Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4: Complex Functions of a Complex V a r i a b l e
1.4.1(L)
C l e a r l y each element of S i s a complex number s i n c e b o t h cos t
and s i n t a r e r e a l f o r a l l 0 5 t I T . Thus, S e x i s t s w i t h o u t any
r e f e r e n c e t o a p i c t u r e . The p o i n t i s , however, t h a t i f we use
t h e Argand diagram, w e view x + i y a s t h e p o i n t z [ = ( x , y ) 1 i n
t h e xy-plane.
I n o t h e r words i f we i d e n t i f y t h e p o s i t i o n v e c t o r R w i t h t h e
complex number z , we s e e t h a t t h e ."graph" o f S (by which w e
mean t h e s e t of p o i n t s i n t h e Argand diagram which r e p r e s e n t s
S ) i s t h e curve whose v e c t o r e q u a t i o n i s
R ( t ) = cos t i
+
sin t j, 0 < t < IT.
T h i s , a s we a l r e a d y know from o u r s t u d y o f v e c t o r s , i s t h e curve
whose p a r a m e t r i c form i s
x = cos t
O < t < I T
- y = sin t which w e r e c o g n i z e a s t h e upper h a l f of t h e u n i t c i r c l e c e n t e r e d
a t the origin.
Pictorially,
.\
,L*
A
\
I
,
.
t -t- -4 /
y,
II
> ,&
P (cos t, s i n t ) = cos t f i s i n t
2.
There i s a 1-1 correspondence between complex numbers i n S and
The correspondence i s d e f i n e d
p o i n t s on t h e above s e m i - c i r c l e .
by ( c o s t , s i n t )
++
cos t
+ i s i n t.
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4: Complex F u n c t i o n s o f a Complex V a r i a b l e
1.4.1(L) continued
More g e n e r a l l y , e v e r y c u r v e i n t h e z-plane has t h e e q u a t i o n of
t h e form
( w e s a y more a b o u t t h i s i n E x e r c i s e 1.4.9) and because t h e
Argand diagram h a s t h e s t r u c t u r e of a 2-dimensional v e c t o r s p a c e ,
w e see t h a t e q u a t i o n ( 2 ) i s e q u i v a l e n t t o t h e v e c t o r f u n c t i o n
of a s c a l a r v a r i a b l e , d e f i n e d by
Summarized p i c t o r i a l l y
1. I n v e c t o r form, C i s g i v e n by $ ( t )= x ( t ) f f
(t)j.
+
+
2.
I n t h e Argand diagram R r e p r e s e n t s z , and C i s t h e n t h e s e t o f complex numbers, {z: z = x ( t ) + i y ( t )
.
b. L e t w d e n o t e t h e image of z w i t h r e s p e c t t o f .
I n t h i s case w = z
S i n c e b o t h z and w a r e complex, f i s a c t u a l l y a mapping of a
2-dimensional v e c t o r s p a c e ( t h e z-plane)
v e c t o r s p a c e ( t h e w-plane)
.
i n t o a 2-dimensional
I f w e now i d e n t i f y t h e z - p l a n e w i t h t h e xy-plane and t h e w-plane
w i t h t h e uv-plane, w e see t h a t w = z a c t u a l l y i s e q u i v a l e n t
t o mapping t h e xy-plane i n t o t h e uv-plane ( a t o p i c w e have a l -
ready s t u d i e d f a i r l y thoroughly).
2
.
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s o f a Complex V a r i a b l e
U n i t 4: Complex Functions of a Complex V a r i a b l e
1.4.1 ( L ) c o n t i n u e d Pictorially w-plane
More s p e c i f i c a l l y , i f z = x
+
i y t h e n z 2 = (x2
-
y2
+
i 2xy;
so t h a t
w = u
+
u = x2 - y 2
i v , with
v = 2xy
N o t i c e t h a t w e have a l r e a d y d i s c u s s e d t h e mapping g i v e n by (3)
i n Blocks 3 and 4 of P a r t 2.
