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Elementary Function
JunFeng Yin
§1 Polynomial and Rational functions
Polynomial Functions
pn (z) = a0 + a1 z + a2 z 2 + · · · + an z n
whose degree is n if an 6= 0.
Rational Functions
Rm,n (z) =
a0 + a1 z + a2 z 2 + · · · + am z m
b0 + b1 z + b2 z 2 + · · · + bn z n
has numerator degree m and denominator degree n if am 6= 0
and bn 6= 0.
§1 Polynomial and Rational functions
Polynomial Functions
pn (z) = a0 + a1 z + a2 z 2 + · · · + an z n
whose degree is n if an 6= 0.
Rational Functions
Rm,n (z) =
a0 + a1 z + a2 z 2 + · · · + am z m
b0 + b1 z + b2 z 2 + · · · + bn z n
has numerator degree m and denominator degree n if am 6= 0
and bn 6= 0.
• Polynomials are entire, and rational function are analytic
everywhere except for the zeros of their denominators.
Factored form
Consider
p3 (z) = 12 + 10z − 4z 2 − 2z 3 = −2z 3 − 4z 2 + 10z + 12
whose zeros are
2, −1, −3
thus, its factored form is
p3 (z) = −1(z − 2)(z + 1)(z + 3)
In general,
dividend = divisor × quotient + remainder
in other words,
pn (z) = (z − z1 )pn−1 (z) + constant
where pn−1 (z) has degree n − 1.
If z1 is the zero of pn (z), then
pn (z) = (z − z1 )pn−1 (z)
Example 1
Carry out the deflation of the polynomial z 3 + (2 − i)z 2 − 2iz.
Solution Obviously, z1 = 0 is a zero.
z 3 + (2 − i)z 2 − 2iz = z(z 2 + (2 − i)z − 2i)
Further, −2 and i are the zeros of z 2 + (2 − i)z − 2i.
Thus,
z 3 + (2 − i)z 2 − 2iz = z(z + 2)(z − i)
Fundamental Theorem of Algebra
How do we know pn (z) have any zero?
Gauss in his doctoral dissertation of 1799 proved that
Theorem 2
Every nonconstant polynomial with complex coefficients has at
least one zero in C.
• Consequently, a polynomial of degree n has n zeros.
Thus, a complete factorization of any polynomial is
pn (z) = an (z − z1 )(z − z2 ) · · · (z − zn )
If z0 is a zero of p(z) of multiplicity precisely k if and only if
pn (z) = (z − z0 )k q(z)
where q(z) is a polynomial with q(z0 ) 6= 0.
Example 3
Show that if the polynomial p(z) has real coefficients, it can be
expressed as a product of linear and quadratic factors, each having
real coefficients.
Solution First, the nonreal zeros of a polynomial with real
coefficients occur in complex conjugate pairs, e.g. pn (z1 ) = 0 and
pn (z1 ) = 0.
Thus,
pn (z) = an (z − z1 )(z − z1 ) · · · (z − zn )
= an (z 2 − (z1 + z1 )z + z1 z1 ) · · · (z − zn )
= an (z 2 − 2(Re(z1 ))z + |z1 |2 ) · · · (z − zn )
The pair of complex factors of degree one has been replaced by a
real factor of degree two.
Of course, the Fundamental Theorem only tells us that there are
zeros; it does not tell use how to find them.
We know the algorithm for finding zeros of polynomials of degree
three and four.
For higher order polynomials, Newton’s method is an excellent
zero-finder for any analytic function, if a good initial estimate is
available.
Example 4
Express p3 (z) = 12 + 10z − 4z 2 − 2z 3 in terms of (z − 1).
In order to find di satisfy
p3 (z) = 12 + 10z − 4z 2 − 2z 3
= d0 + d1 (z − 1) + d2 (z − 1)2 + d3 (z − 1)3
1. A brute-force solution is to expand the above formula
2. An alternative: note that z = (z − 1) + 1, denote ζ = z − 1,
and expand p3 (z) = p3 (ζ + 1)
p3 (ζ + 1) = 12 + 10(ζ + 1) − 4(ζ + 1)2 − 2(ζ + 1)3
= 16 − 4ζ − 10ζ 2 − 2ζ 3
p3 (z) = 16 − 4(z − 1) − 10(z − 1)2 − 2(z − 1)3
Taylor form
Moreover,
pn (z0 ) pn0 (z0 )
p 00 (z0 )
+
(z − z0 ) + n
(z − z0 )2
0!
