Staff Site Universitas Negeri Yogyakarta fungsi bil kompleks
1.
Functions of a Complex Variable
Let E be a set of complex numbers. A function f defined on E is a rule that assigns to each
z in E a complex number w. The number w is called the value of f at z and is denoted by f ( z ) ;
that is, w = f ( z ) . The set E is called the domain of definition of f, written D ( f ) ,
D ( f ) = { z ∈ : f ( z ) defined } .
Determine the domain definition of f ( z ) =
z +1
and g ( z ) = z 2 + z + 1
z + z +1
2
Solution:
D ( f ) = {z ∈
: f ( z ) defined }
D ( g ) = { z ∈ : g ( z ) defined }
= { z ∈ : z 2 + z + 1 ≠ 0}
= z∈
2.
=
1 1
:z ≠− ± i 3
2 2
Limits
Let a function f be defined at all points z in some deleted neighborhood of z0 . The statement
that the limit of f ( z ) as z approaches z0 is a number w0 , or
lim f ( z ) = w0
(2)
z → z0
means that the point w = f ( z ) can be made arbitrarily close to w0 if we choose the point z close
enough to z0 but distinct from it. Equation (2) means that, for each positive number ε , there is a
positive number δ such that
f ( z ) − w0 < ε
whenever
0 < z − z0 < δ .
(3)
Figure 4. Limit function
Show that lim z 2 − 1 = 2
z →i
Solution:
From the properties of modulli, we have
(z
2
− 1) − ( −2 ) = z 2 + 1 = z − i z + i .
Observe that when ∀z ∈ is in the region z − i < 1 ,
z + i = z − i + 2i ≤ z − i + 2i < 1 + 2 = 3 .
Hence, for ∀z ∈ such that z − i < 1 ,
(z
2
− 1) − ( −2 ) = z 2 + 1 = z − i z + i < 3 z − i .
For any positive number ε , get δ = min
(z
2
ε
3
,1 such that 0 < z − i < δ ,
− 1) − ( −2 ) < 3 z − i < 3
ε
3
=ε .
Note that when a limit function f ( z ) exist at a point z0 , it is unique.
If f ( z ) =
z
, then lim f ( z ) does not exist.
z →0
z
When z = ( x, 0 ) is a nonzero point on the real axis,
f ( z) =
x − i0
=1
x + i0
and when z = ( 0, y ) is a nonzero point on the imaginary axis,
f ( z) =
0 − iy
= −1 .
0 + iy
Thus, by letting z approach the origin along real axis, we would find that the desired limit is 1.
An approach along imaginary axis would, on the other hand, yield the limit -1. Since the limit is
unique, we must conclude that lim
z →0
z
does not exist.
z
Since limits of the latter type are studied in calculus, we use their definition and properties
freely.
Suppose that
f ( z ) = u ( x, y ) + iv ( x, y ) ,
z0 = x0 + iy0 , and
w0 = u0 + iv0 .
Then
lim f ( z ) = w0
(4)
z → z0
if and only if
lim
( x , y )→( x0 , y0 )
Find lim z 2 +
z →1+ i
u ( x, y ) = u0
and
1
1
x
y
2
= ( x + iy ) +
= ( x2 − y2 ) + 2
+ i 2 xy − 2
.
2
z
x + iy
x +y
x + y2
We have u ( x, y ) = ( x 2 − y 2 ) +
( x , y )→(1,i )
(x
2
− y2 ) +
x
y
and v ( x, y ) = 2 xy − 2
. From Theorem 1,
2
x +y
x + y2
2
x
1
=
2
2
x +y
2
and
lim
( x , y )→(1,i )
thus
lim z 2 +
z →1+ i
If lim f ( z ) , lim g ( z ) exist and c ∈
z → z0
v ( x, y ) = v0
1
.
z
Observe that z 2 +
lim
lim
( x , y )→( x0 , y0 )
z → z0
, then
1
1 3
= + i.
z
2 2
2 xy −
y
3
= ,
2
2
x +y
2
(5)
(
)
(
)
(1) lim f ( z ) + g ( z ) exist and lim f ( z ) + g ( z ) = lim f ( z ) + lim g ( z )
z → z0
(
)
z → z0
(
)
z → z0
z → z0
(2) lim cf ( z ) exist and lim cf ( z ) = c lim f ( z )
z → z0
(
z → z0
)
z → z0
(
)
(3) lim f ( z ) g ( z ) exist and lim f ( z ) g ( z ) = lim f ( z ) lim g ( z )
z → z0
z → z0
z → z0
z → z0
lim f ( z )
f ( z)
f (z)
z → z0
, whenever lim g ( z ) ≠ 0
=
(4) lim
exist and lim
z → z0 g ( z )
z → z0
z → z0 g ( z )
lim g ( z )
z → z0
Limits Involving the Point at Infinity
We have three point about limits that is involving the point at infinity:
1
= 0.
f ( z)
lim f ( z ) = ∞
if and only if
lim f ( z ) = w0
if and only if
lim f
lim f ( z ) = ∞
if and only if
lim
z → z0
z →∞
z →∞
lim
z → z0
z →0
z →0
1
= w0 .
z
1
f 1
( z)
= 0.
i+3
z +i
= ∞ since lim
= 0.
z →− i z + i
z →− i i + 3
Observe that lim
( )
( )
3 1 −i
3z − i
3 − iz
z
= 3 since lim
= lim
= 3.
z →∞ z + 2
z →0
z
→
0
1 +2
1 + 2z
z
Observe that lim
( )
( )
1 3 +2
3z 4 − i
z + 2z 4
z
= lim
= 0.
