Download + z

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  iy1 and z2  x2  iy2 are
equal, z1  z2, if Re( z1 )  Re( z2 ) and Im( z1 )  Im( z2 )
x + iy = 0 iff x = 0 and y = 0.
Ch17_3
Arithmetic Operations
Suppose z1  x1  iy1, z2  x2  iy2
z1  z2  ( x1  x2 )  i ( y1  y2 )
z1  z2  ( x1  x2 )  i ( y1  y2 )
z1  z2  ( x1x2  y1 y2 )  i ( y1x2  x1 y2 )
z1 x1x2  y1 y2 y1x2  x1 y2
 2
i 2
2
2
z2 x2  y2
x2  y2
Ch17_4
Complex Conjugate
Suppose z  x  iy, z  x  iy, and
z1  z2  z1  z2
z1  z2  z1  z2
z1 z2  z1 z2
 z1

 z2
 z1

 z2
Ch17_5
Two important equations
z  z  ( x  iy )  ( x  iy )  2 x
(1)
zz  ( x  iy)( x  iy)  x 2  i 2 y 2  x 2  y 2
(2)
z  z  ( x  iy )  ( x  iy )  2iy
and
(3)
zz
zz
Re( z ) 
, Im( z ) 
2
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 2  y 2  zz
Ch17_8
Example 3
If z = 2 − 3i, then
z  22  (3)2  13
As in Fig 17.2, the sum of the vectors z1 and z2 is the
vector z1 + z2. Then we have
z1  z2  z1  z2
(5)
The result in (5) is also known as the triangle inequality
and extends to any finite sum:
z1  z2  ...  zn  z1  z2  ...  zn
(6)
Using (5),
z1  z2  ( z2 )  z1  z2  z2
z1  z2  z1  z2
(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
Express1  3i in polar form.
Solution
See Fig 17.4 that the point lies in the fourth quarter.
r  z  1  3i  1  3  2
 3
5
tan  
,  arg( z ) 
1
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
 Suppose z1  r1 (cos 1  i sin 1 )
z2  r2 (cos  2  i sin  2 )
Then
z1 z2  r1r2 [(cos 1 cos  2  sin 1 sin  2 )
 i (sin 1 cos  2  cos 1 sin  2 )]
for z2  0,
z1 r1
 [(cos 1 cos  2  sin 1 sin  2 )
z2 r2
 i (sin 1 cos  2  cos 1 sin  2 )]
(2)
(3)
Ch17_16
From the addition formulas from trigonometry,
z1z2  r1r2[cos(1   2 )  i sin(1   2 )]
(4)
z1 r1
 [cos(1   2 )  i sin(1   2 )]
z2 r2
(5)
Thus we can show
z1
| z1 |
| z1z2 |  | z1 | | z2 | ,
,
(6)
z2
| z2 |
 z1 
arg ( z1z2 )  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 n r n (cos n  i sin n )
(8)
Ch17_18
Demoivre’s Formula
When r = 1, then (8) becomes
(cos   i sin  )n  cos n   i sin 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
 n (cos n  i sin n )  r (cos   i sin  )
 n  r ,   r1 / n
cos n  cos  , sin n  sin 
  2k

, 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  n   i sin  n 
1/ n 
(10)
where k = 0, 1, 2, …, n – 1.
Ch17_21
17.3 Sets in the Complex Plane
Terminology
z  x  iy, z0  x0  iy0
z  z0  ( x  x0 ) 2  ( y  y0 ) 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 2  ( x  iy ) 2
u ( x, y )  x 2  y 2 , v( x, y )  2 xy
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  z0 |   .
Ch17_37
THEOREM 17.1
Limit of Sum, Product, Quotient
Suppose limzz0 f ( z )  L1 and limzz0 g ( z )  L2 .
Then
(i)
(ii)
lim [ f ( z )  g ( z )]  L1  L2
z  z0
lim f ( z ) g ( z )  L1L2
z  z0
f ( z ) L1
(iii) lim

, L2  0
z  z0 g ( z )
L2
Ch17_38
DEFINITION 17.5
Continuous Function
A function f is continuous at a point z0 if
lim f ( z )  f ( z0 )
z  z0
A function f defined by
f ( z )  an z n  an1z n1    a2 z 2  a1z  a0 , an  0 (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  z )  f ( z0 )
f ( z0 )  lim
(3)
z 0
z
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
d
c0,
cf ( z )  cf ( z )
dz
dz
Sum Rules:
d
[ f ( z )  g ( z )]  f ( z )  g ( z )
dz
Product Rule:
d
[ f ( z ) g ( z )]  f ( z ) g ( z )  g ( z ) f ( z )
dz
(4)
(5)
(6)
Ch17_41
Quotient Rule:
d  f ( z )  g ( z ) f ( z )  f ( z ) g ( z )



d z  g ( z) 
[ g ( z )]2
Chain Rule:
d
f ( g ( z ))  f ( g ( z )) g ( z )
dz
Usual rule
d n
z  nz n1 , n an integer
dz
(7)
(8)
(9)
Ch17_42
Example 3
2
z
4
3
.
Differentiate (a ) f ( z )  3 z 5 z 2 z , (b) f ( z ) 
4z  1
Solution
(a) f '( z )  12 z 3  15 z 2  2
(4 z  1)2 z  z 4 4 z  2 z
(b) f '( z ) 

