hidden for two thousand years. It hides in this chapter: the five fifth roots of unity (Theorem 28.15) sum to zero, and that one line of algebra squeezes cos⁡72∘\cos 72^\circ out of a quadratic equation — square roots only, hence constructible, and signed, of course, by the golden ratio. Around this jewel: fluency, de Moivre’s shortcut to trigonometry, and multiplication unmasked as geometry.

Part I — Fluency.

  1. Compute (2+i)(3−i)(2 + \iu)(3 - \iu), 1+i1−i\dfrac{1 + \iu}{1 - \iu} (Method 28.5), and ∣3+4i∣\abs{3 + 4\iu}.
  2. Put in exponential form: 1+i1 + \iu; −2-2; 1−i31 - \iu\sqrt3.
  3. Compute (1+i)8(1 + \iu)^8 using the exponential form.
  4. Solve z2=iz^2 = \iu, then z2−2z+5=0z^2 - 2z + 5 = 0 (Theorem 28.13).
  5. Geometric dictionary (Proposition 28.17): describe the map z↦izz \mapsto \iu z, and the set ∣z−(1+i)∣=2\abs{z - (1 + \iu)} = 2.

Part II — Five roots, one pentagon. Let ω=e2iπ/5\omega = \eu^{2\iu\pi/5} and consider 1,ω,ω2,ω3,ω41, \omega, \omega^2, \omega^3, \omega^4.

  1. Check these are exactly the solutions of z5=1z^5 = 1, and describe the figure they draw in the plane.
  2. Prove that their sum is zero. (Geometric progression — or multiply the sum by 1−ω1 - \omega.)

  3. Take real parts and deduce

    1+2cos⁡2π5+2cos⁡4π5=0.1 + 2\cos\frac{2\pi}{5} + 2\cos\frac{4\pi}{5} = 0 .

  4. Set c=cos⁡2π5c = \cos\frac{2\pi}{5} and use the double-angle formula for cos⁡4π5\cos\frac{4\pi}{5} to convert question 8 into a quadratic equation for cc; solve it and conclude

    cos⁡72∘=5−14.\cos 72^\circ = \frac{\sqrt5 - 1}{4} .
  5. Verify numerically, then let the golden ratio sign its work: show c=12φc = \frac{1}{2\varphi} where φ=1+52\varphi = \frac{1 + \sqrt5}{2} (Problem 2.1) — and state the famous geometric echo: in a regular pentagon, the diagonal is φ\varphi times the side.

Part III — De Moivre’s shortcuts.

  1. The construction verdict: cos⁡72∘\cos 72^\circ needs only square roots (question 9), so the pentagon is ruler-and-compass constructible; cos⁡20∘\cos 20^\circ satisfies an unbreakable cubic (Problem 24.1), so the 60∘60^\circ angle cannot be trisected. State the emerging criterion (square roots buildable, cube roots not), whose full theory — Gauss, age 19, and the 17-gon — lives in the university volumes.
  2. Solve z4=−4z^4 = -4 by exponential form and recover Exercise 28.8’s factorization z4+4=(z2−2z+2)(z2+2z+2)z^4 + 4 = (z^2 - 2z + 2)(z^2 + 2z + 2) by pairing conjugate roots.
  3. Expand (cos⁡θ+isin⁡θ)3(\cos\theta + \iu\sin\theta)^3 with the binomial theorem and de Moivre, and re-derive cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta in three lines — compare with the addition-formula route of Problem 24.1.
  4. Cross-check with Exercise 28.9: for θ=2π5\theta = \frac{2\pi}{5} and n=4n = 4, what does the sum 1+cos⁡θ+⋯+cos⁡4θ1 + \cos\theta + \dots + \cos 4\theta equal, and why is that consistent with question 8?
  5. Compute the three cube roots of 8i8\iu in algebraic form.

Part IV — Multiplication is geometry.