Of c o u r s e , we have something "going f o r u s " now t h a t w e d i d n ' t
have t h e n . Namely, we a r e now a b l e t o view mappings o f t h e
xy-plane i n t o t h e uv-plane ( a c o n c e p t which c e r t a i n l y e x i s t s
i n d e p e n d e n t l y of t h e i n v e n t i o n of complex numbers) a s complex
v a l u e d f u n c t i o n s of a complex v a r i a b l e which map t h e z-plane
i n t o t h e w-plane.
With t h i s i n t e r p r e t a t i o n , w e a r e now a b l e t o d i s c u s s a v e c t o r
p r o d u c t t h a t was undefined b e f o r e ( a l t h o u g h w i t h h i n d s i g h t we
c o u l d have gone back t o Blocks 2 , 3 , and 4 o f P a r t 2 and i n v e n t e d
t h e v e c t o r p r o d u c t which c o r r e s p o n d s t o t h e p r o d u c t of two comp l e x numbers) and we may conclude t h a t z2 i s t h e complex number
whose magnitude i s t h e s q u a r e of t h e magnitude of z and whose
argument Is t w i c e t h e argument of z .
I n p a r t i c u l a r , t h e n , s i n c e each p o i n t i n S h a s u n i t magnitude,
i t s image under t h e s q u a r i n g f u n c t i o n a l s o h a s u n i t magnitude.
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4 : Complex F u n c t i o n s of a Complex V a r i a b l e
1.4.1(L)
continued
Moreover, s i n c e t h e argument of t h e image i s t w i c e t h e argument
o f t h e p o i n t , we see t h a t s i n c e t h e arguments o f t h e p o i n t s i n
S v a r y from 0 t o 1 8 0 ° , t h e arguments of t h e images r a n g e from
.
I n summary, t h e n , t h e mapping w = z2 c a r r i e s t h e
s e t S i n t o t h e whole u n i t c i r c l e c e n t e r e d a t t h e o r i g i n . I n
O 0 t o 360 O
p a r t i c u l a r , t h e p o i n t ( r , B ) maps o n t o ( r , 2 8 ) .
H e r e w e s e e , a s an i m p o r t a n t a s i d e , how t h e t h e o r y of mapping
t h e complex p l a n e i n t o t h e complex p l a n e g i v e s us new i n s i g h t t o
r e a l mappings.
I n p a r t i c u l a r , with r e s p e c t t o equation ( 3 )
w e now have t h a t t h i s mapping, i n terms of what i t means t o
m u l t i p l y complex numbers, i s e a s y t o e x p l a i n p i c t o r i a l l y ,
S p e c i f i c a l l y , t h e image of a g i v e n p o i n t i n t h e xy-plane i s
found by d o u b l i n g t h e argument o f t h e p o i n t ( v e c t o r ) and
s q u a r i n g i t s magnitude.
Again w e h a s t e n t o p o i n t o u t t h a t we could have i n v e n t e d t h e
p r o d u c t o f two v e c t o r s t o be t h e v e c t o r i n t h e same p l a n e
e q u i v a l e n t , t o t h e p r o d u c t of t h e two given v e c t o r s a s complex
numbers.
That i s ,
(a
1+
b
5) ( c
+
d
3) =
(ac
-
bd)
f +
(bc
+
ad)
5.
But n o t i c e how much more n a t u r a l t h i s d e f i n i t i o n becomes i n
terms o f t h e language of complex numbers.
I n o t h e r words, one major r e a l a p p l i c a t i o n of t h e t h e o r y of
complex f u n c t i o n s o f a complex v a r i a b l e i s t o t h e r e a l problem
o f mapping t h e xy-plane i n t o t h e uv-plane.
These problems c a n
be t a c k l e d w i t h o u t r e f e r e n c e t o t h e complex numbers, b u t a
knowledge of t h e complex numbers g i v e s us a c o n s i d e r a b l e qnount
o f " n e a t " n o t a t i o n which i s h e l p f u l i n o b t a i n i n g r e s u l t s
f a i r l y quickly.
