Complex Numbers

Excerpts

Equations

Complex Numbers
i^2 = -1
The imaginary unit i is defined so that its square is minus one.
Complex Numbers
(\sqrt{p})^2 i^2 = -p
The square of either root of x^2 = -p, written as ±√p i, equals -p.
Complex Numbers
(a+bi) + (c+di) = (a+c) + (b+d)i
Two complex numbers are added by adding their real parts and their imaginary parts separately.
Complex Numbers
(a+bi) - (c+di) = (a-c) + (b-d)i
Subtraction of complex numbers is defined by subtracting real parts and imaginary parts separately; it is the inverse of addition.
Complex Numbers
(a+bi)(c+di) = ac-bd+(ad+bc)i
Multiplication of complex numbers follows formal algebra, with i^2 replaced by -1.
Complex Numbers
(a+bi)(a-bi) = a^2-b^2i^2 = a^2+b^2
A complex number times its conjugate equals the sum of the squares of its real and imaginary parts.
Complex Numbers
\frac{e+fi}{a+bi} = \frac{(e+fi)(a-bi)}{a^2+b^2} = \frac{ae+bf}{a^2+b^2} + \frac{af-be}{a^2+b^2} i
Division of complex numbers is performed by multiplying numerator and denominator by the conjugate of the denominator.
Complex Numbers
a^2+b^2 = 0
For real a and b, a^2+b^2 = 0 forces a = b = 0, so division by a nonzero complex number is always possible.
Complex Numbers
a+bi=0
A complex number is zero if and only if both its real and imaginary parts are zero.
Complex Numbers
r = \sqrt{a^2+b^2}
The modulus (absolute value) of a+bi is the positive length r of the segment from the origin to the point (a, b).
Complex Numbers
\cos \theta = a/r
The amplitude θ of a+bi has cosine equal to the real part divided by the modulus.
Complex Numbers
\sin \theta = b/r
The amplitude θ of a+bi has sine equal to the imaginary coefficient divided by the modulus.
Complex Numbers
a+bi = r(\cos\theta + i\sin\theta)
A complex number equals its modulus times the cosine plus i times the sine of its amplitude (trigonometric form).
Complex Numbers
\omega = -\tfrac{1}{2} + \tfrac{1}{2}\sqrt{3}i
The complex cube root of unity ω has real part -1/2 and imaginary part √3/2.
Complex Numbers
x^3-1 = (x-1) (x^2+x+1)
The difference of cubes factors as (x-1) times (x^2+x+1), so the roots of x^3 = 1 are 1 and the roots of x^2+x+1 = 0.
Complex Numbers
(x + \tfrac{1}{2})^2 = -\tfrac{3}{4}
Completing the square in x^2+x+1 = 0 gives this relation, whose roots are the two complex cube roots of unity.
Complex Numbers
\omega^2 + \omega+1 = 0
The complex cube root of unity ω satisfies ω^2 + ω + 1 = 0.
Complex Numbers
\omega^3 = 1
The complex cube root of unity ω cubed equals one.
Complex Numbers
\omega \omega' = 1
The two complex cube roots of unity ω and ω' have product one.
Complex Numbers
\omega' = \omega^2
The second complex cube root of unity equals the square of ω.
Complex Numbers
\cos \theta + i \sin \theta
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Complex Numbers
(\cos \theta + i \sin \theta) (\cos \alpha + i \sin \alpha) = \cos (\theta + \alpha) + i \sin (\theta + \alpha)
The product of two complex numbers in trigonometric form has modulus 1 and amplitude θ + α; with unit moduli this gives the addition formula for cosine and sine.
Complex Numbers
\frac{\cos \beta + i \sin \beta} {\cos \theta + i \sin \theta} = \cos(\beta - \theta) + i \sin(\beta - \theta)
The quotient of two complex numbers in trigonometric form has amplitude β - θ, the difference of the amplitudes.
Complex Numbers
\frac{1}{\cos\theta + i \sin\theta} = \cos\theta - i \sin\theta
The reciprocal of a unit complex number cos θ + i sin θ is its conjugate cos θ - i sin θ.
Complex Numbers
(\cos\theta + i \sin\theta)^n = \cos n\theta + i \sin n\theta
For any positive whole number n, the nth power of cos θ + i sin θ equals cos nθ + i sin nθ.
Complex Numbers
4\sqrt{2} + 4\sqrt{2} i = 8(\cos 45° + i \sin 45°)
The complex number 4√2 + 4√2 i has modulus 8 and amplitude 45° in trigonometric form.
Complex Numbers
3 \theta = 45°+ k·360°
Equal cubes of complex numbers have amplitudes differing by a whole multiple of 360°, so 3θ = 45° + k·360° for an integer k.
Complex Numbers
\theta = 15°+k·120°
The amplitudes of the cube roots of 8(cos 45° + i sin 45°) are 15° + k·120° for integer k.
Complex Numbers
r^n(\cos n\theta + i \sin n\theta) = \cos A + i \sin A
The nth power of r(cos θ + i sin θ), by de Moivre's theorem, must equal cos A + i sin A.
Complex Numbers
n\theta = A + k·360°
The amplitude of an nth root satisfies nθ = A + k·360° for an integer k.
Complex Numbers
\cos\left(\frac{A + k·360°}{n}\right) + i \sin\left(\frac{A + k·360°}{n}\right)
The nth roots of cos A + i sin A are given by this expression for integer k; only k = 0, 1, ..., n-1 give distinct roots.
Complex Numbers
\cos\frac{2k \pi}{n} + i \sin\frac{2k \pi}{n}
The n distinct nth roots of unity are given by this expression for k = 0, 1, ..., n-1.
Complex Numbers
R = \cos\frac{2\pi}{n} + i \sin\frac{2\pi}{n}
R is the primitive nth root of unity with amplitude 2π/n radians, the case k = 1.
Complex Numbers
R,\ R^2,\ R^3,\dotsc,\ R^{n-1},\ R^n = 1
The n distinct nth roots of unity are the powers of R, and the last power R^n equals 1.
Complex Numbers
R = \cos\pi/2 + i \sin\pi/2 = i
For n = 4, R equals cos(π/2) + i sin(π/2), which is the imaginary unit i.
Complex Numbers
\rho^n=1
A primitive nth root of unity ρ satisfies ρ^n = 1.
Complex Numbers
\rho^l \neq 1
A primitive nth root of unity has no positive integral power l < n that equals 1.
Complex Numbers
(R^k)^{\frac{n}{d}} = (R^n)^{\frac{k}{d}} = 1
If k and n have a common divisor d > 1, then R^k is not a primitive nth root of unity, since its power n/d equals 1.
Complex Numbers
R^{kl} = \cos\frac{2kl\pi}{n} + i \sin\frac{2kl\pi}{n}
By de Moivre's theorem, the power R^{kl} equals cos and sin of 2klπ/n.