1!
2!
(n)
pn (z0 )
+··· +
(z − z0 )n
2!
n
(k)
X
pn (z0 )
=
(z − z0 )k
k!
pn (z) =
k=0
is the Taylor form of the polynomial pn (z), centered at z0 .
Taylor form
Moreover,
pn (z0 ) pn0 (z0 )
p 00 (z0 )
+
(z − z0 ) + n
(z − z0 )2
0!
1!
2!
(n)
pn (z0 )
+··· +
(z − z0 )n
2!
n
(k)
X
pn (z0 )
=
(z − z0 )k
k!
pn (z) =
k=0
is the Taylor form of the polynomial pn (z), centered at z0 .
If z0 = 0, the form is called Maclaurin form.
For Rational Functions:
Rm,n (z) =
a0 + a1 z + a2 z 2 + · · · + am z m
b0 + b1 z + b2 z 2 + · · · + bn z n
We assume that common zeros have been canceled.
Zeros of the denominator are called poles of Rm,n (z).
Clearly, the magnitude of Rm,n (z) grows without bound as z
approaches a pole.
Example 5
Find all the poles and their multiplicities for
R(z) =
(3z + 3i)(z 2 − 4)
(z − 2)(z 2 + 1)2
Example 5
Find all the poles and their multiplicities for
R(z) =
(3z + 3i)(z 2 − 4)
(z − 2)(z 2 + 1)2
Solution The zeros of the denominator are 2, i, and −i, which are
candidates for the poles.
It is deduced that
3(z + i)(z − 2)(z + 2)
3(z + 2)
(3z + 3i)(z 2 − 4)
=
=
(z − 2)(z 2 + 1)2
(z − 2)(z + i)2 (z − i)2
(z + i)(z − i)2
The poles of R(z) are at z = i of multiplicity 2 and z = −i of
multiplicity 1.
The poles of Rm,n (z) enable us to express Rm,n (z) in terms of
partial fractions.
For instance,
1
1
1
3z 2 + 4z − 5
=
+
+
(z − 2)(z + 1)(z + 3)
z −2 z +1 z +3
4z + 4
−1
8
−7
6
=
+
+
+
2
z(z − 1)(z − 2)
z
z − 1 z − 2 (z − 2)2
Theorem 6
If
Rm,n (z) =
a0 + a1 z + a2 z 2 + · · · + am z m
bn (z − ζ1 )d1 (z − ζ2 )d2 · · · (z − ζr )dr
is a rational function with n = d1 + d2 + · · · + dr > m, then,
Rm,n (z) has a partial fraction decomposition of the form
(1)
Rm,n (z) =
(1)
(1)
A
A0
A1
+
+ · · · + d1 −1
d
d
−1
(z − ζ1 )
(z − ζ1 ) 1
(z − ζ1 ) 1
(2)
=
(2)
A
A0
+ · · · + d2 −1
d
2
(z − ζ1 )
(z − ζ1 )
=
(r )
(r )
Adr −1
A1
A0
+
+
·
·
·
+
(z − ζ1 )
(z − ζ1 )dr
(z − ζ1 )dr −1
(r )
where
(j)
As = lim
z−ζj
1 ds
[(z − ζj )dj Rm,n (z)]
s! dz s
are constants. (The ζk ’s are assumed distinct.)
Example 7
Reproduce the partial fraction decomposition of
4z+4
.
z(z−1)(z−2)2
Example 7
Reproduce the partial fraction decomposition of
Solution The desired form is
(2)
(1)
R(z) =
4z+4
.
z(z−1)(z−2)2
(3)
(3)
A
A0
A
A
4z + 4
= 0 + 0 +
+ 1
2
2
z(z − 1)(z − 2)
z
z − 1 (z − 2)
z −2
where
(1)
A0 = lim zR(z) =
z→0
4·0+4
= −1
(0 − 1)(0 − 2)2
(2)
A0 = lim (z − 1)R(z) =
z→0
(3)
4·1+4
=8
1(1 − 2)2
A0 = lim (z − 2)2 R(z) =
z→0
and
(3)
A1
4·2+4
=6
2(2 − 1)
d d
= lim
(z − 2)2 R(z) = lim
z→0 dz
z→0 dz
=
(22 − 2)4 − (4 · 2 + 4)(2 · 2 − 1)
2
2
4z + 4
z2 − z
= −7
Residue
Let R = P/Q be a rational function with degP < degQ. If ζ is a
pole of R, then the coefficient of 1/(z − ζ) in the partial fraction
expansion of R is called the residue of R(z) at ζ, and is denoted
by Res(ζ).
