Transcript Functions of a Complex Variable

Contents
17.1 Complex Numbers
17.2 Powers and Roots
17.3 Sets in the Complex Plane
17.4 Functions of a Complex Variable
17.5 Cauchy-Riemann Equations
17.6 Exponential and Logarithmic Functions
17.7 Trigonometric and Hyperbolic Functions
17.8 Inverse Trigonometric and Hyperbolic Functions
Ch17_1
17.1 Complex Numbers
DEFINITION 17.1
Complex Number
A complex number is any number of the z = a + ib
where a and b are real numbers and i is the imaginary
units.
z = x + iy, the real number x is called the real part and
y is called the imaginary part:
Re(z) = x, Im(z) = y
Ch17_2
DEFINITION 17.2
Complex Number
Complex number z1  x1  iy 1 and z 2  x 2  iy 2 are
equal, z1  z 2, if Re( z1 )  Re( z 2 ) and Im( z1 )  Im( z 2 )
x + iy = 0 iff x = 0 and y = 0.
Ch17_3
Arithmetic Operations
z1  x1  iy 1 , z 2  x 2  iy 2
Suppose
z1  z 2  ( x1  x 2 )  i ( y1  y 2 )
z1  z 2  ( x1  x 2 )  i ( y1  y 2 )
z1  z 2  ( x1 x 2  y1 y 2 )  i ( y1 x 2  x1 y 2 )
z1
z2

x1 x 2  y1 y 2
2
x2  y2
2
i
y1 x 2  x1 y 2
2
x2  y2
2
Ch17_4
Complex Conjugate
S u p p o se z  x  iy , z  x  iy , an d
z1  z 2  z1  z 2
z1  z 2  z1  z 2
z1 z 2  z1 z 2
 z1 
z1


z2
 z2 
Ch17_5
Two important equations
z  z  ( x  iy )  ( x  iy )  2 x
z z  ( x  iy )( x  iy )  x  i y  x  y
2
2
2
2
z  z  ( x  iy )  ( x  iy )  2 iy
and
Re( z ) 
z z
2
, Im( z ) 
(1)
2
(2)
(3)
zz
2i
Ch17_6
Geometric Interpretation
Fig 17.1 is called the complex plane and a complex
number z is considered as a position vector.
Ch17_7
DEFINITION 17.3
Modulus or Absolute Values
The modulus or absolute value of z = x + iy, denoted
by │z│, is the real number
|z|
x  y 
2
2
zz
Ch17_8
Example 3
If z = 2 − 3i, then
z 
2  (  3) 
2
2
13
As in Fig 17.2, the sum of the vectors z1 and z2 is the
vector z1 + z2. Then we have
z1  z 2  z 1  z 2
(5)
The result in (5) is also known as the triangle inequality
and extends to any finite sum:
z1  z 2  ...  z n  z1  z 2  ...  z n
(6)
Using (5),
z1  z 2  (  z 2 )  z1  z 2  z 2
z1  z 2  z1  z 2
(7)
Ch17_9
Fig 17.2
Ch17_10
17.2 Powers and Roots
Polar Form
Referring to Fig 17.3, we have
z = r(cos  + i sin )
(1)
where r = |z| is the modulus of z and  is the
argument of z,  = arg(z). If  is in the interval − < 
 , it is called the principal argument, denoted by
Arg(z).
Ch17_11
Fig 17.3
Ch17_12
Example 1
Express 1 
3i in polar form.
Solution
See Fig 17.4 that the point lies in the fourth quarter.
r  z  1
tan  

3i 
3
1 3  2
,   arg( z ) 
1
5
3
5
5 

z  2 cos
 i sin

3
3 
Ch17_13
Example 1 (2)
In addition, choose that − <   , thus  = −/3.