Problems

Exercise Page2

Exercise Page2, problem 1, p. 2
$\sqrt{-9}$.
Printed answer:
- $3i$.
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- identity: FLAG-PARSE 3*I
Exercise Page2, problem 10, p. 2
Prove that the conjugate of the sum of two complex numbers is equal to the sum of their conjugates. Does the result hold true if each word sum is replaced by the word difference?
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- Yes.
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Exercise Page2, problem 11, p. 2
Prove that the conjugate of the product (or quotient) of two complex numbers is equal to the product (or quotient) of their conjugates.
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Exercise Page2, problem 12, p. 2
Prove that, if the product of two complex numbers is zero, at least one of them is zero.
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Exercise Page2, problem 13, p. 2
Find two pairs of real numbers $x$, $y$ for which (x+yi)^2 = -7+24i.
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- $3$, $4$ and $-3$, $-4$.
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Exercise Page2, problem 14, p. 2
$-11+60i$.
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- $±(5 + 6i)$.
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- solve: the printed answer does not match the problem [5 + 6*I, -5 - 6*I]
Exercise Page2, problem 15, p. 2
$5-12i$.
Printed answer:
- $±(3 - 2i)$.
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- solve: the printed answer does not match the problem [3 - 2*I, -3 + 2*I]
Exercise Page2, problem 16, p. 2
$4cd+(2c^2-2d^2)i$.
Printed answer:
- $±\bigl[c + d + (c - d)i\bigr]$.
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- solve: the printed answer does not match the problem [c + d + (c - d)*I, -c - d - (c - d)*I]
Exercise Page2, problem 2, p. 2
$\sqrt{4}$.
Printed answer:
- $2$.
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- identity: passes 2
Exercise Page2, problem 3, p. 2
$(\sqrt{25} + \sqrt{-25})\sqrt{-16}$.
Printed answer:
- $-20 + 20i$.
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- identity: FLAG-PARSE -20 + 20*I
Exercise Page2, problem 4, p. 2
$-\frac{2}{3}$.
Printed answer:
- $-\frac{2}{3}$.
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- identity: passes -Rational(2, 3)
Exercise Page2, problem 5, p. 2
$8 + 2\sqrt{3}\vphantom{\dfrac{1}{1}}$.
Printed answer:
- $(8 + 2\sqrt{3})$.
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- identity: passes 8 + 2*sqrt(3)
Exercise Page2, problem 6, p. 2
$\dfrac{3 + \sqrt{-5}}{2 + \sqrt{-1}}$.
Printed answer:
- $\frac{1}{5}(6 + \sqrt{5}) + \frac{1}{5}(2\sqrt{5} - 3)i$.
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- identity: FLAG-PARSE (6 + sqrt(5))/5 + (2*sqrt(5) - 3)*I/5
Exercise Page2, problem 7, p. 2
$\dfrac{3 + 5i}{2 - 3i}$.
Printed answer:
- $\dfrac{-9}{13} + \dfrac{19}{13} i$.
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- identity: FLAG-PARSE -Rational(9, 13) + Rational(19, 13)*I
Exercise Page2, problem 8, p. 2
$\dfrac{a + bi}{a - bi}$.
Printed answer:
- $\dfrac{a^2 - b^2}{a^2 + b^2} + \dfrac{2ab}{a^2 + b^2}i$.
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- identity: FLAG-PARSE (a**2 - b**2)/(a**2 + b**2) + 2*a*b*I/(a**2 + b**2)
Exercise Page2, problem 9, p. 2
Prove that the sum of two conjugate complex numbers is real and that their difference is a pure imaginary.
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Exercise Page6

