Contents / Algebra / Complex Numbers & Sequences

Chapter 8

Complex Numbers & Sequences

Complex numbers, the imaginary unit, and arithmetic and geometric sequences.

Introduction

Complex numbers, the imaginary unit, and arithmetic and geometric sequences.

A complex number is a number of the form a + b i, where a and b are real numbers and i is the imaginary unit satisfying i² = -1.

Complex numbers extend the real numbers so that every polynomial equation has solutions (Fundamental Theorem of Algebra). They are essential in electrical engineering, quantum mechanics, signal processing, and higher mathematics. We can add, subtract, multiply, and divide them just like reals, with a few new rules.

8.1Definition and the Imaginary Unit

Definition 8.1 (Complex number). The imaginary unit $i$ is defined by

i^{2} = - 1,

and a complex number is any $z = a + bi$ with $a$ and $b$ real. Here $a$ is the real part $Re (z)$ and $b$ is the imaginary part $Im (z)$ .

If $b = 0$ then $z$ is real; if $a = 0$ and $b \neq = 0$ then $z$ is pure imaginary.

The complex numbers can be visualized as the plane with real axis horizontal, imaginary vertical.

Intuition. i is a new "number" we invent so that negatives have square roots: √(-1)=i.

A complex number is like a pair (real, imaginary) that we treat as a single entity with its own arithmetic.

Example 8.2. $3 + 4 i$ has real part $Re (z) = 3$ , imaginary part $Im (z) = 4$ . (Note: the imaginary part is the real number $4$ , not $4 i$ .)

Solution.

$- 2 i$ is $0 - 2 i$ , so $Re = 0$ and $Im = - 2$ : pure imaginary.
$7$ is $7 + 0 i$ , so $Im = 0$ : real. This shows the real numbers sit inside the complex numbers as the case $b = 0$ .

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Exercises for §8.1 →

8.2Arithmetic Operations

Proposition 8.3 (Addition and multiplication).

(a + bi) + (c + d i) = (a + c) + (b + d) i,

(a + bi) (c + d i) = a c + a d i + b c i + b d i^{2} = (a c - b d) + (a d + b c) i,

the last step using $i^{2} = - 1$ . Subtraction works like addition.

To divide, multiply numerator and denominator by the conjugate of the denominator, then simplify.

Intuition. Add and subtract like combining like terms (reals with reals, i's with i's).

Multiplication uses the distributive law and replaces i² with -1.

Example 8.4.

(1 + 2 i) (3 - i) = 1 \cdot 3 + 1 \cdot (- i) + 2 i \cdot 3 + 2 i \cdot (- i) = 3 - i + 6 i - 2 i^{2} = 3 + 5 i + 2 = 5 + 5 i

Example 8.5. Compute $(2 + i) / (1 - i)$ . Multiply top and bottom by the conjugate of the denominator, $1 + i$ :

Solution.

$\frac{2 + i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{( 2 + i ) ( 1 + i )}{( 1 - i ) ( 1 + i )}$ .
Denominator: $(1 - i) (1 + i) = 1^{2} - i^{2} = 1 - (- 1) = 2$ (a real number — that is the whole point of the conjugate).
Numerator: $(2 + i) (1 + i) = 2 + 2 i + i + i^{2} = 2 + 3 i - 1 = 1 + 3 i$ .
So the quotient is $\frac{1 + 3 i}{2} = \frac{1}{2} + \frac{3}{2} i$ .
Sanity check: multiply back by $1 - i$ : $(\frac{1}{2} + \frac{3}{2} i) (1 - i) = \frac{1}{2} - \frac{1}{2} i + \frac{3}{2} i - \frac{3}{2} i^{2} = \frac{1}{2} + \frac{3}{2} + i = 2 + i$ . ✓

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Exercises for §8.2 →

8.3Complex Conjugates and Magnitude

Definition 8.6 (Conjugate and modulus). The conjugate of $z = a + bi$ is $\overset{z}{ˉ} = a - bi$ , and its modulus (or magnitude) is

∣ z ∣ = a^{2} + b^{2} .