 

z  2 cos(  )  i sin(  )

3
3 
Ch17_14
Fig 17.4
Ch17_15
Multiplication and Division
z1  r1 (cos  1  i sin  1 )
 Suppose
z 2  r2 (cos  2  i sin  2 )
Then
z1 z 2  r1 r2 [(cos  1 cos  2  sin  1 sin  2 )
 i (sin  1 cos  2  cos  1 sin  2 )]
(2)
for z2  0,
z1
z2

r1
r2
[(cos  1 cos  2  sin  1 sin  2 )
 i (sin  1 cos  2  cos  1 sin  2 )]
(3)
Ch17_16
From the addition formulas from trigonometry,
z1z2  r1r2 [cos(1   2 )  i sin(1   2 )]
z1
z2

r1
r2
[cos(1   2 )  i sin(1   2 )]
(4)
(5)
Thus we can show
| z1 z2 |  | z1 | | z2 | ,
z1
z2
,
| z1 |
| z2 |
(6)
 z1 
arg ( z1 z2 )  arg z1  arg z2 , arg    arg z1  arg z2 (7)
 z2 
Ch17_17
Powers of z
z  r (cos 2  i sin 2 )
2
2
z  r (cos 3  i sin 3 )
3
3
z  r (cos n   i sin n  )
n
n
(8)
Ch17_18
Demoivre’s Formula
When r = 1, then (8) becomes
(cos   i sin  )  cos n   i sin n
n
(9)
Ch17_19
Roots
A number w is an nth root of a nonzero number z if
wn = z. If we let w =  (cos  + i sin ) and
z = r (cos  + i sin ), then
 (cos n   i sin n  )  r (cos   i sin  )
n
  r,   r
n
1/ n
cos n   cos  , sin n   sin 
 
  2 k
, k  0 ,1, 2 ,..., n  1
n
The root corresponds to k=0 called the principal nth root.
Ch17_20
Thus the n roots of a nonzero complex number
z = r (cos  + i sin ) are given by
wk  r
   2k 
   2k
cos 
  i sin 

n
n




1/ n 



(10)
where k = 0, 1, 2, …, n – 1.
Ch17_21
17.3 Sets in the Complex Plane
Terminology
z  x  iy , z 0  x 0  iy 0
z  z0 
( x  x0 )  ( y  y 0 )
2
2
If z satisfies |z – z0| = , this point lies on a circle of
radius  centered at the point z0.
Ch17_22
Example 1
(a) |z| = 1 is the equation of a unit circle centered at the
origin.
(b) |z – 1 – 2i|= 5 is the equation of a circle of radius 5
centered at 1 + 2i.
Ch17_23
If z satisfies |z – z0| < , this point lies within (not on)
a circle of radius  centered at the point z0. The set is
called a neighborhood of z0, or an open disk.
A point z0 is an interior point of a set S if there exists
some neighborhood of z0 that lies entirely within S.
If every point of S is an interior point then S is an
open set. See Fig 17.7.
Ch17_24
Fig 17.7
Ch17_25
Fig 17.8
The graph of |z – (1.1 + 2i)| < 0.05 is shown in Fig
17.8. It is an open set.
Ch17_26
Fig 17.9
The graph of Re(z)  1 is shown in Fig 17.9. It is not
an open set.
Ch17_27
Example 2
Fig 17.10 illustrates some additional open sets.
Ch17_28
Example 2 (2)
Ch17_29
If every neighborhood of z0 contains at least one point
that is in a set S and at least one point that is not in S,
z0 is said to be a boundary point of S. The boundary
of S is the set of all boundary points.
If any pair of points z1 and z2 in an open set S can be
connected by a polygonal line that lies entirely in S is
said to be connected. See Fig 17.11. An open
connected set is called a domain.
Ch17_30
Fig17.11
Ch17_31
A region is a domain in the complex plane with all,
some or none of its boundary points. Since an open
connected set does not contain any boundary points, it
is a region. A region containing all its boundary
points is said to be closed.
Ch17_32
17.4 Functions of a Complex Variable
Complex Functions
w  f ( z )  u ( x , y )  iv ( x , y )
(1)
where u and v are real-valued functions.
Also, w = f(z) can be interpreted as a mapping or
transformation from the z-plane to the w-plane. See
Fig 17.12.
Ch17_33
Fig 17.12
Ch17_34
Example 1
Find the image of the line Re(z) = 1 under f(z) = z2.
Solution
f ( z )  z  ( x  iy )
2
2
u ( x , y )  x  y , v ( x , y )  2 xy
2
2
Now Re(z) = x = 1, u = 1 – y2, v = 2y.
y  v / 2 , then
u 1 v /4
2
See Fig 17.13.
Ch17_35
Fig 17.13
Ch17_36
DEFINITION 17.4
Limit of a Function
Suppose the function f is defined in some neighborhood
of z0, except possibly at z0 itself. Then f is said to
possess a limit at z0, written
lim f ( z )  L
z  z0
if, for each  > 0, there exists a  > 0 such that
f ( z )  L   whenever 0  | z  z 0 |   .
Ch17_37
THEOREM 17.1
Suppose lim
Then
(i)
(ii)
z  z0
Limit of Sum, Product, Quotient
f ( z )  L1 and lim z  z 0 g ( z )  L 2 .
lim [ f ( z )  g ( z )]  L1  L 2
z  z0
lim f ( z ) g ( z )  L1 L 2
z  z0
(iii) lim
z  z0
f (z)
g (z)