Compute Res(i) for R(z) =
2z+3
(z−i)(z 2 +1)
§2 The Exponential, Trigonometric, and Hyperbolic
function
The complex exponential function
e z = e x (cos y + sin yi)
play a prominent role in analytic function, fundamental
requirement for trigonometric and hyperbolic function.
d z
dz e
I
entire function:
I
|e z | = e x
I
arge z = y + 2kπ,
= ez
k = 0, ±1, ±2, . . .
Theorem 8
(i) The equation e z = 1 holds if, and only if, z = 2kπi, where k
is an integer.
(ii) The equation e z1 = e z2 holds if, and only if, z1 = z2 + 2kπi,
where k is an integer.
An important consequence is that e z is periodic and satisfy
e z+2πi = e z
A function is said to be periodic in a domain D if there exists a
nonzero constant λ, s.t. f (z + λ) = f (z).
From the identity
e iy = cos y + sin yi
we have
sin y =
e iy − e −iy
,
2i
cos y =
e iy + e −iy
2
Definition 9
Given any complex number z, we define
sin z =
e iz − e −iz
,
2i
cos z =
e iz + e −iz
2
Since e iz is entire, so does sin z and cos z.
Since e iz is entire, so does sin z and cos z.
d
d
sin z =
dz
dz
e iz − e iz
2i
=
1 iz
(ie − (−i)e −iz ) = cos z
2i
d
cos z = − sin z
dz
sin(z + 2π) = sin z,
sin(−z) = − sin z,
2
cos(z + 2π) = cos z
cos(−z) = cos z
2
sin z + cos z = 1
sin(z1 ± z2 ) = sin z1 cos z2 ± sin z2 cos z1
cos(z1 ± z2 ) = cos z1 cos z2 ∓ sin z1 sin z2
sin 2z = 2 sin z cos z,
cos 2z = cos2 z − sin2 z
Example 10
Prove that sin z = 0 if and only if z = kπ, where k is an integer.
Example 10
Prove that sin z = 0 if and only if z = kπ, where k is an integer.
Solution If z = kπ, then clearly sin z = 0, i.e.,
e iz − e −iz
= 0,
2i
that is
e iz = e −iz
If follows that
iz = −iz + 2kπi
i.e.
z = kπ
for some integer k.
1. cos z = 0 if and only if z = kπ + π/2,
2.
tan z =
3.
sin z
cos z
1
1
, cot z =
, sec z =
, csc z =
,
cos z
sin z
cos z
sin z
d
d
tan z = sec2 z,
sec z = sec z tan z
dz
dz
d
d
cot z = csc2 z,
csc z = − csc z cot z
dz
dz
4.
| cos x| 6 1,
but
∀x ∈ R
−y
e + ey = cosh y > 1,
| cos(iy )| = 2
Definition 11
∀z, define
sinh z :=
e z − e −z
,
2
cosh z :=
e z + e −z
2
satisfy
d
d
sinh z = cosh z,
cosh z = sinh z,
dz
dz
sin iz = i sinh z, sinh iz = i sin z, cos iz = cosh z, cosh iz = i cos z,
cosh2 z − sinh2 z = 1
tanh z :=
sinh z
cosh z
1
1
, coth z :=
, sechz :=
, cschz :=
.
cosh z
sinh z
cosh z
sinh z
§3 The Logarithmic Function
w = f (z)
Single-valued functions
w = z 2,
w = ez
Multiple-valued functions
w = arg z,
w = z 1/2
Define log z as the inverse of the exponential function
w = log z,
if
z = ew
where z 6= 0 since e w 6= 0.
Note that
z = re iθ = e u+iv = e u e iv
we have
u = Logr ,
v = arg z = θ
Definition 12
If z 6= 0, then we define log z to be the set of infinitely many values
log z
= Log|z| + i arg z
= Log|z| + iArgz + i2kπ
(k = 0, ±1, ±2, . . .)
Definition 12
If z 6= 0, then we define log z to be the set of infinitely many values
log z
= Log|z| + i arg z
= Log|z| + iArgz + i2kπ
(k = 0, ±1, ±2, . . .)