As a f i n a l o b s e r v a t i o n , l e t u s observe t h a t a s a f u n c t i o n f
h a s t h e s a m e s t r u c t u r e ( b u t a d i f f e r e n t domain) w h e t h e r w e
I n e i t h e r c a s e w e have a f u n c t i o n
w r i t e f ( x ) = x2 o r f ( z ) = z 2
.
machine i n which t h e o u t p u t i s t h e s q u a r e o f t h e i n p u t .
b i g d i f f e r e n c e i s from t h e g e o m e t r i c a l p o i n t of view.
The
In the
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4 : Complex Functions of a Complex V a r i a b l e
1 . 4 . 1 (L) continued
e x p r e s s i o n f ( x ) = x2 s e may view b o t h t h e domain and t h e image of
f a s b e i n g 1-dimensional ( s i n c e x i s assumed t o be r e a l ) .
A c c o r d i n g l y , w e may graph t h e f u n c t i o n i n t h e 2-dimensional xyp l a n e i n t e r m s of t h e curve y = x 2
.
On t h e o t h e r hand, i n t h e e x p r e s s i o n w = f ( z ) = z 2 , t h e domain
2
and t h e image of f must be 2-dimensional s i n c e n e i t h e r z n o r z
i s r e q u i r e d t o be r e a l . Thus, w e would r e q u i r e a $-dimensional
s p a c e t o graph t h i s f u n c t i o n i f w e wanted a graph which was t h e
a n a l o g of t h e graph y = f ( x ) , S i n c e we c a n n o t , i n t h e usual.
g e o m e t r i c s e n s e , draw a 4-dimensional s p a c e , o u r g e o m e t r i c
i n t e r p r e t a t i o n must i n v o l v e viewing t h e z-plane ( t h e domain of
f ) a s b e i n g mapped i n t o t h e w-plane ( t h e r a n g e of f 1
.
a. I f w e look a t z a s b e i n g t h e p o i n t ( r , 8 ) i n t h e z - p l a n e , t h e n
3
Thus, under f e a c h p o i n t i n t h e z-plane
z = ( r , 0 ) = ( r 3,38)
i s mapped i n t o t h e p o i n t ( p , r $ ) i n t h e w-plane where p=r3and
.
$I= 38 [ i . e . ,
argument]
.
t h e mapping cubes t h e magnitude and t r i p l e s t h e
-
I n p a r t i c u l a r t h e p o i n t ( 1 , 8 ) where 0 < 8 < 90° i s mapped o n t o
Thus,
(I', 30) = ( 1 , 3 8 ) and s i n c e O 0 < 8 < 90°, O 0 < 8 < 270'.
t h e f i r s t q u a d r a n t S of t h e u n i t c i r c l e i s mapped o n t o t h e
f i r s t t h r e e q u a d r a n t s of t h e u n i t c i r c l e .
Again p i c t o r i a l l y ,
z-plane w-plane
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4: Complex F u n c t i o n s of a Complex V a r i a b l e
--
1.4.2
-
continued
By t h e same t o k e n , e a c h p o i n t i n T I w r i t t e n i n p o l a r c o o r d i n a t e s ,
.
[ I f t h e l i n e extended i n t o t h e t h i r d
h a s t h e form ( r, 45O )
q u a d r a n t , t h e p o i n t s on t h i s p a r t would be r e p r e s e n t e d a s (r,225O).]
Hence "cubing" such a p o i n t y i e l d s ( r 3, 135O)
.
I n o t h e r words,
t h e mapping d e f i n e d by f ( z ) = z3 maps t h e r a y 0 = 45'
onto the
r a y 0 = 135O i n s u c h a way t h a t t h e image of each p o i n t h a s t h e
cube of t h e magnitude of t h e p o i n t .
Pictorially,
w-plane
iy)3
3
2
= x3 + 3x ( i y ) + 3 x ( i Y l 2 + ( i y )
2
i y3 = x3 + 3x y i - 3xy2
3
2
2
= (x3 - 3 x y ) + i ( 3 x y - y 1.
= (x
+
-
Hence,
Again, by way of r e v i e w , e q u a t i o n (1) d e f i n e s a r e a l mapping o f
2-space i n t o 2-space, b u t from o u r knowledge of complex v a r i a b l e s ,
w e know t h a t t h e r a t h e r cumbersome system (1) i s e q u i v a l e n t t o
mapping e a c h p o i n t ( v e c t o r ) i n t h e xy-plane i n t o t h e p o i n t whose
magnitude i s t h e cube of t h e given magnitude and whose argument
i s t r i p l e t h a t of t h e g i v e n argument.
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s o f a Complex V a r i a b l e
U n i t 4: Complex F u n c t i o n s o f a Complex V a r i a b l e
1.4.3 a.
z = x
+
iy
+
22 = 2x
+
i2y.
T h e r e f o r e , w = 22 = 2x
+
i2y.
u d e n o t e t h e r e a l p a r t o f w and v t h e i m a g i n a r y p a r t w e h a v e
b
-
w = f (2) = z
Hence,
+
w = x
-
iy.
Hence,
Letting
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4 : Complex F u n c t i o n s of a Complex V a r i a b l e
1.4.3
continued
Hence,
2
x -Fy2#0
( s i n c e z # 0)
Again, a s a r e m i n d e r , t h i s problem shows us t h a t w e may view t h e
mapping
a s f ( 2 ) = z2
+
22
+
i; and
1.4.4 (L) Our main a i m i n t h i s e x e r c i s e i s t o g e t a b e t t e r f e e l i n g f o r t h e
" r e a l i t y " o f complex f u n c t i o n s of a complex v a r i a b l e . P a r t s ( b )
and ( c ) a r e concerned w i t h e x t e n d i n g t h e a n a l o g s of f ( x ) = x + c
and f ( x ) = cx where c and x a r e r e a l numbers t o f ( z ) = z + c
and f ( z ) = c z where c and z a r e now complex numbers. A s w e s h a l l
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s o f a Complex V a r i a b l e
U n i t 4: Complex Functions of a Complex V a r i a b l e
1 . 4 . 4 (L) continued
see, t h e a l g e b r a of t h e s e f u n c t i o n s i s t h e same a s t h a t of t h e i r
r e a l a n a l o g s , b u t t h e g e o m e t r i c i n t e r p r e t a t i o n i s a b i t more
s o p h i s t i c a t e d ( t h e r e s u l t o f b o t h o u r domain and image s p a c e
b e i n g 2-dimensional r a t h e r t h a n 1 - d i m e n s i o n a l ) .
In p a r t (a)
w e want t o emphasize t h e f a c t t h a t what l o o k s l i k e a new f u n c t i o n
t o u s i s r e a l l y an o l d f u n c t i o n t h a t we h a n d l e d i n a v e r y r e a l
situation,
In particular,
a . R e c a l l i n o u r t r e a t m e n t of t h e double i n t e g r a l t h a t when we
wanted t o r e v e r s e t h e o r d e r of i n t e g r a t i o n , t h e t e c h n i q u e was
g e o m e t r i c a l l y expressed by t h e mapping o f t h e xy-plane i n t o
t h e uv-plane given by
I t s h o u l d be c l e a r t h a t we do n o t have t o know a n y t h i n g about
complex numbers t o t a l k a b o u t t h e mapping d e f i n e d by e q u a t i o n
(1). I f , however, we want t o view t h e mapping a s b e i n g from
t h e z - p l a n e i n t o t h e w-plane,
o u r procedure i s t o w r i t e (1) i n
t h e form u + i v , which i n t h i s c a s e means t h a t we s t u d y t h e
complex f u n c t i o n of a complex v a r i a b l e d e f i n e d by
f ( z ) = x
= x
+ i(-y) -
iy. I f w e now r e c a l l t h a t z i s x
definition
z.