L1
L2
, L2  0
Ch17_38
DEFINITION 17.5
Continuous Function
A function f is continuous at a point z0 if
lim f ( z )  f ( z 0 )
z  z0
A function f defined by
f ( z )  a n z  a n 1 z
n
n 1
   a 2 z  a1 z  a 0 , a n  0
2
(2)
where n is a nonnegative integer and ai, i = 0, 1, 2, …,
n, are complex constants, is called a polynomial of
degree n.
Ch17_39
DEFINITION 17.6
Derivative
Suppose the complex function f is defined in a
neighborhood of a point z0. The derivative of f at z0 is
f ( z0 )  lim
f ( z0  z )  f ( z0 )
z 0
z
(3)
provided this limit exists.
If the limit in (3) exists, f is said to be differentiable at
z0. Also,
if f is differentiable at z0, then f is continuous at z0.
Ch17_40
Rules of differentiation
Constant Rules:
d
c0,
dz
d
cf ( z )  cf ( z )
(4)
dz
Sum Rules:
d
[ f ( z )  g ( z )]  f ( z )  g ( z )
(5)
dz
Product Rule:
d
[ f ( z ) g ( z )]  f ( z ) g ( z )  g ( z ) f ( z )
(6)
dz
Ch17_41
Quotient Rule:
d  f ( z )  g ( z ) f ( z )  f ( z ) g ( z )

2


d z  g ( z) 
[ g ( z )]
(7)
Chain Rule:
d
f ( g ( z ))  f ( g ( z )) g ( z )
(8)
dz
Usual rule
d
z  nz
n
n 1
, n an integer
(9)
dz
Ch17_42
Example 3
Differentiate ( a ) f ( z )  3 z 4  5 z 3  2 z , ( b ) f ( z ) 
z
2
4z  1
.
Solution
( a ) f '( z )  12 z  15 z  2
3
2
(4 z  1)2 z  z 4
2
( b ) f '( z ) 
(4 z  1)
2
4z  2z
2

(4 z  1)
2
Ch17_43
Example 4
Show that f(z) = x + 4iy is nowhere differentiable.
Solution
With  z   x  i  y , we have
f ( z  z)  f ( z)
 ( x   x )  4 i ( y   y )  x  4 iy
And so
lim
z 0
f ( z  z)  f ( z)
z
 lim
z 0
 x  4 i y
 x  i y
(10)
Ch17_44
Example 4 (2)
Now if we let z0 along a line parallel to the x-axis
then y=0 and the value of (10) is 1. On the other hand,
if we let z0 along a line parallel to the y-axis then
x=0 and the value of (10) is 4. Therefore f(z) is not
differentiable at any point z.
Ch17_45
DEFINITION 17.7
Analyticity at a Point
A complex function w = f(z) is said to be analytic at
a point z0 if f is differentiable at z0 and at every point
in some neighborhood of z0.
A function is analytic at every point z is said to be an
entire function. Polynomial functions are entire
functions.
Ch17_46
17.5 Cauchy-Riemann Equations
THEOREM 17.2
Cauchy-Riemann Equations
Suppose f(z) = u(x, y) + iv(x, y) is differentiable at a
point z = x + iy. Then at z the first-order partial
derivatives of u and v exists and satisfy the
Cauchy-Riemann equations
u
x