  1. Describe completely the map z↦(1+i)zz \mapsto (1 + \iu)z: by what angle does it rotate, by what factor does it scale, and what does it do to the unit square?
  2. Prove ∣z1z2∣=∣z1∣ ∣z2∣\abs{z_1 z_2} = \abs{z_1}\,\abs{z_2} with conjugates (Proposition 28.7), then compute (1+i)n(1+\iu)^n for n=1,2,3,4n = 1, 2, 3, 4 and describe the spiral the powers trace.
  3. The number j=e2iπ/3\mathrm{j} = \eu^{2\iu\pi/3}: show 1+j+j2=01 + \mathrm{j} + \mathrm{j}^2 = 0, and verify on the triangle (1,j,j2)\left(1, \mathrm{j}, \mathrm{j}^2\right) the classical criterion: a+jb+j2c=0a + \mathrm{j} b + \mathrm{j}^2 c = 0 holds for (one orientation of) an equilateral triangle abcabc.
  4. The fundamental theorem of algebra (admitted: d’Alembert–Gauss): every polynomial factors completely over C\C. Deduce the real-world corollary — every real polynomial factors into real linear and quadratic pieces — and point to question 12 as an instance.
  5. Finale — C\C’s portrait, one sentence each: numbers become points; multiplication becomes rotation-and-scaling; the roots of unity become regular polygons; de Moivre turns powers into trigonometric identities; and algebraic closure guarantees that no equation ever needs a bigger world. Coda: which two celebrated numbers met in question 10?
Solution

Solution of Problem 28.1.

1. (2+i)(3−i)=6−2i+3i+1=7+i(2 + \iu)(3 - \iu) = 6 - 2\iu + 3\iu + 1 = 7 + \iu. 1+i1−i=(1+i)22=2i2=i\frac{1 + \iu}{1 - \iu} = \frac{(1 + \iu)^2}{2} = \frac{2\iu}{2} = \iu. ∣3+4i∣=5\abs{3 + 4\iu} = 5.

2. 1+i=2 eiπ/41 + \iu = \sqrt2\,\eu^{\iu\pi/4}; −2=2eiπ-2 = 2\eu^{\iu\pi}; 1−i3=2e−iπ/31 - \iu\sqrt3 = 2\eu^{-\iu\pi/3}.

3. (1+i)8=(2)8e8iπ/4=16 e2iπ=16(1 + \iu)^8 = \left(\sqrt2\right)^8 \eu^{8\iu\pi/4} = 16\,\eu^{2\iu\pi} = 16.

4. z2=eiπ/2z^2 = \eu^{\iu\pi/2}: z=±eiπ/4=±22(1+i)z = \pm\eu^{\iu\pi/4} = \pm\frac{\sqrt2}{2}(1 + \iu). And Δ=4−20=−16\Delta = 4 - 20 = -16: z=1±2iz = 1 \pm 2\iu.

5. z↦izz \mapsto \iu z: rotation by a quarter turn about the origin. ∣z−(1+i)∣=2\abs{z - (1 + \iu)} = 2: the circle of center 1+i1 + \iu and radius 22.

6. (ωk)5=e2iπk=1\left(\omega^k\right)^5 = \eu^{2\iu\pi k} = 1, and the five points e2ikπ/5\eu^{2\iu k\pi/5} are distinct: all five solutions of z5=1z^5 = 1. They sit on the unit circle at equal angles of 72∘72^\circ: a regular pentagon with one vertex at 11.

7. S=1+ω+⋯+ω4S = 1 + \omega + \dots + \omega^4 satisfies (1−ω)S=1−ω5=0(1 - \omega)S = 1 - \omega^5 = 0, and ω≠1\omega \neq 1: S=0S = 0. (The pentagon’s balance point is its center.)

8. ω4=ω‾\omega^4 = \overline\omega and ω3=ω2‾\omega^3 = \overline{\omega^2}: pairing conjugates, the real part of the sum reads 1+2cos⁡2π5+2cos⁡4π5=01 + 2\cos\frac{2\pi}{5} + 2\cos\frac{4\pi}{5} = 0.

9. cos⁡4π5=2c2−1\cos\frac{4\pi}{5} = 2c^2 - 1, so 1+2c+4c2−2=01 + 2c + 4c^2 - 2 = 0, i.e. 4c2+2c−1=04c^2 + 2c - 1 = 0: c=−1+54c = \frac{-1 + \sqrt5}{4} (the positive root, since 72∘72^\circ is acute): cos⁡72∘=5−14\cos 72^\circ = \frac{\sqrt5 - 1}{4}.

10. Numerically c≈0.30902c \approx 0.30902: matches the calculator. And 12φ=11+5=5−14\frac{1}{2\varphi} = \frac{1}{1 + \sqrt5} = \frac{\sqrt5 - 1}{4} (rationalize): c=12φc = \frac{1}{2\varphi}. The pentagon is golden through and through: its diagonal cuts its side in the ratio φ\varphi — the pentagram’s endless self-similarity.

11. Square roots are compass-constructible (the half-circle machine of the Middle School volume’s geometric mean), so a number built from rationals by square roots alone — like cos⁡72∘\cos 72^\circ — is constructible; cube roots are not, and cos⁡20∘\cos 20^\circ needs one: the pentagon yields, trisection resists. Gauss’s criterion settles every regular polygon — his 17-gon at nineteen made him choose mathematics.