For instance
log 3 = Log3 + i arg 3 = (1.908) + i2kπ
1
π
log(1 + i) = Log|1 + i| + i arg(1 + i) = Log2 + i( + 2kπ)
2
4
Some properties:
1. If z 6= 0, we have
z = e log z
but
log e z = z + i2kπ
(k = 0, ±1, ±2, . . .)
2.
log(z1 z2 ) = log z1 + log z2
3.
log
z1
= log z1 − log z2
z2
Recall that a branch cut: Argz ∈ (−π, π], principal value of arg z.
Recall that a branch cut: Argz ∈ (−π, π], principal value of arg z.
The principal value of the logarithm Logz is
Logz := Log|z| + iArgz
Recall that a branch cut: Argz ∈ (−π, π], principal value of arg z.
The principal value of the logarithm Logz is
Logz := Log|z| + iArgz
Other branches argτ z ∈ (−τ, τ + 2π]
Theorem 13
The function Logz is analytic in the domain D ∗ consisting of all
points of the complex plane except those lying on the nonpositive
real axis. Furthermore,
1
d
Logz = ,
dz
z
forz ∈ D ∗ .
Theorem 13
The function Logz is analytic in the domain D ∗ consisting of all
points of the complex plane except those lying on the nonpositive
real axis. Furthermore,
1
d
Logz = ,
dz
z
forz ∈ D ∗ .
For instance, find f 0 (z) when
f (z) := Log(3z − i)
Theorem 13
The function Logz is analytic in the domain D ∗ consisting of all
points of the complex plane except those lying on the nonpositive
real axis. Furthermore,
1
d
Logz = ,
dz
z
forz ∈ D ∗ .
For instance, find f 0 (z) when
f (z) := Log(3z − i)
f 0 (z) =
d
1 d
3
Log(3z − i) =
(3z − i) =
dz
3z − i dz
3z − i
Corollary 14
The function Argz is harmonic in D ∗ .
Corollary 15
The real function Log|z| is harmonic in the entire plane with the
exception of the origin.
Definition 16
F (z) is said to be a branch of a multiple-valued function f (z) in a
domain D if F (z) is single-valued and continuous in D and has the
property that, ∀z ∈ D, the value F (z) is one of the values of f (z).
Definition 16
F (z) is said to be a branch of a multiple-valued function f (z) in a
domain D if F (z) is single-valued and continuous in D and has the
property that, ∀z ∈ D, the value F (z) is one of the values of f (z).
Determine a branch of f (z) = log(z 3 − 2) that is analytic at z = 0,
and find f (0) and f 0 (0)
Definition 16
F (z) is said to be a branch of a multiple-valued function f (z) in a
domain D if F (z) is single-valued and continuous in D and has the
property that, ∀z ∈ D, the value F (z) is one of the values of f (z).
Determine a branch of f (z) = log(z 3 − 2) that is analytic at z = 0,
and find f (0) and f 0 (0)
Define g (z) = z 3 − 2, choose any branch of the logarithm that
is analytic at g (0) = −2
Thus F (z) = L−π/4 (g (z)),
F (0) = L−π/4 (03 − 2) = Log2 + iπ
F 0 (0) = L0−π/4 (g (0))g 0 (0) =
g 0 (0)
=0
g (0)
§4 Washer, Wedges, and Wall
log z, when suitably restricted, is analytic.
§4 Washer, Wedges, and Wall
log z, when suitably restricted, is analytic.
Harmonic functions
1. Polynomial function, Rational function
2. Exponential function, Trigonometric function
3. Logarithmic function
governed by Laplace’s equation φxx + φyy = 0.
Example 1
Find a function φ(x, y ) that is harmonic in the washer-shaped
region between |z| = 1 and |z| = 2, and takes the values φ = 20
on the inner circle and φ = 30 on the outer circle.
Example 1
Find a function φ(x, y ) that is harmonic in the washer-shaped
region between |z| = 1 and |z| = 2, and takes the values φ = 20
on the inner circle and φ = 30 on the outer circle.
Solution Denote the harmonic function is
φ(x, y ) = ALog|z| + B
we have
ALog|1| + B = 20,
and ALog|2| + B = 30.
So B = 20, and A = 10/Log2, and the solution is
φ(z)10
Log|z|
+ 20
Log2
Example 2
Find a function φ(x, y ) that is harmonic in the wedge-shaped
region, and takes the values φ = 20 on the upper side and φ = 30
on the lower side.