v
y
and
u
y

v
x
(1)
Ch17_47
THEOREM 17.2 Proof
Proof
Since f ’(z) exists, we know that
f  ( z )  lim
f ( z  z)  f ( z)
z
(2)
By writing f(z) = u(x, y) + iv(x, y), and z = x + iy,
form (2)
z 0
f ( z )
 lim
z 0
(3)
u ( x   x , y   y )  iv ( x   x , y   y )  u ( x , y )  iv ( x , y )
 x  i y
Ch17_48
THEOREM 17.2 Proof (2)
Since the limit exists, z can approach zero from any
direction. In particular, if z0 horizontally, then z =
x and (3) becomes
f  ( z )  lim
u ( x  x, y )  u ( x, y )
x
x 0
 i lim
(4)
v( x  x, y )  v( x, y )
x
x 0
By the definition, the limits in (4) are the first partial
derivatives of u and v w.r.t. x. Thus
f ( z ) 
u
x
i
v
x
(5)
Ch17_49
THEOREM 17.2 Proof (3)
Now if z0 vertically, then z = iy and (3) becomes
f  ( z )  lim
u ( x, y  y )  u ( x, y )
i y
y  0
 lim
iv ( x , y   y )  iv ( x , y )
(6)
i y
y  0
which is the same as
f ( z )   i
u
y

v
y
(7)
Then we complete the proof.
Ch17_50
Example 1
The polynomial f(z) = z2 + z is analytic for all z and
f(z) = x2 − y2 + x + i(2xy + y). Thus u = x2 − y2 + x, v
= 2xy + y. We can see that
u
x
u
y
 2x 1 
 2 y  
v
y
v
x
Ch17_51
Example 2
Show that f(z) = (2x2 + y) + i(y2 – x) is not analytic at
any point.
Solution
u
x
u
y
 4 x and
1
and
v
y
v
x
 2y
 1
We see that u/y = −v/x but u/x = v/y is
satisfied only on the line y = 2x. However, for any z on
this line, there is no neighborhood or open disk about z
in which f is differentiable. We conclude that f is
nowhere analytic.
Ch17_52
THEOREM 17.3
Criterion for Analyticity
Suppose the real-valued function u(x, y) and v(x, y) are
continuous and have continuous first-order partial
derivatives in a domain D. If u and v satisfy the
Cauchy-Riemann equations at all points of D, then the
complex function f(z) = u(x, y) + iv(x, y) is analytic in D.
Ch17_53
Example 3
f (z) 
For the equation
u
x
u
y
y x
2

2
(x  y )
 
2
2 2

2 xy
(x  y )
2
2 2
x
x  y
2
2
i
y
x  y
2
2
, we have
v
y
 
v
x
That is, the Cauchy-Riemann equations are satisfied
except at the point x2 + y2 = 0, that is z = 0. We
conclude that f is analytic in any domain not containing
the point z = 0.
Ch17_54
From (5) and (7), we have
f ( z ) 
u
x
i
v
x

v
y
i
u
y
(8)
This is a formula to compute f ’(z) if f(z) is
differentiable at the point z.
Ch17_55
DEFINITION 17.8
Harmonic Functions
A real-valued function (x, y) that has continuous
second-order partial derivatives in a domain D and
satisfies Laplace’s equation is said to be harmonic in D.
THEOREM 17.4
A Source of Harmonic Functions
Suppose f(z) = u(x, y) + iv(x, y) is analytic in a domain D.
Then the functions u(x, y) and v(x, y) are harmonic
functions.
Ch17_56
THEOREM 17.4
Proof
we assume u and v have continuous second order derivative
u
v u
v

,
 
, then
x y y
x
 u
2
x
2
 v

2
and
xy
 u
2
Thus
 u
2
x
2
 u
y
2
 v
2
 
yx
2

y
2
0
Similarly we have
 v
2
x
2
 v
2

y
2
0
Ch17_57
Conjugate Harmonic Functions
If u and v are harmonic in D, and u(x,y)+iv(x,y) is an
analytic function in D, then u and v are called the
conjugate harmonic function of each other.
Ch17_58
Example 4
(a) Verify u(x, y) = x3 – 3xy2 – 5y is harmonic in the
entire complex plane.
(b) Find the conjugate harmonic function of u.
Solution
(a )
u
x
2
 3x  3 y ,
2
 u
2
x
 u
2
2
 u
x
2
 6 x,
u
y
 u
2
  6 xy  5 ,
y
2
 6 x
2