Proposition 8.7 (Properties of the conjugate and modulus).

z \overset{z}{ˉ} = ∣ z ∣^{2} (real and non-negative), z + \overset{z}{ˉ} = 2 Re (z),

∣ z w ∣ = ∣ z ∣ ∣ w ∣, \frac{z}{w} = \frac{∣ z ∣}{∣ w ∣} (w \neq = 0) .

Geometrically, |z| is the distance from origin in the complex plane.

Intuition. Conjugate is the "mirror image" across the real axis.

Modulus is how "far from zero" the number is, generalizing absolute value.

Example 8.8. For $z = 3 - 4 i$ , $\overset{z}{ˉ} = 3 + 4 i$ , $∣ z ∣ = 5$ , $z \overset{z}{ˉ} = 9 + 16 = 25 = 5^{2}$ .

Example 8.9. Compute $∣ (1 + i) (2 - i) ∣$ using $∣ z w ∣ = ∣ z ∣∣ w ∣$ : $∣1 + i ∣ = 1^{2} + 1^{2} = 2$ and $∣2 - i ∣ = 2^{2} + (- 1)^{2} = 5$ , so $∣ (1 + i) (2 - i) ∣ = 2 \cdot 5 = 10$ .

Solution.

Sanity check by multiplying first: $(1 + i) (2 - i) = 2 - i + 2 i - i^{2} = 2 + i + 1 = 3 + i$ , and $∣3 + i ∣ = 9 + 1 = 10$ . ✓

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Exercises for §8.3 →

8.4Solving Equations with Complex Numbers

Theorem 8.10 (Complex roots come in conjugate pairs). A quadratic with real coefficients and negative discriminant has two complex roots, and they are conjugates of one another.

Higher degree polynomials factor completely into linear factors over the complex numbers.

To solve, use quadratic formula (it works for complex), or factor using complex numbers.

Intuition. When the "discriminant under the square root" goes negative, the answers are no longer on the real number line; they have an imaginary component.

The math still works; we just accept i as part of the answer.

Example 8.11.

x^{2} + 4 = 0 \to x^{2} = - 4 \to x = \pm - 4 = \pm 2 i

Solution.

Sanity check: $(2 i)^{2} + 4 = 4 i^{2} + 4 = - 4 + 4 = 0$ . ✓ The two roots $2 i$ and $- 2 i$ are a conjugate pair, as expected for a real-coefficient quadratic.

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Example 8.12. Solve $x^{2} - 2 x + 5 = 0$ .

Solution.

$x = [2 \pm 4 - 20] /2 = [2 \pm - 16] /2 = [2 \pm 4 i] /2 = 1 \pm 2 i$ .
Check: $(1 + 2 i)^{2} - 2 (1 + 2 i) + 5 = 1 + 4 i + 4 i^{2} - 2 - 4 i + 5 = (1 - 2 + 5) + (4 i - 4 i) + 4 (- 1) = 4 - 4 = 0$ . Yes.

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Exercises for §8.4 →

Common Mistakes

Writing i² = +1 or forgetting to replace i² with -1 in multiplication: Leads to wrong real/imag parts.
Treating i as a variable: i is a constant; you cannot "solve for i".
Forgetting the conjugate when dividing: Denominator must be rationalized to a real number.
Assuming all complex solutions are pure imaginary: Most have both real and imag parts.
Sign error on the imaginary part when finding conjugates or when distributing minus.
Computing modulus without the square root: |z| is the sqrt, not a² + b².
Forgetting that for real-coeff polynomials, complex roots come in conjugate pairs: If 1+2i is a root, so is 1-2i.

The imaginary unit i is the number defined by the equation i² = -1.

It is not a "real" number in the usual sense, but a new mathematical object that we adjoin to the reals to form the complex numbers. With i we can take square roots of negative numbers and solve every polynomial equation.

Although called "imaginary," i has profound real-world applications in physics, engineering, and mathematics.

8.5Definition and Powers of i

Recall that $i$ is defined so that $i \cdot i = - 1$ .