Exercise Page6, problem 1, p. 6
Verify that $R_2 = \omega R_1$, $R_3 = \omega^2 R_1$. Verify that $R_1$ is a cube root of $8 (\cos 45°+ i \sin 45°)$ by cubing $R_1$ and applying De Moivre’s theorem. Why are the new expressions for $R_2$ and $R_3$ evidently also cube roots?
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Exercise Page6, problem 2a, p. 6
Find the three cube roots of $-27$; those of $-i$; those of $\omega$.
Printed answer:
- $-3$, $-3\omega$, $-3\omega^2$
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- solve: the printed answer does not match the problem [-3, -3*exp(2*pi*I/3), -3*exp(4*pi*I/3)]
Exercise Page6, problem 2b, p. 6
Find the three cube roots of $-27$; those of $-i$; those of $\omega$.
Printed answer:
- $i$, $\omega i$; $\omega^2 i$
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- solve: the printed answer does not match the problem [I, exp(2*pi*I/3)*I, exp(4*pi*I/3)*I]
Exercise Page6, problem 2c, p. 6
Find the three cube roots of $-27$; those of $-i$; those of $\omega$.
Printed answer:
- $R = \cos 40° + i\sin 40°$, $\omega R$, $\omega^2 R$
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How it was checked
- solve: the printed answer does not match the problem [exp(2*pi*I/9), exp(2*pi*I/9)*exp(2*pi*I/3), exp(2*pi*I/9)*exp(4*pi*I/3)]
Exercise Page6, problem 3a, p. 6
Find the two square roots of $i$; those of $-i$; those of $\omega$.
Printed answer:
- $±(1 + i)/\sqrt{2}$
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- solve: the printed answer does not match the problem [(1 + I)/sqrt(2), -(1 + I)/sqrt(2)]
Exercise Page6, problem 3b, p. 6
Find the two square roots of $i$; those of $-i$; those of $\omega$.
Printed answer:
- $±(1 - i)/\sqrt{2}$
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- solve: the printed answer does not match the problem [(1 - I)/sqrt(2), -(1 - I)/sqrt(2)]
Exercise Page6, problem 3c, p. 6
Find the two square roots of $i$; those of $-i$; those of $\omega$.
Printed answer:
- $±\omega^2$
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- solve: the printed answer does not match the problem [exp(4*pi*I/3), -exp(4*pi*I/3)]
Exercise Page6, problem 4, p. 6
Prove that the numbers $\cos\theta + i \sin\theta$ and no others are represented by points on the circle of radius unity whose center is the origin.
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Exercise Page6, problem 5, p. 6
If $a+bi$ and $c+di$ are represented by the points $A$ and $C$ in Fig. 3, prove that their sum is represented by the fourth vertex $S$ of the parallelogram two of whose sides are $OA$ and $OC$. Hence show that the modulus of the sum of two complex numbers is equal to or less than the sum of their moduli, and is equal to or greater than the difference of their moduli.
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- (none printed)
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- other: not a kind the checker handles
Exercise Page6, problem 6, p. 6
Let $r$ and $r'$ be the moduli and $\theta$ and $\alpha$ the amplitudes of two complex numbers represented by the points $A$ and $C$ in Fig. 4. Let $U$ be the point on the $x$-axis one unit to the right of the origin $O$. Construct triangle $OCP$ similar to triangle $OUA$ and similarly placed, so that corresponding sides are $OC$ and $OU, CP$ and $UA$, $OP$ and $OA$, while the vertices $O$, $C$, $P$ are in the same order (clockwise or counter-clockwise) as the corresponding vertices $O$, $U$, $A$. Prove that $P$ represents the product (§5) of the complex numbers represented by $A$ and $C$.
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Exercise Page6, problem 7, p. 6
If $a+bi$ and $e+fi$ are represented by the points $A$ and $S$ in Fig. 3, prove that the complex number obtained by subtracting $a+bi$ from $e+fi$ is represented by the point $C$. Hence show that the absolute value of the difference of two complex numbers is equal to or less than the sum of their absolute values, and is equal to or greater than the difference of their absolute values.
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- (none printed)
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Exercise Page6, problem 8, p. 6
By modifying Ex. 6, show how to construct geometrically the quotient of two complex numbers.
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Exercise Page9