y
2
 6x  6x  0
Ch17_59
Example 4 (2)
(b )
v
y

u
x
 3x  3 y
2
2
and
v
x
 
u
y
 6xy  5
Integratin g the first one, v ( x, y )  3 x y  y  h ( x )
2
and
v
x
3
 6 xy  h ' ( x ), h ' ( x )  5 , h ( x )  5 x  C
Thus v ( x,y )  3 x y  y  5 x  C
2
3
Ch17_60
17.6 Exponential and Logarithmic Functions
Exponential Functions
Recall that the function f(x) = ex has the property
f  ( x )  f ( x ) and
f ( x1  x 2 )  f ( x1 ) f ( x 2 )
(1)
and the Euler’s formula is
e
iy
 cos y  i sin y ,
Thus
e
x  iy
y : a real num ber
(2)
 e (cos y  i sin y )
x
Ch17_61
DEFINITION 17.9
e e
z
Exponential Functions
x iy
 e (cos y  i sin y )
x
(3)
Ch17_62
Example 1
Evaluate e1.7+4.2i.
Solution
e
1 .7  4 .2 i
e
1 .7
(cos 4 . 2  i sin 4 . 2 )
  2 . 6873  4 . 7710 i
Ch17_63
Also we have
de
z
e
z
dz
z1
e e
z2
e
z1  z2
,
e
z1
e
z2
e
z1  z2
Ch17_64
Periodicity
e
z  i 2
e e
z
i 2
 e (cos 2   i sin 2  )  e
z
z
Ch17_65
Polar From of a Complex number
z  r (cos   i sin  )  re
i
Ch17_66
Logarithm Function
Given a complex number z = x + iy, z  0, we define
w = ln z if z = ew
(5)
Let w = u + iv, then
x  iy  e
u  iv
 e (cos v  i sin v )  e cos v  ie sin v
u
u
u
We have
x  e cos v , y  e sin v
u
u
and also
e
2u
 x  y  r  | z | , u  log e | z |
tan v 
2
2
y
2
2
, v    2 n ,   arg z , n  0,  1,  2,...
x
Ch17_67
DEFINITION 17.10
Logarithm of a Complex Number
For z  0, and  = arg z,
ln z  log e | z | i (  2n ) , n  0,  1,  2, 
(6)
Ch17_68
Example 2
Find the values of (a) ln (−2) (b) ln i, (c) ln (−1 – i ).
Solution
( a )   arg(  2 )   , log
e
|  2 | 0 . 6932
ln(  2 )  0 . 6932  i (  2 n  )
( b )   arg( i ) 

2
, log e 1  0

 2n )
2
5
( c )   arg(  1  i ) 
, log
4
ln( i )  i (
ln(  1  i )  0 . 3466  i (
5
4
e
|  1  i | log
e
2  0 . 3466
 2n )
Ch17_69
Example 3
Find all values of z such that e z 
Solution
z  ln(
3  i ), |
3  i | 2 , arg(
3  i.
3  i) 

6
z  ln(
3  i )  log e 2  i (
 0 . 6931  i (


 2n )
6
 2n )
6
Ch17_70
Principal Value

Ln z  log e | z | iArg z
(7)
Since Arg z  (   ,  ]is unique, there is only one value
of Ln z for which z  0.
Ch17_71
Example 4
The principal values of example 2 are as follows.
( a ) Arg (  2 )  
Ln (  2 )  0 . 6932  i 
( b ) Arg ( i ) 

, Ln ( i )  i
2
( c ) Arg (  1  i ) 

2
5
is not the principal
value.
4
Let n   1, then Ln (  1  i )  0 . 3466 
3
i
4
Ch17_72
Example 4 (2)
Each function in the collection of ln z is called a
branch. The function Ln z is called the principal
branch or the principal logarithm function.
Some familiar properties of logarithmic function hold
in complex case:
ln( z1 z 2 )  ln z1  ln z 2
ln(
z1
z2
)  ln z1  ln z 2
(8)
Ch17_73
Example 5
Suppose z1 = 1 and z2 = −1. If we take ln z1 = 2i,
ln z2 = i, we get
ln( z1 z 2 )  ln(  1)  ln z1  ln z 2  3 i
ln(
z1
z2
)  ln(  1)  ln z1  ln z 2   i
Ch17_74
Analyticity
The function Ln z is not analytic at z = 0, since Ln 0
is not defined. Moreover, Ln z is discontinuous at all
points of the negative real axis. Since Ln z is the
principal branch of ln z, the nonpositive real axis is
referred to as a branch cut. See Fig 17.19.
Ch17_75
Fig 17.91
Ch17_76
It is left as exercises to show
d
dz
Ln z 
1
z
(9)
for all z in D (the complex plane except those on the
non-positive real axis).
Ch17_77
Complex Powers
In real variables, we have x   e  ln x .
If  is a complex number, z = x + iy, we have

 ln z
z e
, z0
(10)
Ch17_78
Example 6
Find the value of i2i.
Solution
With z  i , arg z   / 2 ,   2 i , from (9),
i
2i
e
2 i [log e 1  i (  / 2  2 n  )]
e
 (1  4 n ) 
where n  0 ,  1,  2 ,...
Ch17_79
17.7 Trigonometric and Hyperbolic Functions
Trigonometric Functions
From Euler’s Formula, we have
e e
ix
sin x 
2i
ix
e e
ix
cos x 
ix
(1)
2
Ch17_80
DEFINITION 17.11
Trigonometric Sine and Cosine
For any complex number z = x + iy,
sin z 
e
iz
e
 iz
e e
iz
cos z 
2i
 iz
(2)
2
Four additional trigonometric functions:
tan z 
sec z 
sin z
, cot z 
1
cos z
tan z
1
1
cos z
, csc z 
,
(3)
sin z
Ch17_81
Analyticity
Since eiz and e-iz are entire functions, then sin z and
cos z are entire functions. Besides, sin z = 0 only for
the real numbers z = n and cos z = 0 only for the real
numbers z = (2n+1)/2. Thus tan z and sec z are
analytic except z = (2n+1)/2, and cot z and
csc z are analytic except z = n.
Ch17_82
Derivatives