Proposition 8.13 (Powers of $i$ cycle with period four).

i^{1} = i, i^{2} = - 1, i^{3} = - i, i^{4} = 1, i^{5} = i, \dots

Proof. The pattern repeats because $i^{4} = (i^{2})^{2} = (- 1)^{2} = 1$ , so multiplying by $i^{4}$ changes nothing.∎

To compute i^k for large k, reduce the exponent modulo 4.

Intuition. i is a "90-degree turn" button in number land. Pressing it twice gets you to -1.

The cycle of 4 comes from going around the circle of directions: up, left, down, right, repeat.

Example 8.14.

i^{7} = i^{4 + 3} = i^{4} \cdot i^{3} = 1 \cdot (- i) = - i

Example 8.15. $i^{100}$ : $100 mod 4 = 0$ , so $i^{100} = (i^{4})^{25} = 1^{25} = 1$ .

Exercises for §8.5 →

8.6Geometric Interpretation

In the complex plane, the real numbers lie on the horizontal axis and pure imaginaries on the vertical axis.

Proposition 8.16 (Multiplication by $i$ is a quarter turn). Multiplying any complex number $z$ by $i$ rotates $z$ counterclockwise by $9 0^{\circ}$ around the origin, without changing its distance from zero.

Multiplying by i again rotates another 90° (total 180°, which multiplies by -1), and so on.

Intuition. i itself sits at (0,1) — one unit straight up.

Multiplying by i is like turning the arrow representing the number 90° left (counterclockwise).

Example 8.17. Start with $1$ (point $(1, 0)$ ). $\cdot i \to i$ (point $(0, 1)$ ). $\cdot i \to - 1$ (point $(- 1, 0)$ ). $\cdot i \to - i$ (point $(0, - 1)$ ). $\cdot i \to 1$ again.

Example 8.18. Take $2 + 3 i$ (arrow to $(2, 3)$ ). Multiply by $i$ : $i (2 + 3 i) = 2 i + 3 i^{2} = - 3 + 2 i$ (arrow to $(- 3, 2)$ — rotated $9 0^{\circ}$ CCW).

Exercises for §8.6 →

8.7Role in the Complex Numbers

Every complex number can be written using real coefficients times 1 and i: a + b i.

The introduction of i lets us solve x^2 + 1 = 0 (roots ±i) and, more generally, makes the complex numbers algebraically closed.

It also appears in Euler's formula,

e^{i θ} = cos θ + i sin θ,

which connects exponentials, trigonometry, and complex numbers.

Intuition. With i we "complete" the number system so nothing is missing when we try to solve equations.

It is the secret ingredient that makes many formulas in science and engineering work beautifully.

Example 8.19. Solve $x^{2} + 9 = 0$ : $x^{2} = - 9$ , so $x = \pm 3 i$ .

Solution.

Sanity check: $(3 i)^{2} + 9 = 9 i^{2} + 9 = - 9 + 9 = 0$ . ✓

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Example 8.20. Simplify $- 9$ using $i$ . Split off the negative sign first: $- 9 = 9 \cdot - 1 = 3 i$ (the principal root).

Solution.

Note the caution: the rule $a b = ab$ only holds for non-negative $a, b$ . So we must factor out $- 1 = i$ before multiplying; writing $- 9 = - 1 \cdot 9$ is fine, but $- 1 \cdot - 1 \neq = (- 1) (- 1) = 1$ — it equals $i^{2} = - 1$ .

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Exercises for §8.7 →

Common Mistakes

Thinking i is a variable: It is a fixed constant, like π or e.
Computing i^2 as +1: The whole point is that it equals -1.
Forgetting the cycle when reducing large powers: Always mod 4.
Writing √(-4) = ± 2i without specifying principal: The principal square root is the one with non-negative real part or the convention 2i.
Assuming all powers of i are imaginary: i^4=1 is real, i^2=-1 is real.
Confusing i with the engineering j: Same thing, different letter in some fields.
Trying to order complex numbers on the line: They live in a plane; there is no total order compatible with the arithmetic like on the reals.