Example 2
Find a function φ(x, y ) that is harmonic in the wedge-shaped
region, and takes the values φ = 20 on the upper side and φ = 30
on the lower side.
Solution Instead Argz, θ ∈ (−π, π], the harmonic function is
chosen as
φ(x, y ) = Aarg0 z + B, θ ∈ (0, 2π]
we have
5π
3π
+ B = 20, and A
+ B = 30.
4
4
So B = 5, and A = 20/π, and the solution is
A
φ(z) =
20
arg0 z + 20
π
The more sophisticated boundary value problem is an alignment of
several 1800 wedges, which can be written as
φ(x, y ) = A1 Arg(z − x1 )+A2 Arg (z −x2 )+· · ·+An Arg (z −xn )+B.
Example 3
Find a function φ(x, y ) that is harmonic in the upper half-plane,
and takes

 φ(x, 0) = 0, forx > 1;
φ(x, 0) = 1, for − 1 < x < 1;

φ(x, 0) = 0, forx < −1;
.
Example 3
Find a function φ(x, y ) that is harmonic in the upper half-plane,
and takes

 φ(x, 0) = 0, forx > 1;
φ(x, 0) = 1, for − 1 < x < 1;

φ(x, 0) = 0, forx < −1;
.
Solution Denote the harmonic function is of the form
φ(x, 0) = A1 Arg(z + 1) + A2 Arg(z − 1) + B
For z = x > 1, Arg(z − 1) = Arg(z + 1) = 0, so
A1 · 0 + A2 · 0 + B = 0.
For z = x, −1 < x < 1, Arg(z − 1) = π, Arg(z + 1) = 0, so
A1 · 0 + A2 · π + B = 1.
For z = x < −1, Arg(z − 1) = Arg(z + 1) = π, so
A1 · π + A2 · π + B = 0.
Therefore B = 0, A1 = −1/π, A2 = 1/π, so we have
1
1
φ(z) = − Arg(z + 1) + Arg(z − 1).
π
π
§5 Complex Powers and inverse Trigonometric functions
One important theoretical use of logarithmic function is to define
complex powers of z.
§5 Complex Powers and inverse Trigonometric functions
One important theoretical use of logarithmic function is to define
complex powers of z.
The definition is motivated by the identity
z n = (e log z )n = e n log z ,
∀z
§5 Complex Powers and inverse Trigonometric functions
One important theoretical use of logarithmic function is to define
complex powers of z.
The definition is motivated by the identity
z n = (e log z )n = e n log z ,
∀z
Definition 17
If α is a complex constant and z 6= 0, then we define z α by
z α := e α log z
Example 18
Find all the value of (−2)i .
Example 18
Find all the value of (−2)i .
Solution Since log(−2) = Log2 + (π + 2kπ)i, we have
(−2)i = e i log(−2) = e Log2+(π+2kπ)i ,
k = 0, ±1, ±2, . . .
Thus, (−2)i has infinitely many different values.
Note that
z α = e α(Log|z|+iArgz+2kπi) = e α(Log|z|+iArgz e α2kπi
Thus,
e α2k1 πi = e α2k2 πi
only if
α2k1 πi = α2k2 πi + 2mπi
e.g., α = m/(k1 − k2 ), when m is an integer.
z m/n = exp
m
m
Log|z| exp i (Arg|z| + 2kπ) ,
n
n
(k = 0, 1, . . . , n−1)
In summary
I
z α is a single-valued when α is a real integer;
I
z α takes finitely many values when α is a real rational number;
I
z α takes infinitely many values in all other cases;
The principal branch of z α is also analytic in D ∗ = C \(−∞, 0], we
have
d αLogz
d
α
(e
) = e αLogz (αLogz) = e αLogz
dz
dz
z
similar to
1
d α
(z )αz α
dz
z
Example 19
Define a branch of (z 2 − 1)1/2 that is analytic in the exterior of the
unite circle, |z| > 1.
Consider w 2 = z 2 − 1, and the principal branch of (z 2 − 1)1/2 is
e 1/2 Log(z 2 − 1)
will not work; it has branch cutis wherever z 2 − 1 is negative real
on y axis and [-1,1].
Consider z(1 − 1/z 2 )1/2 , and the principal branch of (1 − 1/z 2 )1/2
is
e 1/2 Log(1 − 1/z 2 )
has cuts where 1 − 1/z 2 is negative real in [−1, 1].
Thus, w = f (z) = ze 1/2 Log(1 − 1/z 2 ).