Observe that lim 3
= ∞ since lim
z →0
z → 0 3 − iz 4
z →∞ z + 2
3 1 4 −i
z
Exercises
1. For each of the functions below, describe the domain of definition that is understood
(a) f ( z ) =
1
z +4
2
(c) f ( z ) = cos ( z 2 − i )
(6)
(7)
(8)
z + 2i
z+z
(b) f ( z ) =
2. Write the function f ( z ) = z 3 + 2 z − i in the form f ( z ) = u ( x, y ) + iv ( x, y ) .
3. Let z0 , c denote complex constant. Use definition (3) to prove that
(b) lim ( x + i 2 y ) = 1 − 2i
(a) lim c = c
z → z0
z →1− i
z2
4. Let f ( z ) =
z
2
a. Find lim f ( z ) along the line y = x
z →0
b. Find lim f ( z ) along the line y = 2 x
z →0
c. Find lim f ( z ) along the parabola y = x 2
z →0
d. What can you conclude about the limit of f ( z ) along z → 0
5. Using (6), (7) and (8) of limits, show that
(a) lim
z 4 − z3 + 2z
( z + 1)
z →∞
(b) lim
z → 2i
4
z
( z − 2i )
2
=1
z2 −1
=∞
z →∞ z + 1
(c) lim
=∞
6. Find the value of limits below
(a) lim z 2 + 2 z − 1
z →1+ 2 i
z2 + 2z −1
z →−2 i z 2 − 2 z + 4
(c)
(b) lim
z3 + 8
lim 4
2
z →(1+ i 3 ) z + 4 z + 16
lim
(d)
z →i
z4 −1
z −i
lim
(e)
z →i
z2 + 4z + 2
z +1
z 2 + z − 1 − 3i
z →1+ i z 2 − 2 z + 2
(f) lim
Functions of a Complex Variable
Let E be a set of complex numbers. A function f defined on E is a rule that assigns to each
z in E a complex number w. The number w is called the value of f at z and is denoted by f ( z ) ;
that is, w = f ( z ) . The set E is called the domain of definition of f, written D ( f ) ,
D ( f ) = { z ∈ : f ( z ) defined } .
Determine the domain definition of f ( z ) =
z +1
and g ( z ) = z 2 + z + 1
z + z +1
2
Solution:
D ( f ) = {z ∈
: f ( z ) defined }
D ( g ) = { z ∈ : g ( z ) defined }
= { z ∈ : z 2 + z + 1 ≠ 0}
= z∈
2.
=
1 1
:z ≠− ± i 3
2 2
Limits
Let a function f be defined at all points z in some deleted neighborhood of z0 . The statement
that the limit of f ( z ) as z approaches z0 is a number w0 , or
lim f ( z ) = w0
(2)
z → z0
means that the point w = f ( z ) can be made arbitrarily close to w0 if we choose the point z close
enough to z0 but distinct from it. Equation (2) means that, for each positive number ε , there is a
positive number δ such that
f ( z ) − w0 < ε
whenever
0 < z − z0 < δ .
(3)
Figure 4. Limit function
Show that lim z 2 − 1 = 2
z →i
Solution:
From the properties of modulli, we have
(z
2
− 1) − ( −2 ) = z 2 + 1 = z − i z + i .
Observe that when ∀z ∈ is in the region z − i < 1 ,
z + i = z − i + 2i ≤ z − i + 2i < 1 + 2 = 3 .
Hence, for ∀z ∈ such that z − i < 1 ,
(z
2
− 1) − ( −2 ) = z 2 + 1 = z − i z + i < 3 z − i .
For any positive number ε , get δ = min
(z
2
ε
3
,1 such that 0 < z − i < δ ,
− 1) − ( −2 ) < 3 z − i < 3
ε
3
=ε .
Note that when a limit function f ( z ) exist at a point z0 , it is unique.
If f ( z ) =
z
, then lim f ( z ) does not exist.
z →0
z
When z = ( x, 0 ) is a nonzero point on the real axis,
f ( z) =
x − i0
=1
x + i0
and when z = ( 0, y ) is a nonzero point on the imaginary axis,
f ( z) =
0 − iy
= −1 .
0 + iy
Thus, by letting z approach the origin along real axis, we would find that the desired limit is 1.