Exercise Page9, problem 1, p. 9
Simplify the trigonometric forms $(6)$ of the four fourth roots of unity. Check the result by factoring $x^4-1$.
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Exercise Page9, problem 2, p. 9
For $n=6$, show that $R = -\omega^2$. The sixth roots of unity are the three cube roots of unity and their negatives. Check by factoring $x^6-1$.
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Exercise Page9, problem 3, p. 9
From the point representing $a+bi$, how do you obtain that representing $-(a+bi)$? Hence derive from Fig. 2 and Ex. 2 the points representing the six sixth roots of unity. Obtain this result another way.
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Exercise Page9, problem 4, p. 9
Find the five fifth roots of $-1$.
Printed answer:
- $-1$, $\cos A + i \sin A$ ($A=36°$, $108°$, $252°$, $324°$).
unverified: no computed check settled this one (yet)
How it was checked
- solve: the printed answer does not match the problem [-1, cos(pi/5) + I*sin(pi/5), cos(3*pi/5) + I*sin(3*pi/5), cos(7*pi/5) + I*sin(7*pi/5), cos(9*pi/5) + I*sin(9*pi/5)]
Exercise Page9, problem 5, p. 9
Obtain the trigonometric forms of the nine ninth roots of unity. Which of them are cube roots of unity?
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Exercise Page9, problem 6, p. 9
Which powers of a ninth root $(7)$ of unity are cube roots of unity?
Printed answer:
- $R^3$, $R^6$, $R^9$.
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- other: not a kind the checker handles

Exercise Page10

Exercise Page10, problem 1, p. 10
Show that the primitive cube roots of unity are $\omega$ and $\omega^{2}$.
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Exercise Page10, problem 2, p. 10
For $R$ given by $(7)$, prove that the primitive $n$th roots of unity are (i) for $n=6$, $R$, $R^5$; (ii) for $n=8$, $R$, $R^3$, $R^5$, $R^7$; (iii) for $n=12$, $R$, $R^5$, $R^7$, $R^{11}$.
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Exercise Page10, problem 3, p. 10
When $n$ is a prime, prove that any $n$th root of unity, other than $1$, is primitive.
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Exercise Page10, problem 4, p. 10
Let $R$ be a primitive $n$th root $(7)$ of unity, where $n$ is a product of two different primes $p$ and $q$. Show that $R, \dotsc, R^n$ are primitive with the exception of $R^p$, $R^{2p}, \dotsc, R^{qp}$, whose $q$th powers are unity, and $R^q$, $R^{2q}, \dotsc, R^{pq}$, whose $p$th powers are unity. These two sets of exceptions have only $R^{pq}$ in common. Hence there are exactly $pq - p - q + 1$ primitive $n$th roots of unity.
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Exercise Page10, problem 5, p. 10
Find the number of primitive $n$th roots of unity if $n$ is a square of a prime $p$.
Printed answer:
- $p(p-1)$.
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- other: not a kind the checker handles p*(p-1)
Exercise Page10, problem 6, p. 10
Extend Ex. 4 to the case in which $n$ is a product of three distinct primes.
Printed answer:
- $(p-1)(q-1)(r-1)$ if $n=pqr$.
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- other: not a kind the checker handles (p-1)*(q-1)*(r-1)
Exercise Page10, problem 7, p. 10
If $R$ is a primitive $15$th root $(7)$ of unity, verify that $R^3$, $R^6$, $R^9$, $R^{12}$ are the primitive fifth roots of unity, and $R^5$ and $R^{10}$ are the primitive cube roots of unity. Show that their eight products by pairs give all the primitive $15$th roots of unity.
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Exercise Page10, problem 8, p. 10
If $\rho$ is any primitive $n$th root of unity, prove that $\rho$, $\rho^2, \dots, \rho^n$ are distinct and give all the $n$th roots of unity. Of these show that $\rho^k$ is a primitive $n$th root of unity if and only if $k$ is relatively prime to $n$.
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Exercise Page10, problem 9, p. 10
Show that the six primitive $18$th roots of unity are the negatives of the primitive ninth roots of unity.
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- (none printed)
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