d
d e e
iz
sin z 
dz
dz
 iz
e e
iz

2i
 iz
 cos z
2
Similarly we have
d
dz
d
dz
d
dz
sin z  cos z
tan z  sec z
2
sec z  sec z tan z
d
dz
d
dz
d
cos z   sin z
cot z   csc z
2
(4)
csc z   csc z cot z
dz
Ch17_83
Identities
sin(  z )   sin z
cos
2
z  sin
2
cos (  z )  cos z
z 1
sin( z1  z 2 )  sin z1 cos z 2  cos z1 sin z 2
cos( z1  z 2 )  cos z1 cos z 2  sin z1 sin z 2
sin 2 z  2 sin z cos z
cos 2 z  cos
2
z  sin
2
z
Ch17_84
Zeros
If y is real, we have
e e
y
sinh y 
y
e e
y
and
cosh y 
2
y
(5)
2
From Definition 11.17 and Euler’s formula
sin z 
e
i ( x  iy )
e
 i ( x  iy )
2i
e e
y
 sin x (
2
y
e e
y
)  i cos x (
y
)
2
Ch17_85
Thus we have
sin z  sin x cosh y  i cos x sinh y
(6)
cos z  cos x cosh y  i sin x sinh y
(7)
and
From (6) and (7) and cosh2y = 1 + sinh2y
| sin z |  sin
2
| cos z |  cos
2
x  sinh
2
2
x  sinh
2
y
2
y
(8)
(9)
Ch17_86
Example 1
From (6) we have
sin( 2  i )  sin 2 cosh 1  i cos 2 sinh 1
 1 . 4301  0 . 4891 i
Ch17_87
Example 2
Solve cos z = 10.
Solution
e e
iz
cos z 
 iz
 10
2
e
2 iz
 20 e  1  0 , e
iz
iz
 10  3 11
iz  log e (10  3 11 )  2 n  i
Since log e (10  3 11 )   log e (10  3 11 )
we have
z  2 n   i log e (10  3 11 )
Ch17_88
DEFINITION 17.12
Hyperbolic Sine and Cosine
For any complex number z = x + iy,
e e
z
sinh z 
z
e e
z
cosh z 
2
z
(10)
2
Additional functions are defined as
tanh z 
sinh z
cosh z
sec h z 
1
cosh z
1
coth z 
tanh z
csc h z 
(11)
1
sinh z
Ch17_89
Similarly we have
d
dz
sinh z  cosh z and
d
cosh z  sinh z
(12)
dz
sin z   i sinh( iz ) , cos z  cosh( iz )
(13)
sinh z   i sin( iz ) , cosh z  cos( iz )
(14)
Ch17_90
Zeros
 sinh z   i sin iz   i sin(  y  ix )
  i[sin(  y ) cosh x  i cos(  y ) sinh x ]
Since sin(−y) = − sin y, cos(−y) = cos y, then
sinh z  sinh x cos y  i cosh x sin y
(15)
cosh z  cosh x cos y  i sinh x sin y
(16)
It also follows from (14) that the zeros of sinh z and
cosh z are respectively,
z = ni and z = (2n+1)i/2, n = 0, 1, 2, ….
Ch17_91
Periodicity
From (6),
sin( z  2  i )
 sin( x  iy  2  )
 sin( x  2  ) cosh y  i cos( x  2  ) sinh y
 sin x cosh y  i cos x sinh y  sin z
The period is then 2.
Ch17_92
17.8 Inverse Trigonometric and Hyperbolic
Functions
Inverse Sine
We define
z  sin w
w  sin
if
1
(1)
z
From (1),
e
iw
e
 iw
 z, e
2 iw
 2 ize
iw
1  0
2i
e
iw
 iz  (1  z )
2 1/ 2
(2)
Ch17_93
Solving (2) for w then gives
sin
1
z  i ln[iz  (1  z )
2 1/ 2
(3)
]
Similarly we can get
cos
1
z  i ln[ z  i (1  z )
2 1/ 2
tan
1
z
i
2
ln
iz
iz
]
(4)
(5)
Ch17_94
Example 1
Find all values of sin
Solution
From (3),
sin
1
(1  ( 5 ) )
1
5.
5   i ln[
2 1/ 2
sin
1
 (4)
1/ 2
5   i ln[(
5i  (1  ( 5 ) )
2 1/ 2
]
  2i
5  2 )i ]
  i[log e ( 5  2 )  (

 2 n  ) i ],
2
n  0 ,  1,  2 ,...
Ch17_95
Example 1 (2)
Noting that
log e ( 5  2 )  log
1
e
52
  log e ( 5  2 ).
Thus for n  0 ,  1,  2 ,...
sin
1
5

2
 2 n   i log e ( 5  2 )
(6)
Ch17_96
Derivatives
If we define w = sin-1z, z = sin w, then
d
dz
z 
d
sin w
g ives
dz
dw
dz
U sin g co s w  sin w  1,
2
 (1  z )
2
1/ 2
,
2
1

co s w
co s w  (1  sin w )
2
1/ 2
th u s
d
sin
1
z
dz
d
cos
dz
d
dz
1
tan
1
z
z
1
(1  z )
2 1/ 2
1
2 1/ 2
(8)
2
(9)
(1  z )
1
1 z
(7)
Ch17_97
Example 2
Find the derivative of w = sin-1 z at z = 5 .
Solution
(1  ( 5 ) )
2 1/ 2
dw
dz
z 5

 (4)
1/ 2
 2i
1
(1  ( 5 ) )
2 1/ 2

1
2i
 
1
i
2
Ch17_98
Inverse Hyperbolic Functions
Similarly we have
sinh
cosh
1
z  ln[ z  ( z  1)
]
1
z  ln[ z  ( z  1)
]
2
1
tanh z 
2
1
ln
2
d
dz
sinh
1
z 
1 z
1 z
1
( z  1)
2
1/ 2
1/ 2
1/ 2
(10)
(11)
(12)
(13)
Ch17_99
d
cosh
1
z 
dz
d
dz
tanh
1
z
1
( z  1)
2
1/ 2
1
1 z
(14)
(15)
2
Ch17_100
Example 3
Find all values of cosh-1(−1).
Solution
From (11),
cosh
1
(  1)  ln(  1)  log e 1  (  2 n  ) i
 (   2 n  ) i  ( 2 n  1)  i
n  0 ,  1,  2 ,...
Ch17_101