An arithmetic sequence (also called arithmetic progression) is a list of numbers in which each term after the first is obtained by adding a fixed constant d (the common difference) to the preceding term.

Examples: 2, 5, 8, 11, ... (d=3); 10, 7, 4, 1, ... (d=-3); the odd numbers 1, 3, 5, 7, ...

Arithmetic sequences model any situation with constant additive change: equal deposits, constant speed (distance each second), evenly spaced fence posts, etc. Their partial sums are arithmetic series.

8.8Definition and General Term

Definition 8.21 (Arithmetic sequence). An arithmetic sequence is determined by its first term $a$ (also written $a_{1}$ ) and its common difference $d$ : each term after the first is the previous term plus $d$ .

Theorem 8.22 (nth term of an arithmetic sequence).

a_{n} = a + (n - 1) d .

This is a linear function of n: the terms lie on an arithmetic (straight-line) progression when plotted against their index.

Intuition. Each step you add exactly the same amount d, like climbing stairs of equal height.

To find the 10th step you don't count one by one; you do first + 9 * d.

Example 8.23. Sequence: 4, 7, 10, 13, ... a=4, d=3.

Solution.

a_5 = 4 + 4*3 = 16.
a_20 = 4 + 19*3 = 61.

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Example 8.24. If a_1 = 100 and a_6 = 70, then d = (70-100)/5 = -6.

Solution.

General term: 100 + (n-1)(-6) = 106 - 6n.

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Exercises for §8.8 →

8.9Sum of a Finite Arithmetic Series

Theorem 8.25 (Sum of a finite arithmetic series). The sum of the first $n$ terms of an arithmetic series is

S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} (a_{1} + a_{n}),

that is, $\frac{n}{2}$ times the sum of the first and last terms.

Proof. Write the sum forward and backward and add the two identical sums: each of the $n$ pairs sums to (first $+$ last). Then divide by $2$ .∎

This formula lets us add long lists without adding term by term.

Intuition. Pair the first and last (they add to the same constant $a_{1} + a_{n}$ ), second and second-last, etc. Writing the sum forwards and backwards gives $n$ copies of that pair-sum, so $2 S_{n} = n (a_{1} + a_{n})$ and thus $S_{n} = \frac{n}{2} (a_{1} + a_{n})$ .

The " $n /2$ pairs" picture is exact when $n$ is even; the forwards-and-backwards trick above works for any $n$ , even or odd, so you never have to worry about a leftover middle term.

Example 8.26. Sum of first 5 terms of 3, 5, 7, 9, 11.

Solution.

First+last=3+11=14, n=5, S=5/2 *14=35.
(Or 5/2 * [23 +42]=5/2*(6+8)=35.)

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Example 8.27. Sum of integers from 1 to 100: S=100/2 * (1+100)= 50*101=5050. (Gauss's famous shortcut.)

Example 8.28. (working backwards to find $d$ )

Solution. A 15-term arithmetic series sums to 270, and its first term is $a = 4$ . Find the common difference $d$ .

Use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ with $n = 15$ , $a = 4$ , $S_{15} = 270$ : $\frac{15}{2} [2 (4) + 14 d] = 270$ . Multiply both sides by $\frac{2}{15}$ : $8 + 14 d = 36$ . So $14 d = 28$ , giving $d = 2$ . Sanity check: the last term is $a_{15} = 4 + 14 (2) = 32$ , and $S_{15} = \frac{15}{2} (4 + 32) = \frac{15}{2} (36) = 15 \cdot 18 = 270$ . ✓□

Exercises for §8.9 →

8.10Applications and Word Problems

Arithmetic sequences appear whenever a quantity increases or decreases by a constant amount per time period or per step:

Savings: fixed amount deposited each month.
Linear motion at constant velocity (position each second).
Seats in a theater or bleachers with constant added per row.
Number of handshakes or blocks in a pattern that grows linearly.

Method 8.29 (Solving an arithmetic word problem).

Identify $a$ and $d$ from the description.
Decide whether you need a particular term or the sum.
Use the corresponding formula.
Interpret the answer in context.

