Download 1 Complex numbers and the complex plane

MM Vercelli.
L1: Complex numbers and complex-valued functions.
Contents:
• The field of complex numbers.
• Real and imaginary part. Conjugation and modulus or absolute valued.
• Inequalities: The triangular and the Cauchy.
• Polar representation: the argument. The exponential notation and the De Moivre
theorem.
• Circles and lines by using complex numbers.
• Complex valued functions and its differential.
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Complex numbers and the complex plane
It is completely obvious that the equation x2 + 1 = 0 cannot be solved using the real
numbers.
Mathematicians have therefore invented the imaginary number ”i” 1 . In other words one
declares i2 = −1, and naturally also (− i)2 = −1.
Once ”i” is defined as a number, we must make sure we can compute 2 i, 4 i, 1i , i2 5,
4 i +2 i3 5 and the like. So mathematicians have invented the number ”i”, but this in
turn immediately generates many other numbers, namely those that can be written as
a0 +a1 i +a2 i2 +···+an in
with a1 , · · · , an , b1 , · · · , bm real.
b0 +b1 i +b2 i2 +···+bm im
2
n
+a1 i +a2 i +···+an i
Once ”i” is born, a whole set C, called set of complex numbers ba00+b
2
m
1 i +b2 i +···+bm i
with a1 , · · · , an , b1 , · · · , bm comes along. Notice R is a subset of C, for if a ∈ R then
a = −a i2 ∈ C, and R ⊂ C.
An important theorem is the following.
Theorem 1.1. If z ∈ C, i.e. z =
be written uniquely as:
a0 +a1 i +a2 i2 +···+an in
b0 +b1 i +b2 i2 +···+bm im
, a1 , · · · , an , b1 , · · · , bm , then z can
z = x + iy
with x, y ∈ R.
The real number x is called the real part of z , Re(z) := x, while the real number y
is the imaginary part and we write Im(z), so that z = Re(z) + i Im(z).
1
Sometimes
√ √ the symbol
same as −1 −1 .
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√
−1 is used to indicate i, but be careful because
1
p
(−1)(−1) is not the
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MM Vercelli.
a0 +a1 i +a2 i2 +···+an in
A+B i
,
= C+D
i
b0 +b1 i +b2 i2 +···+bm im
(AC+BD)+(BC−AD) i
. Therefore z
C 2 +D2
proof. From i2 = −1 it follows that z =
where A, B, C, D ∈
A+B i C−D i
Besides, z = ( C+D
)(
) =
i C−D i
AC+BD
BC−AD
and y = C 2 +D2 . Uniqueness
C 2 +D2
0
0
0
R.
= x + y i, with
x=
is easy to prove. If there were z such that
z = x + y i = x + y i and y 6= y , we would have
i=
x − x0
y0 − y
which implies i is real! Contradiction. Hence y = y 0 and necessarily x = x0 . 2
The inverse
1
z
and the conjugate z
Let z denote a complex number z = x + i y . The previous proof provides us with the
idea of how to find the inverse z1 of a complex number. It is convenient to define first the
conjugate z = x − y i of z as the complex number whose imaginary part is opposite to
the imaginary part of z . Geometrically the conjugate is the symmetric point to z with
respect to the x-axis.
Two important properties are:
Proposition 1.2. Let z, w be complex numbers. Then:
zw = zw
z+w =z+w
Using the first property repeatedly we have z n = z n .
Observe z = z iff Im(z) = 0, i.e. iff z is real.
Now the important remark:
z.z = x2 + y 2
The product of a number by its conjugate equals the distance squared of the point
from the origin, i.e.the square of the modulus of the vector z . Since the modulus is |z|,
z.z = |z|2 .
From this the inverse is easy.
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MM Vercelli.
Proposition 1.3. The inverse
1
z
of z 6= 0 is
1
z
x
y
= 2 = 2
− 2
i
2
z
|z|
x +y
x + y2
Proof . From the product z |z|z 2 =
2
zz
|z|2
Example 1.4. The inverse of i is
=
|z|2
|z|2
= 1 we see that
1
z
=
z
|z|2
, by definition.