2
(4 z  1)
(4 z  1) 2
2
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)  4i ( y  y )  x  4iy
And so
f ( z  z )  f ( z )
x  4iy
lim
 lim
z 0
z 0 x  iy
z
(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 v
u
v

and

(1)
x y
y
x
Ch17_47
THEOREM 17.2 Proof
Proof
Since f ’(z) exists, we know that
f ( z  z )  f ( z )
f ( z )  lim
z 0
(2)
z
By writing f(z) = u(x, y) + iv(x, y), and z = x + iy,
form (2)
f ( z )
(3)
u ( x  x, y  y )  iv ( x  x, y  y )  u ( x, y )  iv ( x, y )
 lim
z 0
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
u ( x  x, y )  u ( x, y )
f ( z )  lim
x 0
x
(4)
v( x  x, y )  v( x, y )
 i lim
x 0
x
By the definition, the limits in (4) are the first partial
derivatives of u and v w.r.t. x. Thus
u v
f ( z )   i
(5)
x x
Ch17_49
THEOREM 17.2 Proof (3)
Now if z0 vertically, then z = iy and (3) becomes
u ( x , y  y )  u ( x , y )
f ( z )  lim
y  0
i y
iv( x, y  y )  iv( x, y )
 lim
y  0
i y
which is the same as
u v
f ( z )  i 
y y
(6)
(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
v
 2x  1 
x
y
u
v
 2 y  
y
x
Ch17_51
Example 2
Show that f(z) = (2x2 + y) + i(y2 – x) is not analytic at
any point.
Solution
u
v
 4 x and
 2y
x
y
u
v
 1 and
 1
y
x
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
x
y
For the equation f ( z )  2
i 2
, we have
2
2
x y
x y
u
y 2  x2
v
 2

2 2
x ( x  y )
y
u
2 xy
v
 2

2 2
y
x
(x  y )
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
u v v u
f ( z )   i   i
x x y 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
 2u  2 v

2
xy
x
 2u
 2v
and

2
yx
y
 2u  2u
Thus
 2 0
2
x y
Similarly we have
 2v  2v
 2 0
2
x y
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
2
2
u

u

u

u
2
2
(a)
 3 x  3 y , 2  6 x,
 6 xy  5, 2  6 x
x
y
x
y
 2u  2u
 2  6x  6x  0
2
x y
Ch17_59
Example 4 (2)
v u
v
u
2
2
(b)

 3 x  3 y and

 6xy  5
y x
x
y
Integratin g the first one, v( x, y )  3 x 2 y  y 3  h( x)
v
and
 6 xy  h' ( x), h' ( x)  5, h( x)  5 x  C
x
Thus v( x,y )  3 x 2 y  y 3  5 x  C
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  x2 )  f ( x1 ) f ( x2 )
and the Euler’s formula is
eiy  cos y  i sin y,
Thus
y : a real number
(1)
(2)
e xiy  e x (cos y  i sin y)
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
e1.7 4.2i  e1.7 (cos 4.2  i sin 4.2)
 2.6873  4.7710i
Ch17_63
Also we have
dez
 ez
dz
e e e
z1 z2
z1  z2
z1
e
, z2  e z1  z2
e
Ch17_64
Periodicity
e z i 2  e z ei 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  eu iv  eu (cos v  i sin v)  eu cos v  ieu sin v
We have
x  eu cos v, y  eu sin v
and also
e2u  x 2  y 2  r 2 | z |2 , u  log e | z |
y
tan v  , v    2n ,   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 (  2n )
(b)   arg( i ) 


2
, log e 1  0
ln(i )  i (  2n )
2
5
(c)   arg( 1  i )  , log e | 1  i | log e 2  0.3466
4
5
ln( 1  i )  0.3466  i (  2n )
4
Ch17_69
Example 3
Find all values of z such that e z  3  i.
Solution
z  ln( 3  i ), | 3  i | 2, arg( 3  i ) 

6

z  ln( 3  i )  log e 2  i (  2n )
6

 0.6931  i (  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 ) 