Intuition. "Constant step size" is the signature of arithmetic.

The formulas turn "how many steps" and "size of each step" into "where you are" or "total accumulated."

Example 8.30. A worker earns $50 on day 1 and gets a $3 raise each day. How much total after 30 days?

Solution.

a=50, d=3, n=30. Last day pay=50+29*3=137.
Total S_30 =30/2 * (50+137)=15*187=2805.

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Example 8.31. An arithmetic sequence of temperatures: starts at 20°C and drops 1.5° each hour. What is the temperature after 8 hours?

Solution.

a_9 =20 +8*(-1.5)=20-12=8°C.

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Exercises for §8.10 →

Common Mistakes

Using n instead of (n-1) in the general term: $a_{n} = a + n d$ would be off by one (gives the $(n + 1)$ th term).
Forgetting the /2 in the sum formula: Easy to double the sum.
Confusing sequence (the list of terms) with series (the sum of terms): The problem may ask for the 10th term or the sum of first 10.
Wrong sign for d: If the sequence is decreasing, d is negative; the formulas still work.
Applying the arithmetic sum formula to a geometric sequence: Different pattern (common ratio instead of difference).
Not identifying the first term correctly: Sometimes the problem gives the second term or a later one; solve for a first.
Units or context errors: The answer 2805 dollars makes sense; 2805 days probably does not.

A geometric sequence (geometric progression) is a list in which each term after the first is obtained by multiplying the preceding term by a fixed constant r called the common ratio.

Examples: 2, 6, 18, 54, ... (r=3); 100, 50, 25, 12.5, ... (r=1/2); 3, -6, 12, -24, ... (r=-2).

Geometric sequences model multiplicative growth or decay: compound interest, population growth, bouncing balls (heights), half-life decay, and many fractal or scaling patterns. Their partial sums are geometric series; when |r|<1 the infinite sum converges to a finite value.

8.11Definition and General Term

Definition 8.32 (Geometric sequence). A geometric sequence is determined by its first term $a$ and its common ratio $r$ : each term after the first is the previous term multiplied by $r$ .

Theorem 8.33 (nth term of a geometric sequence).

a_{n} = a \cdot r^{n - 1} .

This is an exponential function of the index $n$ .

If r > 1 the terms grow without bound (if a ≠ 0).

If 0 < r < 1 the terms shrink toward zero.

If r < 0 the signs alternate.

Intuition. Each step you multiply by the same "growth factor" r, like a photocopier set to enlarge or shrink by a fixed percentage.

The term number tells how many multiplications by r have happened since the start.

Example 8.34. 3, 6, 12, 24, ... a=3, r=2.

Solution.

a_6 = 3 * 2^5 = 3*32=96.

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Example 8.35. A quantity starts at 1000 and is multiplied by 0.9 each day.

Solution. The "start" value 1000 is the first term $a_{1}$ (day 1, $n = 1$ ). The value on day 10 is the 10th term $a_{10}$ , so the exponent is $n - 1 = 9$ : $a_{10} = 1000 \cdot (0.9)^{9}$ . Numerically, $(0.9)^{9} \approx 0.3874$ , so $a_{10} \approx 387.4$ . Why $n - 1$ and not $n$ ? On day 1 no multiplication has happened yet ( $0. 9^{0} = 1$ ); each later day adds one factor of $0.9$ . By day 10 exactly nine multiplications have occurred.□

Exercises for §8.11 →

8.12Sum of a Finite Geometric Series

Theorem 8.36 (Sum of a finite geometric series). For $r \neq = 1$ , the sum of the first $n$ terms of a geometric series is

S_{n} = a \frac{r ^{n} - 1}{r - 1} = a \frac{1 - r ^{n}}{1 - r} .

Proof. Write $S_{n}$ , multiply it by $r$ , subtract the two equations — almost everything cancels — and solve for $S_{n}$ .∎

When r=1 the sum is simply n a (all terms equal a).