Thus,
+
i y , we s e e t h a t x
-
i y i s by
( 2 ) becomes
Of c o u r s e we a r r i v e d a t ( 3 ) r a t h e r i n v e r s e l y t o t h e wording
o f t h e e x e r c i s e i n which w e w e r e t o b e g i n w i t h ( 3 ) and d e r i v e
(1). Our purpose f o r doing t h i s was simply t o s t a r t t h e
e x e r c i s e emphasizing t h e r e l a t i o n s h i p between complex f u n c t i o n s
of a complex v a r i a b l e and r e a l mappings. Had we begun w i t h
( 3 ) , w e would have merely r e v e r s e d o u r s t e p s t o o b t a i n :
,
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4 : Complex F u n c t i o n s o f a Complex V a r i a b l e
1 . 4 . 4 (L) c o n t i n u e d
and from t h e r e a l and imaginary p a r t s of f ( z ) , w e would have
concluded t h a t t h e graph of f was e q u i v a l e n t t o t h e mapping def i n e d by
T h i s mapping i s e q u i v a l e n t t o r e f l e c t i n g t h e xy-plane a b o u t t h e
x-axis
(i.e.
, we
l e a v e x a l o n e and change t h e s i g n o f y.
Pictorially,
z-plane
w-plane
f(z) = F=(x,-y)
( F i g u r e 1)
But s i n c e the w-plane i s a r e p l i c a of t h e z-plane w e may s u p e r impose t h e two p l a n e s i n F i g u r e 1 t o o b t a i n
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4: Complex F u n c t i o n s of a Complex V a r i a b l e
1 . 4 . 4 (L) continued
Thus, t h e e f f e c t o f f on s e t S i n t h e Argand diagram i s t o produce t h e m i r r o r image of S w i t h r e s p e c t t o t h e x - a x i s .
%
p a r t i c u l a r i f S i s any c l o s e d r e g i o n , f ( S ) = S ( i . e . ,
image have t h e same s i z e and s h a p e )
.
.
+
I n t h e r e a l c a s e , we saw t h a t t h e graph of f ( x )
In
S and i t s
c j u s t "raised"
In particular
e a c h p o i n t of t h e curve y = f ( x ) by an amount c.
t h e g r a p h of f ( x ) = x t c , was o b t a i n e d by l i f t i n g each p o i n t on
t h e l i n e y = x by c u n i t s .
Pictorially,
Now, g i v e n f ( z ) = z t c , w e s e e t h a t i n t h e Arqand diagram t h i s
sum must be i n t e r p r e t e d a s a v e c t o r sum. A s a v e c t o r t h e complex
-t
t
number c i s w r i t t e n a s c l i + c 2 1 (where w e a r e assuming t h a t
c = c1 + c 2 i ) . L e t t i n g c d e n o t e c l i + c 2 j r w e s e e t h a t adding
c t o z i s e q u i v a l e n t t o d i s p l a c i n g z by an amount e q u a l t o t h e
magnitude of c i n t h e d i r e c t i o n of c .
F o r example, t h e mapping d e f i n e d by f ( z ) = z
+
3
+
4 i maps t h e
p o i n t z i n t o t h e p o i n t 5 u n i t s from z i n t h e d i r e c t i o n
4 t
Solutions
Block 1: A n I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i b
U n i t 4 : Complex F u n c t i o n s of a Complex V a r i a b l e
1 . 4 . 4 (L) c o n t i n u e d
G e o m e t r i c a l l y , adding 3
+
4 i o n t o z s h i f t s ( t r a n s l a t e s ) P t o Q.
That i s , P i s t r a n s l a t e d 5 u n i t s i n t h e d i r e c t i o n 3 1
+
43.
c. Here w e invoke t h e f a c t t h a t we have a very c o n v e n i e n t way of
m u l t i p l y i n g complex numbers u s i n g p o l a r c o o r d i n a t e s .
In parti-
cular i f c = (r
8 ) t h e n c z has a s i t s magnitude ro t i m e s t h e
magnitude of z and a s i t s argument Q0 p l u s t h e argument of z.