1
= −i
i
The inverse z1 , together with the sum and the product allows to see C as a number
field 2 . Two complex numbers z = x + y i, w = a + b i are multiplied as follows:
z.w = (x + y i)(a + b i) = xa + xb i +ya i +yb i2 = xa + (xb + ya) i +yb(−1)
so z.w = (xa − yb) + (xb + ya) i. A famous formula:
|z|2 |w|2 = (x2 + y 2 )(a2 + b2 ) = (xa − yb)2 + (xb + ya)2 = |zw|2 .
This was Euler’s starting point for proving Fermat’s theorem3 , which states that the
natural numbers of the form 4k + 1 are the sum of two squares.
Notice at last that the real and imaginary parts can be found using conjugates:
Re(z) =
z+z
2
Im(z) =
z−z
2i
The number z is said purely imaginary if Re(z) = 0. Thus z is purely imaginary
iff z = y i, y ∈ R. The condition z = −z is necessary and sufficient for z to be purely
imaginary.
2
3
First noticed by the mathematician Raffaele Bombelli from Bologna, in 1572.
http://it.wikipedia.org/wiki/Teorema di Fermat sulle somme di due quadrati
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Inequalities
Here are two important inequalities:
(triangle)
|z + w| ≤ |z| + |w|
(1)
and
(Cauchy 0 s)
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|a1 b1 + · · · + an bn |2 ≤ (|a1 |2 + · · · + |an |2 )(|b1 |2 + · · · + |bn |2 )
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(2)
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The exponential, polar coordinates
z = ρei θ := ρ (cos(θ) + i sin(θ))
where ρ = |z| is called modulus or norm.
The angle θ , i.e. a real number determined up to integer multiples of 2π , is called
argument or amplitude and denoted by
arg(z)
by using the trigonometric identities we have
ei θ ei ψ = ei θ+ψ
and so
arg(zw) = arg(z) + arg(w)
z
arg( ) = arg(z) − arg(w)
w
we have also the so called de Moivre’s formula
(cos(θ) + i sin(θ))n = (cos(nθ) + i sin(nθ))
which is useful in order to obtain the n solutions of
zn = a
namely
z=
√
n
µ
¶
ψ
2π
ψ
2π
r cos( + k ) + i sin( + k )
n
n
n
n
k = 0, 1, · · · , n − 1
where a = rei ψ .
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Circles and lines by using complex numbers
Here is the equation of the circle of radius r centered at z0 :
|z − z0 |2 = r2
an straight line can be written as
az + az = r
where r is a real number.
Here is an example:
x + 2y = 1
can be written as:
az + az = 1
with a =
1
2
− i.
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Complex valued functions and its differential
A complex valued function is a function from a certain set S to the complex numbers.
Namely,
f :S→C.
So if p ∈ S then
f (p) = u(p) + i v(p)
where u(p), v(p) ∈ R.
A complex valued function can be also regarded as function to R2 .
If S ⊂ Rn is an open subset then the function f : S → C is called smooth if all
partial derivatives of u and v do exists.
We are mainly interested in complex functions. Namely, when S ⊂ C. In this case
we will regard f as a function of (x, y):
f (x, y) = (u(x, y), v(x, y))
Recall that if f has partial derivatives we can form the so called Jacobian matrix Jf :
µ ∂u
Jf =
∂x
∂v
∂x
∂u ¶
∂y
∂v
∂y
µ
=
ux uy
vx vy
¶
µ
When the Jacobian matrix Jf is multiplied by the column
dx
dy
¶
we get the the
differential df of the function f . Namely,
µ
df = Jf
dx
dy
¶
= (ux dx + uy dy, vx dx + vy dy)
By using complex numbers we write:
df = (ux dx + uy dy) + i (vx dx + vy dy)
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