Intuition. The trick of multiplying the sum by r shifts every term; subtracting cancels most terms, leaving only the first and the "next after last".

Dividing by (r-1) "normalizes" to give the closed form.

Example 8.37. Sum of first 5 terms of 4, 12, 36, ... (a=4, r=3).

Solution.

S_5 =4*(3^5 -1)/(3-1) =4*(243-1)/2 =4*121=484.
Sanity check by adding directly: the five terms are 4, 12, 36, 108, 324, and 4+12+36+108+324 = 484. ✓

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Example 8.38. Sum of geometric series 1 + 1/2 + 1/4 + ... + 1/2^{9} (10 terms, a=1 r=1/2).

Solution.

S= (1 - (1/2)^{10} ) / (1 - 1/2) = (1 - 1/1024)/(1/2) = 2*(1023/1024) ≈ 1.998 .

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Exercises for §8.12 →

8.13Infinite Series and Applications

Theorem 8.39 (Sum of an infinite geometric series). When $∣ r ∣ < 1$ , the infinite geometric series sums to the finite value

S_{\infty} = \frac{a}{1 - r} .

Proof. This is the limit of the finite sum formula as $n \to \infty$ , since $r^{n} \to 0$ when $∣ r ∣ < 1$ .∎

This is the foundation for many repeating decimal representations and for the mathematics of fractals, probability (infinite trials), and physics (infinite reflections, damping).

Intuition. If each step you keep only a fraction r <1 of the previous, the total "stuff" you ever get is finite even if you continue forever.

The formula a/(1-r) is the "grand total" after all the infinite shrinking steps.

Example 8.40. 0.3333... = 3/10 + 3/100 + 3/1000 + ... = (0.3) / (1 - 0.1) = 0.3 / 0.9 = 1/3.

Example 8.41. A ball dropped from 10 m bounces to 60% of previous height each time. Total distance traveled (down + up + down + ...):

Solution.

First drop 10 m (this leg happens only once, so keep it separate).
After that first drop, every bounce is traveled twice — once up, once back down. The peak heights are $6, 6 \cdot 0.6, 6 \cdot 0. 6^{2}, \dots$ (a geometric sequence with $a = 6$ , $r = 0.6$ , and $∣ r ∣ < 1$ so the infinite sum converges).
Then $2 \cdot (6 + 6 \cdot 0.6 + 6 \cdot 0. 6^{2} + \dots) = 2 \cdot \frac{6}{1 - 0.6} = \frac{12}{0.4} = 30$ m.
Grand total $10 + 30 = 40$ m.

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Example 8.42. (converting a repeating decimal, general recipe)

Solution. Write $0. \overline{27} = 0.272727 \dots$ as a fraction.

It is the geometric series $0.27 + 0.27 (0.01) + 0.27 (0.01)^{2} + \dots$ with $a = 0.27$ and $r = 0.01$ . Since $∣ r ∣ = 0.01 < 1$ , use $S_{\infty} = \frac{a}{1 - r} = \frac{0.27}{1 - 0.01} = \frac{0.27}{0.99} = \frac{27}{99} = \frac{3}{11}$ . Sanity check: $3 \div 11 = 0.2727 \dots$ ✓□

Exercises for §8.13 →

Common Mistakes

Using the arithmetic sum formula on a geometric series (or vice versa): Different closed forms; check whether the problem adds or multiplies each step.
Forgetting the exponent is n-1: a r^n would be the term after the nth.
Applying the infinite sum formula when |r| ≥ 1: The series diverges (grows without bound or oscillates).
Sign errors with negative r: The sum formula still holds, but terms alternate; the infinite sum may still converge if |r|<1.
Confusing the first term a with S_1: S_1 = a, but the formula a(r^n-1)/(r-1) for n=1 gives a correctly only after algebra.
Writing the sum formula with r in the numerator instead of r^n: a(r-1)/(r^n-1) is backwards.
Forgetting to multiply by 2 for up-and-down bounces in ball problems: The infinite geometric after the first drop is usually doubled.
Domain for r=1 case: Must handle separately; the general formula has division by zero.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter