Complex Numbers

Two pairs of polar coordinates (r1, θ1) and (r2, θ2) are the Cartesian coordinates of the same point if and only if:

1. r1 = r2

2. θ1 - θ2 is an integer multiple of 2π.

A generic complex number in trigonometric form is the number written in polar coordinates:

z = x + iy = r(cos θ + i sin θ)

Complex exponential:

For each z ∈ C, we define:

e^z = e^(x+iy) = e^x(cos(y) + i sin(y)) = e^x cos(y) + i e^x sin(y).

For z ∈ R, z = x + i0, e^z = e^(x+0i) = e^x(cos(0) + i sin(0)) = e^x.

Euler's formula:

If the imaginary number of which we construct the exponential is purely imaginary, then:

e^(iθ) = cos θ + i sin θ, for all θ ∈ R.

Properties of complex exponentials:

1. e^z + e^w = e^(z+w)

2. -z = -1 * z, so e^(-z) = 1/e^z

3. From here it follows that the complex exponential never

vanishesz Rez|e | ∀z ∈= e , C,4. z+i2nπ z∀z ∈ ∀n ∈C, Ze = ethe complex exponential is said to be periodic of ni \pi, a property that the real exponential does not have.⇒ 6. is an integer multiple ofz z ⇐⇒ −e = e z z 2iπ1 2 1 2 33.5 Exponential form of a complex numberThe exponential writing of a complex number is a more compact way than the trigonometric form.Using Euler’s formula a complex number written as in exponential formiθ · ·e = cos θ + i sen θ z = r(cos θ + i sen θ)is written . This form is particularly convenient for operations between complexes because it exploitsiθ·z = r andoperations between powers.3.6 De Moivre’s formulaThis relation is used in the case of exponentiation of a complex number written in exponential form:nix inxe = eC4 Triangular inequality inTheorem: |z | ≤ |z | |z | ∀z ∈+ z + , z C1 2 1 2 1 2Demonstration: 2

1. · · · ·|z | + z ) = (z + z ) (z + z ) = z z + z z + z z + z z =+ z = (z + z ) (z1 2 1 2 1 2 1 1 1 2 2 1 2 21 2 1 2 2 2 2 22 2 |z | |z | · ≤ |z | |z | |z · ||z | |z | z + z z = + + 2 Re (z z ) + + 2 z == + + z 2 1 2 1 2 1 2 1 2 1 21 2 12 2 2 2 2|z | |z | |z | |z | |z | |z | |z | |z | | |z |)= + + 2 = + + 2 = (|z +1 2 1 2 1 2 1 2 1 22 2⇒ |z | ≤ | |z |)+ z (|z +1 2 1 2or |z | ≤ |z | |z |+ z +1 2 1 25 Roots of a complex numberDati e , I am looking for where is of the form (con e ),∗ n iϕ∈ ̸ ∈ ∈ ∈a C, a = 0 n N z C : z = a, a a = re r > 0 ϕ Rso I look for solutions in exponential form of the form iθ ∈z z = ρe (con ρ > 0, θ R).n n inθ iϕ n⇐⇒ · ⇐⇒ −z = a ρ e = r e ρ = r, n θ ϕ = 2kπ2πϕ1⇐⇒ ∃k ∈ρ = r + k, Z : θ =n n nOf distinct solutions are found those ranging from a −k

= 0 k = n 1.5.1 Observations on the nth roots1

11. They all have the same form , therefore they are all on the same circumference centered in of radius .r O rn n

2. They are spaced apart by , so the solutions represent the vertices of a sided polygon.2π nn6

Exemple

We calculate the fifth roots of −32 iπ−32 ·= 32 e r = 32 ϕ = π n = 5π π πi = 2 cosz = 2e + i sen50 5 5( ) ( )π 2π 3πi + iz = 2e = 2e5 5 51 ( )π 4πi + i −2z = 2e = 2e =5 52 ( )π 6π 7πi + i= 2ez = 2e 5 5 53 ( )π 8π 9πi + iz = 2e = 2e5 5 51 4◦27 degree equationsSecond degree equations are always solvable to since it is not an ordered field.C, Moltiplico per2 ∈ ̸az + bz + c = 0 con a, b, c C, a = 0 4aI add and subtract b leading to the second member2 24a2z + 4abz + 4ac = 0 ; b2 2 2 2 −4a = bz + 4abz + b 4ac w = 2az + b| {z }2=(2az+b) 2 2⇒ −(2az + b) = b 4ac2 2 −w = b 4achas two complex roots which are2w w , w1 2 −b +

For I have two solutions so I will also have two inw z:

1. −b+w 12a In the -th root of a strictly positive real number is the only strictly positive real
2. −b+w R nObservation:22anumber such that nx = aIn all complex numbers have roots of any order, moreover, since there is no order relation, it is not possible toC,privilege one root over another.In we speak of the -th root of a real number, while in we speak of the -th roots of a complex number.R n C n

8 Fundamental theorem of algebra

Let be a polynomial with complex coefficients of degree . n n−1P(z) P(z) ·zn, = a z +a +. . .+a z+a

Theorem: n n−1 1 0con ∈ ̸ ≥a C, a = 0, n 1.i n

So they exist such that the polynomial can be written asP(z)z , z , . . . , z1 2 n nYP(z) −= a (z z )n kk=1

In particular are the solutions of the equation P(z)z , z , . . . , z = 0.1 2 n

That is, this theorem tells us that can be factored using the formula for certain complex numbers which areP(z)also the

solutions of the equation P(z) = 0.

Observation:

• Not all are necessarily disjoint zk
• The number of times a factor is repeated in the factorization is called multiplicity. −z α8.

1 Reformulation of the fundamental theorem of algebra

Every algebraic equation of degree in has solutions in order to count each of its solutions ≥n 1 C, n

Theorem:

with its multiplicity.

Relation between factoring and reformulation solutions

Observation:

The that make up in the factorization of the polynomial are all and only the solutions of the corresponding zk equation. The reason is to be found in Ruffini’s theorem which says that a polynomial is divisible by P(z) = 0 if and only if is the solution of the polynomial (it also holds in complexes). −z z z0 0

Furthermore, the factoring is unique, up to the order of the factors.

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