1
COMPLEX FUNCTIONS AND POLYNOMIALS
Lecture 3
1
COMPLEX FUNCTIONS AND POLYNOMIALS
Lecture 3
2
COMPLEX FUNCTIONS
EXTRA
3
Definition of THE EXPONENTIAL FUNCTION
f(z)= e
z
For
z = x + iy 0
e
z
= e
x
(cos y + i∙sin y).
1.The function
e
z
is periodic:
e
z+2ki
= e
z
for k
Integer
2. For a real number
z = x
, we obtain
e
z
= e
x
( cos 0 + i sin 0) = e
x
- the classical exponential function.
3. The modulus and the argument:
|e
z
| = e
x
, arg(e
z
) = y.
EXTRA
4
2
1
2
1
z
z
z
z
e
e
e
1
.
2
1
2
1
z
z
z
z
e
e
e
2
.
1
e
3
0
.
,
.
PROPERTIES OF THE EXPONENTIAL FUNCTION
Examples:
1)
-1·i = e
i
· e
i/2
= e
i3/2
= -i
2) Find the real and imaginary parts of
z = e
-i
:
e
-i
=cos(-1) + i sin(-1) = 0.540 + i
0.841
(
in radians)
5
The Logarithm
The logarithm of a complex variable is of form:
f(z) = log z
and
it is the inverse of the exponential function, i.e.
we want to have
log( e
z
) = z
and
e
log z
= z
Definition
z = r e
i
, r > 0 (z0)
log z = { lnr + i ( + 2k): k Z }
The logarithm
log z
is not in the strict sense a function
because the argument of a complex number is not uniquely defined,
The logarithm has infinitely many branches: every fixed k has one branch
which is a function in the strict meaning of the word.
EXTRA
6
The main branch is a function:
Log z = lnr + i∙Arg z (r > 0), π
Arg z ≤ π
Log z = lnr + i∙ for z = r e
i
PROPERTIES
Log (z
1
∙z
2
) = Logz
1
+
Logz
2
,
Log1 = 0,
e
Logz
= z.
Examples:
2
i
= ?
i
i
= ?
EXTRA
7
2 = e
Log2
(the main branch of the logarithm
Log z
)
2
i
= e
i Log 2 =
cos(ln2) + i sin (ln2) ≈ 0.769239.. + i 0.638961..
The multivalued logarithm:
log i = ln 1 + i arg i = 0 + i (2k + ½ )π, k = 0, ±1, ±2, ±3, ...
The main branch:
Log i = ln 1 + i Arg i = 0 + i π/2,
i
i
= e
ii
Log i
= e
i
i
π/2
= e
– π/2
≈
0.20788....
FIND
2
i
FIND
i
i
i
i
= e
ii log i
= e
i
i (2k + ½) π
= e
-(2k + ½) π
,
k = 0, ±1, ±2, ±3, ...
ALL
i
i
ARE REAL !!!!!
EXTRA
8
For complex quantities THE RULE:
(x y)
a
= x
a
y
a
doesn't always
apply for complex numbers.
Example;
(2 z + z
2
)
a
≠ z
a
(2 + z)
a
e.g.
z = -3, a = i
(2 (-3) + (-3)
2
)
i
≠ (-3)
i
(2 + (-3))
i
LHS
:
(-6 + 9)
i
= 3
i
= e
iln3
0.454832.. + i 1.09861...
RHS
:
(-3)
i
(2-3)
i
= (-3)
i
(-1)
i
= (3e
iπ
)
i
(e
iπ
)
i
= e
i Log(-3)
e
-π
= e
i
(ln3+iπ
) e
-π
e
iln3
·
0.00186744.. 0.000849374.. + i
0.0016631...
LHS RHS
CAUTION !!!
9
POLYNOMIALS
10
Definitions
(i) A
polynomial in z
with real coefficients is the function W
n
(z)
of
the form:
W
n
(z) = a
n
z
n
+... + a
1
z + a
0
,
defined for zC, when nN.
If a
n
≠ 0 and a
m
= 0 for m>n then W
n
is a polynomial
of degree
n,
and the number n is called the degree of the polynomial W
n
.
e.g. W
4
(z) = 5 z
4
+ 1.05 z
3
+ 2 z
2
– 7 z + 0.5
(ii) The complex number x is called
the root
(zero) of the
polynomial W
n
(x)
iff W
n
(x) = 0, which means that
a
n
x
n
+... + a
1
x + a
0
= 0.
