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Complex numbers.
1. Definition and notations
A complex number z is a formal expression of the form z = x + yi where
x, y ∈ R. Denote by
C = {x + yi | x, y ∈ R}
the set of all complex numbers. For a complex number z = x + yi ∈ C,
where x, y ∈ R put
Re(z) := x,
the real part of z,
Im(z) := y, the imaginary part of z,
p
|z| := x2 + y 2 , the absolute value of z,
z := x − yi, the complex conjugate of z.
By convention, for x ∈ R we identify z = x + 0i ∈ C with x ∈ R and write
x + 0i = x. Thus we have R ⊆ C and, moreover, R = {z ∈ C | Im(z) = 0}.
Also by convention we write i = 0 + 1i, −i
√ = 0 + (−1)i, and, more generally,
yi = 0 + yi where y ∈ R. Thus i, −i, 2i, 5i ∈ C.
Note that for z ∈ C we have |z| = 0 if and only if z = 0.
2. Algebraic operations on complex numbers
2.1. Addition, multiplication and complex conjugation. For z1 =
x1 + y1 i, z2 = x2 + y2 i ∈ C (where x1 , x2 , y1 , y2 ∈ R) define
z1 + z2 := (x1 + x2 ) + i(y1 + y2 )
z1 z2 := x1 x2 − y1 y2 + i(x1 y2 + x2 y1 ),
so that z1 + z2 ∈ C and z1 z2 ∈ C. Note that, following the above definition
of multiplication for complex numbers, we have
i2 = (0 + 1i)(0 + 1i) = (02 − 12 ) + i(0 · 1 + 1 · 0) = −1 + 0i = −1,
so that i2 = −1 in C.
Also, for z = x + iy ∈ C (where x, y ∈ R) put
−z := −x + (−y)i,
so that −z = (−1)z.
2.2. Basic properties of addition, multiplication and complex conjugation. We have:
1
2
z1 + (z2 + z3 ) = (z1 + z2 ) + z3 for any z1 , z2 , z3 ∈ C,
z1 + z2 = z2 + z1 for any z1 , z2 ∈ C,
z + 0 = 0 + z = z for any z ∈ C,
z + (−z) = (−z) + z = 0
for any z ∈ C,
z1 (z2 z3 ) = (z1 z2 )z3 for any z1 , z2 , z3 ∈ C,
z1 z2 = z2 z1 for any z1 , z2 ∈ C,
1 · z = z · 1 = z for any z ∈ C,
0 · z = z · 0 = 0 for any z ∈ C,
z1 (z2 + z3 ) = z1 z2 + z1 z3
2
2
for any z1 , z2 , z3 ∈ C,
2
zz = zz = x + y = |z| ∈ R for any z = x + iy ∈ C, where x, y ∈ R,
zz = 0 ⇐⇒ z = 0 for any z ∈ C,
z=z
for any z ∈ C,
z1 + z2 = z1 + z2 for any z1 , z2 ∈ C,
z1 z2 = z1 z2 for any z1 , z2 ∈ C,
z = z ⇐⇒ z ∈ R where z ∈ C.
2.3. Division of complex numbers. Recall that for any z ∈ C we have
zz = |z|2 and that zz = |z 2 | > 0 whenever z 6= 0.
For z1 = x1 + y1 i, z2 = x2 + y2 i ∈ C where x1 , x2 , y1 , y2 ∈ R such that
z2 6= 0 (and hence |z2 | > 0) we put
z1
1
x1 x2 + y1 y2 y1 x2 − x1 y2
z 1 z2 =
:=
+
i.
2
z2
|z2 |
x22 + y22
x22 + y22
In particular, for z = x + yi 6= 0 (where x, y ∈ R)
z
z
x
−y
1
=
= 2 = 2
+ 2
i.
