1. Chapter 14 Complex Numbers in Algebra
Impossible Numbers
Quadratic Equations
Cubic Equations
Attempt at Geometric Representation
Angle Division
The Fundamental Theorem of Algebra

2. Impossible numbers

3. Geometric representationa b
|z| ϕ -b

4. Roots of unity
n=5

5. 14.1 Quadratic equations
Although quadratic formula was known for centuries, quadratic equations did not lead to the discovery of complex numbers
Reason: quadratic equations appeared from practical needs (e.g. land measuring), so if there were no (positive, let alone complex) solutions, it was assumed the equation had no solutions
Another treatment of solutions of quadratic equations: graphically, if parabola does not intersect the x-axis, there is no solutions

6. 14.2 Cubic equations

7. Birth of Complex Numbers

8. Bombelli’s work

9. Bombelli’s work
Bombelli conjectured that −1 =2+𝑛 −1 and found n by cubing both sides

10. Bombelli’s work Thus 3 2+11 −1 =2+ −1 and 3 2−11 −1 =2− −1 ,
which leads to − −11 −1 =4, a solution of 𝑥3= 15𝑥 +4

11. 14.3 Wallis’s Attempt at Geometric Representation
Wallis: geometric representation of roots of 𝑥2+2𝑏𝑥+𝑐2, where 𝑏,𝑐≥0
Roots: 𝑥 =−𝑏± 𝑏2−𝑐2
Positioning roots on the x-axis we get two points P1 and P2

12. 𝑥 =−𝑏± 𝑏2−𝑐2 c c b b b c P1 P2 -b x x -b P1 P2 O O Real roots
𝑥 =−𝑏± 𝑏2−𝑐2 c c b b b c P1 P2 -b x x -b P1 P2 O O
Real roots
Complex roots

13. 14.5 Angle Division
Recall: Angle-division formula of de Moivre: where x = sin θ and y = sin nθ
Note: this formula involves complex numbers since -1 ≤ y ≤ 1

14. Modern form of de Moivre’s formula
de Moivre did not state it explicitly
The first connection of trigonometric functions and complex numbers was established by Johann Bernoulli (1702)

15. Partial fractions decomposition
which (using integration) may lead to the formula for tan -1z for complex z
This in turn would require definition of complex logarithms
This idea was developed further by Euler

16. Bernoulli’s formula relating tan(nθ) and tanθ
Let y=tan(nθ) and x=tanθ
Then nθ = tan-1 y = n tan -1 x
So dy/(1+y2) = ndx /(1+x2)
Using partial fraction decompositin and integration we get: ln 𝑦+𝑖 𝑦−𝑖 = ln 𝑥+𝑖 𝑥−𝑖 𝑛
Therefore (x-i)n(y+i) = (x+i)n(y-i)
The last formula allows to express y as a rational function of x, e.g if n=4 we get 𝑦= 4𝑥−4𝑥3 𝑥4−6𝑥2+1
Thus results about real number can be obtained using complex numbers

17. Cotes theorem about regular polygons
Roger Cotes (1682 –1716)
Newton-Cotes formula for numerical integration
ix = ln (cos x + i sin x) (similar to Euler’s formula eix = cos x + i sin x)
Cotes assigned point (a,b) to a+b√-1
He was also interested in integrating 1/(1-xn) and 1/(1+xn) using partial fractions

18. Theorem
Let A0, A1, …, An-1 be the vertices of a regular polygon on the unit circle centered at O. If P is a point on OA0 such that OP = x then 1-xn = PA0 PA1 … PAn-1

19. 1-xn = PA0 (PA1)2 … (PA(n-1)/2)2 , n odd
Due to symmetry we get:
1-xn = PA0 (PA1)2 … (PA(n-1)/2)2 , n odd
1-xn = PA0 (PA1)2 … (PAn/2-1)2 (PAn/2)2 , n even
This gives factorization of 1-xn into linear and quadratic factors, sine (PAk)2 = 1 – 2x cos (2𝜋k /n) +x2 A1 O A0
2𝜋/n P x
An-1
An-2