An approach along imaginary axis would, on the other hand, yield the limit -1. Since the limit is
unique, we must conclude that lim
z →0
z
does not exist.
z
Since limits of the latter type are studied in calculus, we use their definition and properties
freely.
Suppose that
f ( z ) = u ( x, y ) + iv ( x, y ) ,
z0 = x0 + iy0 , and
w0 = u0 + iv0 .
Then
lim f ( z ) = w0
(4)
z → z0
if and only if
lim
( x , y )→( x0 , y0 )
Find lim z 2 +
z →1+ i
u ( x, y ) = u0
and
1
1
x
y
2
= ( x + iy ) +
= ( x2 − y2 ) + 2
+ i 2 xy − 2
.
2
z
x + iy
x +y
x + y2
We have u ( x, y ) = ( x 2 − y 2 ) +
( x , y )→(1,i )
(x
2
− y2 ) +
x
y
and v ( x, y ) = 2 xy − 2
. From Theorem 1,
2
x +y
x + y2
2
x
1
=
2
2
x +y
2
and
lim
( x , y )→(1,i )
thus
lim z 2 +
z →1+ i
If lim f ( z ) , lim g ( z ) exist and c ∈
z → z0
v ( x, y ) = v0
1
.
z
Observe that z 2 +
lim
lim
( x , y )→( x0 , y0 )
z → z0
, then
1
1 3
= + i.
z
2 2
2 xy −
y
3
= ,
2
2
x +y
2
(5)
(
)
(
)
(1) lim f ( z ) + g ( z ) exist and lim f ( z ) + g ( z ) = lim f ( z ) + lim g ( z )
z → z0
(
)
z → z0
(
)
z → z0
z → z0
(2) lim cf ( z ) exist and lim cf ( z ) = c lim f ( z )
z → z0
(
z → z0
)
z → z0
(
)
(3) lim f ( z ) g ( z ) exist and lim f ( z ) g ( z ) = lim f ( z ) lim g ( z )
z → z0
z → z0
z → z0
z → z0
lim f ( z )
f ( z)
f (z)
z → z0
, whenever lim g ( z ) ≠ 0
=
(4) lim
exist and lim
z → z0 g ( z )
z → z0
z → z0 g ( z )
lim g ( z )
z → z0
Limits Involving the Point at Infinity
We have three point about limits that is involving the point at infinity:
1
= 0.
f ( z)
lim f ( z ) = ∞
if and only if
lim f ( z ) = w0
if and only if
lim f
lim f ( z ) = ∞
if and only if
lim
z → z0
z →∞
z →∞
lim
z → z0
z →0
z →0
1
= w0 .
z
1
f 1
( z)
= 0.
i+3
z +i
= ∞ since lim
= 0.
z →− i z + i
z →− i i + 3
Observe that lim
( )
( )
3 1 −i
3z − i
3 − iz
z
= 3 since lim
= lim
= 3.
z →∞ z + 2
z →0
z
→
0
1 +2
1 + 2z
z
Observe that lim
( )
( )
1 3 +2
3z 4 − i
z + 2z 4
z
= lim
= 0.
Observe that lim 3
= ∞ since lim
z →0
z → 0 3 − iz 4
z →∞ z + 2
3 1 4 −i
z
Exercises
1. For each of the functions below, describe the domain of definition that is understood
(a) f ( z ) =
1
z +4
2
(c) f ( z ) = cos ( z 2 − i )
(6)
(7)
(8)
z + 2i
z+z
(b) f ( z ) =
2. Write the function f ( z ) = z 3 + 2 z − i in the form f ( z ) = u ( x, y ) + iv ( x, y ) .
3. Let z0 , c denote complex constant. Use definition (3) to prove that
(b) lim ( x + i 2 y ) = 1 − 2i
(a) lim c = c
z → z0
z →1− i
z2
4. Let f ( z ) =
z
2
a. Find lim f ( z ) along the line y = x
z →0
b. Find lim f ( z ) along the line y = 2 x
z →0
c. Find lim f ( z ) along the parabola y = x 2
z →0
d. What can you conclude about the limit of f ( z ) along z → 0
5. Using (6), (7) and (8) of limits, show that
(a) lim
z 4 − z3 + 2z
( z + 1)
z →∞
(b) lim
z → 2i
4
z
( z − 2i )
2
=1
z2 −1
=∞
z →∞ z + 1
(c) lim
=∞
6. Find the value of limits below
(a) lim z 2 + 2 z − 1
z →1+ 2 i
z2 + 2z −1
z →−2 i z 2 − 2 z + 4
(c)
(b) lim
z3 + 8
lim 4
2
z →(1+ i 3 ) z + 4 z + 16
lim
(d)
z →i
z4 −1
z −i
lim
(e)
z →i
z2 + 4z + 2
z +1
z 2 + z − 1 − 3i
z →1+ i z 2 − 2 z + 2
(f) lim