2
z
zz
|z|
x +y
x + y2
2.4. Properties of division. We have:
1
1
· z = z · = 1 for any z 6= 0, z ∈ C,
z
z
1
6= 0 for any z 6= 0, z ∈ C,
z
z1
z1
=
for any z1 , z2 ∈ C, z2 6= 0,
z2
z2
z1 |z1 |
=
z2 |z2 | for any z1 , z2 ∈ C, z2 6= 0,
1
= 1
z |z| for any z ∈ C, z 6= 0.
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2.5. Complex numbers and polar coordinates. Let z = x + yi ∈ C,
where x, y ∈ R and let (r, θ) be the polar coordinates of (x, y) ∈ R2 , where
r ≥ 0, 0 ≤ θ < 2π. In this case we also say that (r, θ) are the polar
coordinates of z.
We have:
(1)
r = |z|, x = r cos θ, y = r sin θ where z = x + yi ∈ C, x, y ∈ R,
(2)
z = r(cos θ+i sin θ), where z ∈ C, and (r, θ) are the polar coordinates of z,
(3)
(r1 (cos θ1 + i sin θ1 )) (r2 (cos θ2 + i sin θ2 )) = r1 r2 (cos(θ1 + θ2 ) + sin(θ1 + θ2 )i)) ,
where r1 , r2 ≥ 0, and θ1 , θ2 ∈ R. Thus for the product of two
complex numbers the polar radius of their product is the product
of their polar radii and the polar angle of their product is the sum
(modulo 2π) of their polar angles.
(4) For any θ ∈ R put
(iθ)2 (iθ)3
(iθ)n
+
+ ...
+ ··· =
2!
3!
n!
θ2 θ3
θ4 θ5
θ6
1 + iθ −
− i+
+ i−
+ ··· =
2!
3! 4! 5!
6!
θ2 θ4 θ6
θ3 θ5
1−
+
−
+ ... + θ −
+
− ... i =
2!
4!
6!
3!
5!
cos θ + i sin θ.
eiθ := 1 + (iθ) +
(5) Thus for any z ∈ C we have
z = r(cos θ + i sin θ) = reiθ
where (r, θ) are the polar coordinates of z. Moreover, If
z1 = r1 eiθ1 , z2 = r2 eiθ2
where r1 , r2 ≥ 0 and θ1 , θ2 ∈ R then
z1 z2 = r1 r2 ei(θ1 +θ2 ) .
(6) If
z = reiθ ,
where r > 0 and θ ∈ R then for every n ∈ Z we have
z n = rn einθ = rn (cos(nθ) + i(nθ)).
In particular, for z 6= 0, we have
1
1
1
= z −1 = e−iθ = (cos θ − i sin θ).
z
r
r
4
Remark. The formula in part (3) above is a good way to derive the
trigonometric identities for cos(θ1 + θ2 ) and sin(θ1 + θ2 ) if you ever forget
these identities.
Indeed, let θ1 , θ2 ∈ R. Then, using the definition of complex multiplication, we have
(cos θ1 + i sin θ1 )(cos θ2 + i sin θ2 ) =
(cos θ1 cos θ2 − sin θ1 sin θ2 ) + i(cos θ1 sin θ2 + sin θ1 cos θ2 ) =
cos(θ1 + θ2 ) + i sin(θ1 + θ2 ),
where the last equality follows from statement (3) above. Hence
cos(θ1 + θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2
and
sin(θ1 + θ2 ) = cos θ1 sin θ2 + sin θ1 cos θ2 .
2.6. Roots of unity. Using the above formulas for complex multiplication
in polar coordinates (specifically, part (6) above) it is easy to deduce that if
n ≥ 1 and
2π+2πk
2π + 2πk
2π + 2πk
) + i sin(
) = ei n ,
n
n
where k = 0, 1, . . . , n − 1, then
ζk = cos(
ζkn = 1.
