MatIntro week 1: sets, polynomial division and complex numbers"
Solved exercises from week 1 of MatIntro at KU — sets, inequalities, polynomial division, complex numbers, polar form and roots.
This post was AI-generated from handwritten notes in a notebook, digitised with a document scanner.
Course: MatIntro 2026, University of Copenhagen. Textbook: Tom Lindstrøm, Kalkulus. "TLO 3.4.1" means section 3.4, exercise 1.
Sets, inequalities and absolute value
TLO 2.1.1
· Expression · Result a) · · b) · · c) · · d) · · e) · · f) · ·
Pitfall: in d) the answer is the set , not the interval . In f) the endpoint stays, because does not contain it.
TLO 2.1.2
, , , .
TLO 2.1.3
· Set · Interval a) · · b) · · c) · · d) · · e) · · f) · ·
Pitfall: d) is not — a large negative number squared is large and positive.
TLO 2.1.5a
. Both sides are non-negative, so square them:
(x−2)² < (x+3)²
x² − 4x + 4 < x² + 6x + 9
−10x < 5
x > −1/2
TLO 2.1.5b
splits into two cases.
case 1: x² − 2x − 8 > 8 ⟹ x² − 2x − 16 > 0 ⟹ roots 1 ± √17
case 2: x² − 2x − 8 < −8 ⟹ x² − 2x < 0 ⟹ roots 0 and 2
Result: or or .
TLO 2.1.9
Show . Insert : , then apply the triangle inequality with , .
TLO 2.1.12
Show for . Start from something true:
Rational and irrational numbers
TLO 2.2.1
· Expression · Result · Method a) · · · b) · · · c) · · · multiply by d) · · · multiply by the conjugate e) · · · multiply by f) · · ·
TLO 2.2.3
a) — rational. b) — irrational (rational minus irrational). c) — rational.
Polynomial division
TLO 1.5.1a
gives quotient and remainder :
TLO 1.5.1b
gives quotient and remainder .
TLO 1.5.3a
. Check , divide by to get , then solve the quadratic:
Arithmetic with complex numbers
TLO 3.1.1
· Expression · Result a) · · b) · · d) · · f) · · h) · ·
Division: multiply numerator and denominator by the conjugate of the denominator.
TLO 3.1.2
f) . h) .
TLO 3.1.5c
Solve .
z − 2 = 3i(z + 1) = 3iz + 3i
z − 3iz = 2 + 3i
z(1 − 3i) = 2 + 3i
z = (2 + 3i)(1 + 3i) / 10 = (−7 + 9i)/10
Result: .
Pitfall: read the fraction the right way up. Solving instead gives — a correct answer to a different equation.
The complex plane
TLO 3.2.1 and 3.2.2b
, , , are the four corners of a square around the origin. lies in the fourth quadrant with modulus .
TLO 3.2.10
· Set · Description a) · · unit circle b) · · open disc, centre , radius c) · · everything outside the disc with centre , radius
is the distance between and .
TLO 3.2.3 — Cartesian to polar
· · · · Polar form a) · · · · b) · · · · c) · · · · d) · · · ·
Method: , , , and read the quadrant from the signs of and .
TLO 3.2.4
a) . b) : , , , second quadrant, , so .
Pitfall: lands in the wrong quadrant. When , add .
TLO 3.2.5 and 3.2.6 — polar to Cartesian
. . .
: reduce the angle first, , giving .
TLO 3.2.7c
, . Moduli multiply, arguments add: . Reduce: .
TLO 3.2.8
With :
Point · Value · Geometry · · twice as long · · reflected through the origin · · half as long · · turned anticlockwise · · turned clockwise · · modulus squared, argument doubled
TLO 3.2.9
, : , , . Addition is coordinatewise — the parallelogram rule.
Complex exponential and de Moivre
TLO 3.3.3b
: , fourth quadrant, .
TLO 3.3.1
a) . c) : reduce , giving .
TLO 3.3.2
Rule: .
a) . b) has modulus and argument (third quadrant): .
TLO 3.3.5
· , · · a) · , · · b) · , · ·
TLO 3.3.8
. Polar form , de Moivre gives . Since :
Plot of for
, so . Each step multiplies the length by and adds — a logarithmic spiral, not a circle.
Step rule without polar form: , so
Eight steps is one full turn and a factor , so all lie on the positive real axis at distances , , .
The points (1+i)^n for n from -8 to 8 spiral outwards
Pitfall: treating as if its imaginary part were gives , , and seventeen copies of the point .
Roots of complex numbers
Recipe: write . The first root is ; the others follow by repeatedly multiplying by the rotation factor . The roots lie evenly spaced on a circle of radius .
Pitfall: the rotation factor is added to 's argument , not to the original .
TLO 3.4.1a
Square roots of : , .
TLO 3.4.1c
Square roots of : , .
TLO 3.4.3a
Cube roots of : , , .
TLO 3.4.3c
Cube roots of . First root . The next angle, , is not a standard angle, so write the rotation factor in Cartesian form, , and multiply:
Pitfall: both terms of the real part of keep their minus sign: , not .
TLO 3.4.9a
. Discriminant , and the coefficients are real, so by Theorem 3.4.8 (iii):
Fundamental theorem of algebra
TLO 3.5.1a
Complex and real factorisation of . Roots of : , , .
A conjugate pair multiplies to the real quadratic .
Pitfall: the real factorisation needs the linear factor too. Degree check: .
TLO 3.5.1c
. Roots of : , , i.e. arguments . The root with argument is the real root .
Conjugate pairs have arguments summing to . On the unit circle each pair gives :
Pitfall: the argument of is , not — points at .
Pitfall: with no in it; it is that equals . And means — write .
Exam practice — MC13-1A-A
A. .
B. : roots , , . Faster: .
Pitfall: the real root is not among the options, and is. But . The correct option is .
C. : discriminant , so .
Deeper exercises
TLO 3.2.16
, . Show is purely imaginary.
On the unit circle , so . Then
and forces .
Geometric consequence: and are the vectors from the ends of the diameter to . Their quotient is purely imaginary, so they are perpendicular — Thales' theorem.
The inscribed angle at z over the diameter from -1 to 1 is a right angle
Pitfall: . The minus goes on the numerator or the denominator, never both.
Pitfall: algebra only excludes . Geometrically must be excluded too — there is no triangle when sits on an endpoint.
TLO 3.2.18
, real. The denominator is the conjugate of the numerator, so with .
a) , so lies on the unit circle.
b) Let , so . Division subtracts arguments: . Hence .
1 + ti lies on the tangent line x = 1, so its height is tan φ
Pitfall: it is , not .
Zero-product rule in
If then or .
Geometric: . These are real, so one factor is , and means .
Algebraic: if , then .
The geometric proof borrows the rule from . The algebraic proof uses only the field axioms, so it holds in any field.