12. −4=4eiπ-4 = 4\eu^{\iu\pi}: the fourth roots have modulus 2\sqrt2 and arguments π4+kπ2\frac{\pi}{4} + k\frac{\pi}{2}: the four numbers ±1±i\pm 1 \pm \iu. Pairing conjugates: (z−1−i)(z−1+i)=z2−2z+2(z - 1 - \iu)(z - 1 + \iu) = z^2 - 2z + 2 and (z+1−i)(z+1+i)=z2+2z+2(z + 1 - \iu)(z + 1 + \iu) = z^2 + 2z + 2: the factorization of Exercise 28.8.

13. (cos⁡θ+isin⁡θ)3=cos⁡3θ+3icos⁡2θsin⁡θ−3cos⁡θsin⁡2θ−isin⁡3θ(\cos\theta + \iu\sin\theta)^3 = \cos^3\theta + 3\iu\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - \iu\sin^3\theta; by de Moivre this equals cos⁡3θ+isin⁡3θ\cos 3\theta + \iu \sin 3\theta. Real parts: cos⁡3θ=cos⁡3θ−3cos⁡θ(1−cos⁡2θ)=4cos⁡3θ−3cos⁡θ\cos 3\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta — and the imaginary parts hand over sin⁡3θ\sin 3\theta for free.

14. It is the real part of 1+ω+⋯+ω4=01 + \omega + \dots + \omega^4 = 0: the sum equals 00 — exactly question 8, seen through Exercise 28.9’s kernel.

15. 8i=8eiπ/28\iu = 8\eu^{\iu\pi/2}: roots 2eiπ/6=3+i2\eu^{\iu\pi/6} = \sqrt3 + \iu; 2e5iπ/6=−3+i2\eu^{5\iu\pi/6} = -\sqrt3 + \iu; 2e3iπ/2=−2i2\eu^{3\iu\pi/2} = -2\iu.

16. 1+i=2 eiπ/41 + \iu = \sqrt2\,\eu^{\iu\pi/4}: the map rotates the plane by 45∘45^\circ and scales it by 2\sqrt2. The unit square becomes a tilted square of side 2\sqrt2 — doubled in area, spun an eighth of a turn.

17. ∣z1z2∣2=z1z2z1z2‾=z1z1‾ z2z2‾=∣z1∣2∣z2∣2\abs{z_1 z_2}^2 = z_1 z_2 \overline{z_1 z_2} = z_1\overline{z_1}\, z_2\overline{z_2} = \abs{z_1}^2\abs{z_2}^2: take square roots. Powers of 1+i1 + \iu: 1+i1 + \iu, 2i2\iu, −2+2i-2 + 2\iu, −4-4: each a 45∘45^\circ turn further and 2\sqrt2 times longer — an outward spiral through moduli 2,2,22,4,…\sqrt2, 2, 2\sqrt2, 4, \dots

18. j\mathrm{j} is a third root of unity distinct from 11, so (question 7’s argument with n=3n = 3) 1+j+j2=01 + \mathrm{j} + \mathrm{j}^2 = 0. Criterion on the standard triangle: a+jb+j2c=1+j⋅j+j2⋅j2=1+j2+j4=1+j2+j=0a + \mathrm{j}b + \mathrm{j}^2 c = 1 + \mathrm{j}\cdot\mathrm{j} + \mathrm{j}^2 \cdot \mathrm{j}^2 = 1 + \mathrm{j}^2 + \mathrm{j}^4 = 1 + \mathrm{j}^2 + \mathrm{j} = 0: satisfied, as the equilateral triangle (1,j,j2)\left(1, \mathrm{j}, \mathrm{j}^2\right) demands.

19. Over C\C, a real polynomial splits into linear factors; its non-real roots come in conjugate pairs (conjugate the equation), and each pair multiplies into a real quadratic z2−2Re⁡(z0) z+∣z0∣2z^2 - 2\operatorname{Re}(z_0)\,z + \abs{z_0}^2: hence the real factorization into degree-11 and degree-22 pieces — question 12 performed it on z4+4z^4 + 4.

20. Points; rotation-and-scaling; regular polygons; trigonometric identities by expansion; and a closed world where every equation of degree nn has its nn roots. Coda: in question 10, π\pi’s pentagon shook hands with the golden ratio — Euclid’s two most famous guests, introduced at last by a sum of five arrows equal to zero.