The numbers ζk , (where k = 0, 1, . . . , n − 1) are called the n-th roots of unity
2π
in C. The number ζ0 = cos( 2π
n ) + i sin( n ) is called the main n-th root of
unity in C.
It is not hard to check that for any n ≥ 1 the numbers ζ0 , . . . , ζn−1 ∈ C
are distinct and that they are the only solutions of the equation
zn = 1
in C.
3. Fundamental theorem of algebra
The Fundamental Theorem of Algebra states the following:
Let f (z) = an z n + . . . a1 z + a0 , (where a0 , . . . , an ∈ C, an 6= 0) be a
polynomial of degree n ≥ 1 with coefficients in C. (Here z is a variable.)
Then there exist complex numbers z1 , . . . zn ∈ C such that f (z) factor as
f (z) = an (z − z1 ) . . . (z − zn ).
Moreover, one can show that in the above situation for z∗ ∈ C we have
f (z∗ ) = 0 if and only if z∗ = zk for some 1 ≤ k ≤ n, that is z1 , . . . , zn are
the only solutions of the equation f (z) = 0 in C.
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In particular, if a polynomial f (z) = an z n + . . . a1 z + a0 of degree n as
above has n distinct roots z1 , . . . , zn in C, then
f (z) = an (z − z1 ) . . . (z − zn ).
Applying this fact to the equation z n − 1 = 0 (where n ≥ 1) we conclude
that for every n ≥ 1 the polynomial z n − 1 factors as
z n − 1 = (z − ζ0 ) . . . (z − ζn−1 )
where ζ0 , . . . , ζn−1 are the n-th roots of unity in C.
3.1. Factorization of polynomials with real coefficients. Let
f (z) = an z n + . . . a1 z + a0
be a polynomial of degree n ≥ 1 with real coefficients ak ∈ R, for k =
0, . . . , n, and an 6= 0.
Suppose that z∗ = x + iy ∈ C is a root of the equation f (z) = 0, so that
f (z∗ ) = 0. Then
0 = 0 = f (z∗ ) = an z∗n + · · · + a1 z∗ + a0 =
an (z∗ )n + . . . a1 z∗ + a0 =
since all ak ∈ R
= an (z∗ )n + · · · + a1 z∗ + a0 = f (z∗ )
Thus z∗ = x − iy is also a root of f (z) = 0. Note that z∗ = z∗ if and only if
z∗ ∈ R.
Hence we see that those complex roots of f (z) = 0 that do not belong to
R, come in pairs of the form z∗ , z∗ .
Suppose that z∗ = x + iy ∈ C is a root of f (z) = 0 where x, y ∈ R and
z∗ 6∈ R, that is y 6= 0. Then
(†) (z−z∗ )(z−z∗ ) = (z−x−iy)(z−x+iy) = (z−x)2 +y 2 = z 2 −2xz+x2 +y 2 .
Since x, y ∈ R, the polynomial (in the variable z) z 2 − 2xz + x2 + y 2 has
coefficients in R. Moreover, the quadratic formula and the assumption that
y 6= 0 imply that the equation z 2 − 2xz + x2 + y 2 = 0 in the variable z has
no solutions in R.
Now let
f (z) = an (z − z1 ) . . . (z − zn )
be the factorization of f (z) provided by the Fundamental Theorem of Algebra. If zk ∈ R is a real number, the term z − zk is a polynomial of degree
1 in z with real coefficients. The roots of f (z) which are not real numbers,
come in conjugate pairs the form z∗ , z∗ . With a small extra argument we can
show that z∗ and z∗ have the same multiplicities in the above factorization
of f (z).
Hence it follows from (†) above that the polynomial f (z) can be written
as a product of polynomials of degree 1 and 2 with real coefficients.
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This fact about factorization of polynomials f (z) with coefficients in R as
a product of linear and quadratic terms (also with coefficients in R) plays
a key role in the theory of partial fractions (which you must have seen
in Calculus II) and in solving linear differential equations with constant
coefficients.