20. 14.6 The Fundamental Theorem of Algebra
The question of factorization of polynomials in connection with integration by partial fractions
Descartes: if z = a is a solution of p(z)=0 then p(z) = (z-a) q(z), where q is a polynomial of deg p - 1
The can be proved using “long division” of polynomial

21. The Fundamental Theorem of Algebra
Every polynomial p(z) (with real or complex coefficients) have at least one complex root
Using induction and Descartes observation, this is equivalent to the following: every polynomial factors completely into the product of linear factors with complex coefficients

22. Properties of complex conjugate
Recall: for 𝑧=𝑎+𝑖𝑏 we let 𝑧 =𝑎−𝑖𝑏
Note that 𝑧+ 𝑧 =2𝑎 and z 𝑧 =𝑎2+𝑏2 are real numbers
𝑧 =𝑧 if and only if z is real
It can also be shown that
𝑧+𝑤 = 𝑧 + 𝑤 and 𝑧𝑤 = 𝑧 𝑤

23. Polynomials with real coefficients
Let 𝑝 𝑧 =𝑎𝑛𝑧𝑛+ 𝑎 𝑛−1 𝑧 𝑛−1 +…+𝑎1𝑧+𝑎0 be a polynomial with real coefficients
Then for any z we have 𝑝 𝑧 = 𝑎𝑛𝑧𝑛+ 𝑎 𝑛−1 𝑧 𝑛−1 +…+𝑎1𝑧+𝑎0 = 𝑎𝑛 𝑧 𝑛+ 𝑎 𝑛−1 𝑧 𝑛−1 +…+ 𝑎1 𝑧 + 𝑎0 = 𝑎𝑛 𝑧 𝑛+ 𝑎 𝑛−1 𝑧 𝑛−1 +…+𝑎1 𝑧 +𝑎0=𝑝( 𝑧 )

24. Polynomials with real coefficients
Thus for any 𝑝 𝑧 with real coefficients we have 𝑝 𝑧 = 𝑝( 𝑧 )
Therefore if such polynomial has a complex root 𝑟, its conjugate 𝑟 is also a root of p: 𝑝 𝑟 = 𝑝 𝑟 = 0 =0
This and the Fundamental Theorem of Algebra imply that a polynomial with real coefficients can be completely factored into linear and irreducible quadratic factors, all with real coefficients

25. 14.7 The Proofs of d’Alembert and Gauss
The Fundamental Theorem of Algebra can be proved in many different ways
The first proofs were given by d’Alembert (1746) and Gauss (1799)
Both proofs are incomplete according to the modern standards
However, the proofs are correct after all necessary details are supplied

26. d’Alembert’s proof
Two ingredients: d’Alembert Lemma and asymptotic behavior of a polynomial
Lemma. If p(z) is non-constant polynomial and p(z0) ≠ 0 then any neighborhood of z0 contains a point z1 such that |p(z1)|<|p(z0)|
d’Alembert proved this lemma using fractional power series (rigorously developed by Puiseux in 1850)
Nevertheless, the lemma can be proved by more elementary methods
Another statement used in the proof is the extreme value theorem (rigorously proved by Weierstrass in 1874)

27. Gauss’ proof Uses asymptotic behavior of a polynomial
Goal: to show that p(z)=0 in some circle of large radius R (similar to d’Alembert’s proof)
Recall: Re(a+ib)=a, Im(a+ib)=b
Gauss looked at the curves Re[p(z)]=0 and Im [p(z)]=0
If there is z s.t. these two curves intersect, we are done, since in this case Re[p(z)]=Im[p(z)]=0 and so p(z)=0
For large |z|, they are close to Re(zn)=0 and Im(zn)=0 (assuming an=1)

28. Gauss’ proof
It can be shown that the curves Re[p(z)]=0 and Im[p(z)]=0 are algebraic
Also, the curves Re[zn]=0 and Im[zn]=0 are collections of straight lines that intersect any circle centered at the origin alternately
Therefore, by continuity, the curves Re[p(z)]=0 and Im[p(z)]=0 intersect a circle (of a sufficiently large radius) alternately
Gauss claimed that it obviously implies that the curves intersect somewhere inside the circle
However, the rigorous proof of this fact is hard and non-elementary (the first proof is due to Ostrowski (1920))