2
, Ln (i )  i

2
5
(c) Arg (1  i ) 
is not the principal value.
4
3
Let n  1, then Ln (1  i )  0.3466  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 z2 )  ln z1  ln z2
z1
ln( )  ln z1  ln z2
z2
(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 z2 )  ln( 1)  ln z1  ln z2  3i
z1
ln( )  ln( 1)  ln z1  ln z2  i
z2
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
1
Ln z 
dz
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
z  e ln z , z  0
(10)
Ch17_78
Example 6
Find the value of i2i.
Solution
With z  i, arg z   / 2,   2i, from (9),
i e
2i
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
eix  eix
sin x 
2i
eix  eix
cos x 
2
(1)
Ch17_80
DEFINITION 17.11
Trigonometric Sine and Cosine
For any complex number z = x + iy,
eiz  e iz
sin z 
2i
eiz  e iz
cos z 
2
Four additional trigonometric functions:
sin z
1
tan z 
, cot z 
,
cos z
tan z
1
1
sec z 
, csc z 
cos z
sin z
(2)
(3)
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 eiz  e iz eiz  e iz
sin z 

 cos z
dz
dz 2i
2
Similarly we have
d
sin z  cos z
dz
d
tan z  sec 2 z
dz
d
sec z  sec z tan z
dz
d
cos z   sin z
dz
d
(4)
cot z   csc 2 z
dz
d
csc z   csc z cot z
dz
Ch17_83
Identities
sin(  z )   sin z
cos ( z )  cos z
cos 2 z  sin 2 z  1
sin( z1  z2 )  sin z1 cos z2  cos z1 sin z2
cos( z1  z2 )  cos z1 cos z2  sin z1 sin z2
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
y
y
y
y
e e
e e
sinh y 
and cosh y 
2
2
From Definition 11.17 and Euler’s formula
ei ( xiy )  e i ( xiy )
sin z 
2i
e y  e y
e y  e y
 sin x(
)  i cos x(
)
2
2
(5)
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 |2  sin2 x  sinh2 y
(8)
| cos z |2  cos 2 x  sinh2 y
(9)
Ch17_86
Example 1
From (6) we have
sin( 2  i )  sin 2 cosh 1  i cos 2 sinh 1
 1.4301  0.4891i
Ch17_87
Example 2
Solve cos z = 10.
iz
iz
Solution
e e
cos z 
 10
2
e 2iz  20eiz  1  0, eiz  10  3 11
iz  log e (10  3 11)  2ni
Since log e (10  3 11)   log e (10  3 11)
we have
z  2n  i log e (10  3 11)
Ch17_88
DEFINITION 17.12
Hyperbolic Sine and Cosine
For any complex number z = x + iy,
e z  e z
sinh z 
2
e z  e z
cosh z 
2
Additional functions are defined as
sinh z
1
tanh z 
coth z 
cosh z
tanh z
1
1
sec h z 
csc h z 
cosh z
sinh z
(10)
(11)
Ch17_89
Similarly we have
d
sinh z  cosh z and d cosh z  sinh z
dz
dz
(12)
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
From (1),
if
w  sin 1 z
eiw  eiw
2 iw
iw
 z, e  2ize  1  0
2i
eiw  iz  (1  z 2 )1/2
(1)
(2)
Ch17_93
Solving (2) for w then gives
sin 1 z  i ln[iz  (1  z 2 )1/ 2 ]
Similarly we can get
(3)
cos 1 z  i ln[z  i(1  z 2 )1/ 2 ]
(4)
i iz
tan z  ln
2 iz
(5)
1
Ch17_94
Example 1
Find all values of sin 1 5.
Solution
From (3),
sin 1 5  i ln[ 5i  (1  ( 5) 2 )1/ 2 ]
(1  ( 5) )
2 1/ 2
 (4)
1/ 2
 2i
sin 1 5  i ln[( 5  2)i ]

 i[log e ( 5  2)  (  2n )i ],
2
n  0,  1,  2,...
Ch17_95
Example 1 (2)
Noting that
1
log e ( 5  2)  log e
  log e ( 5  2).
52
Thus for n  0,  1,  2,...
sin
1
5

2
 2n  i log e ( 5  2)
(6)
Ch17_96
Derivatives
If we define w = sin-1z, z = sin w, then
d
d
dw
1
z
sin w gives

dz
dz
dz cos w
Using cos 2 w  sin 2 w  1, cos w  (1  sin 2 w)1/ 2
 (1  z 2 )1/ 2 , thus
d
1
1
sin z 
dz
(1  z 2 )1/ 2
d
1
1
cos z 
dz
(1  z 2 )1/ 2
d
1
1
tan z 
dz
1  z2
(7)
(8)
(9)
Ch17_97
Example 2
Find the derivative of w = sin-1 z at z = 5.
Solution
(1  ( 5) 2 )1/ 2  (4)1/ 2  2i
dw
dz
z 5
1
1
1

  i
2 1/ 2
2i
2
(1  ( 5) )
Ch17_98
Inverse Hyperbolic Functions
Similarly we have
sinh1 z  ln[z  ( z 2  1)1/ 2 ]
cosh 1 z  ln[z  ( z 2  1)1/ 2 ]
(10)
(11)
1 1 z
tanh z  ln
2 1 z
(12)
d
1
sinh 1 z  2
1/ 2
dz
( z  1)
(13)
1
Ch17_99
1
d
1
cosh z  2
1/ 2
(
z

1
)
dz
(14)
d
1
1
tanh z 
dz
1  z2
(15)
Ch17_100
Example 3
Find all values of cosh-1(−1).
Solution
From (11),
cosh 1 (1)  ln( 1)  log e 1  (  2n )i
 (  2n )i  (2n  1)i
n  0,  1,  2,...
Ch17_101