11
Example
a = -1 is the root of the polynomial W(x) = 3x
5
- 9x
2
- 2x +
10,
because W(-1) = 3(-1)
5
9(-1)
2
2(-1) + 10 = 0.
Please notice that polynomials
do not always have real
roots.
e.g.
W(x) = x
2
+ 1 and W(x) = x
4
+ x
2
+ 2
The theorem of Niels Abel and Evarist Galois tels us that there can
be no general algorithm (finite and involving only arithmetical
operations and radicals) to determine the roots of a polynomial of
degree n 5.
12
POLYNOMIALS WITH INTEGER
COEFFICIENTS
13
RATIONAL ROOT TEST
Consider a polynomial in x R, with only integer coefficients:
f(x) = a
n
x
n
+ ... + a
o
The Rational Root Test tells you that if the polynomial has a rational root x
R
then it must be a fraction x
R
=
p
/
q
,
where
q
is a factor of the leading coefficient a
n
and
p
is a factor of the constant a
o
.
p(x) = 2x
4
− 11x
3
− 6x
2
+ 64x + 32
The factors of the leading coefficient '2' are 2 and 1 (the q 's).
The factors of the constant term '32' are 1, 2, 4, 8, 16, and 32 (the p 's).
Therefore the possible rational roots are ±1, 2, 4, 8, 16, or 32 divided by 2 or 1:
±1/2, 1/1, 2/2, 2/1, 4/2, 4/1, 8/2, 8/1, 16/2, 16/1, 32/2, 32/1
reduced: ± ½, 1, 2, 4, 8, 16, 32
Example:
14
Caution:
1. Don’t make the Rational Root Test out to be more than it is.
2. It doesn’t say that those rational numbers are roots, just that no other
rational
numbers can be roots.
3. And it doesn’t tell you anything about whether some irrational or even
complex roots
exist.
4. The Rational Root Test is only a starting point.
15
POLYNOMIALS WITH COMPLEX COEFFICIENTS
16
Theorem (dividing polynomials)
For every pair of polynomials Q(z) and P(z,) where the degree of Q(z) > 0,
there exist polynomials S(z) and R(z) such that
P(z) = S(z) ∙ Q(z) + R(z),
for z C and the degree of R is smaller that the degree of Q.
The polynomials S(z) and R(z) are uniquely determined by Q(z) and P(z).
Example: (2x
5
+ 2x
4
- x
2
- 1) : (x
2
– x + 2) = (2x
3
+ 4x
2
- 9)(x
2
– x + 2) - 9x+17
17
Theorem
Let the coeffincients of the polynomial W(x)
W(z) = a
n
z
n
+ a
n-1
z
n-1
+ ..... + a
0
,
be real and the complex number z
R
= x + i y be a root of
it,
W(z
R
) = 0,
then the conjugate of z
R
= x - i y, is also a root of this
polynomial
W( z
R
) = 0
The Bezout Theorem
The value a is a root of the polynomial P(z) iff:
P(z) = (z – a)S(z).
Thus the fact that a is a root of the polynomial P(z)
is equivalent to the fact that the binomial (z – a) divides P(z).
18
Example
Let us assume that you know one of the roots of the polynomial
l
,
)
(
10
z
10
z
7
z
2
z
z
W
2
3
4
.
i
5
z
1
which is
.
1
z
19
Fundamental Theorem of Algebra
Every polynomial W
n
(z) of degree n has exactly n complex roots
(taking into account the multiplicity of the roots)
Conclusion I
If the degree of the polynomial is odd, then it always has a real
root.
This is true because the number of
complex roots is even, (they
always appear in pairs ),
only real roots
can have an odd number.
)
,
( z
z
Example
20
Conclusion II
Let z
1
,......,z
m
be all the roots of W
n
(z) of order
respectively
k
1
, k
2
,......, k
m
(k
1
+k
2
+......+k
m
= n).
Then:
m
2
1
k
m
k
2
k
1
n
z
z
z
z
z
z
z
W
)
(
)
(
)
(
)
(
monomials