0' 0
I n o t h e r words w e o b t a i n t h e image of z by r o t a t i n g t h e v e c t o r
z through Q0 d e g r e e s and i n c r e a s i n g i t s magnitude by a f a c t o r o f
r0 '
By way of a n example, i f f
(2)
= (3
+
4 i ) z , t h e n t h e image of a
g i v e n number z i s o b t a i n e d by r o t a t i n g z through an a n g l e e q u a l
4
t o arc tan
and r e p l a c i n g t h e magnitude of z by 5 t i m e s i t s
value. P i c t o r i a l l y ,
1. W e p i c k any p o i n t on OP.
2 . W e e r e c t a p e r p e n d i c u l a r t o OA and l o c a t e B on OA such t h a t
-
4
+ AB = -5 OA.
T h e r e f o r e , t a n # AOB =
4
3 . We mark o f f t h e l e n g t h OP ( i . e . ,
4 . 0% t h e n d e n o t e s ( 3
+
4 i ) $P
= (3
.
121)
+
5 t i m e s a l o n g OB.
4i)z.
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4: Complex Functions o f a Complex V a r i a b l e
1 . 4 . 4 (L) continued
A s a v e r y i n t e r e s t i n g s p e c i a l c a s e , n o t i c e t h a t i f t h e magnitude
o f c i s 1 t h e n t h e mapping f ( z ) = c z simply r o t a t e s z through an
a n g l e e q u a l t o t h e argument of c ( i . e . ,
t h e magnitude i s p r e -
s e r v e d because c has u n i t magnitude).
I f w e l e t 8 denote t h e argument o f c , t h e f a c t t h a t c i s o f u n i t
magnitude means t h a t c = cos 8
+
i s i n 8.
Hence,
c z =
(COS
8
+
= x c o s 8-y
i s i n 8 ) (x
+
iy) s i n 8 t i (x s i n 8
+
y cos 8) ,
and a s d i s c u s s e d i n our e a r l i e r e x e r c i s e s , t h i s i s e q u i v a l e n t
t o t h e r e a l mapping
u = x cos 8
-
y sin 8
v = x s i n 8 + y cos8.
Thus, comparing t h i s r e s u l t w i t h o u r p o l a r c o o r d i n a t e i n t e r p r e t a t i o n , w e s e e t h a t t h e mapping d e f i n e d by e q u a t i o n (1) i s e q u i v a l e n t t o r o t a t i n g t h e xy-plane through e O .
H o p e f u l l y , t h i s shows us s t i l l a n o t h e r way i n which complex
numbers have a r e a l i n t e r p r e t a t i o n .
By t h e way, i n t h e s p e c i a l
c a s e t h a t c i s r e a l , t h e argument of c i s e i t h e r O 0 o r 180°,
depending upon whether c i s p o s i t i v e o r n e g a t i v e . N o t i c e t h e n
t h a t i n t h i s c a s e t h e r e s u l t checks w i t h t h e u s u a l r e s u l t i n
t h e r e a l c a s e ; i . e . , m u l t i p l y i n g by ( r e a l ) c l e a v e s t h e
d i r e c t i o n a l o n e , changes t h e magnitude by a f a c t o r of Icl and
p r e s e r v e s t h e s e n s e i f c > 0 , r e v e r s e s t h e s e n s e if c < 0.
A s a f i n a l n o t e on t h i s e x e r c i s e n o t i c e t h a t t h e l i n e a r mapping
d e f i n e d by f ( z ) = clz
+
c 2 where b o t h cl and c 2 a r e complex-
v a l u e d c o n s t a n t s maps l i n e s through t h e o r i g i n i n t o l i n e s
t h r o u g h t h e o r i g i n ; and c i r c l e s c e n t e r e d a t t h e o r i g i n i n t o
c i r c l e s centered a t the origin.
Namely, t h e mappping f ( z ) i s
a r o t a t i o n (accompanied by a uniform m a g n i f i c a t i o n f a c t o r e q u a l
t o c l ) followed by a t r a n s l a t i o n .
Under a r o t a t i o n , l i n e s
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4 : Complex F u n c t i o n s o f a Complex V a r i a b l e
1.4.4 (L) c o n t i n u e d
through t h e o r i g i n remain l i n e s through t h e o r i g i n , and c i r c l e s
c e n t e r e d a t t h e o r i g i n remain c i r c l e s c e n t e r e d a t t h e o r i g i n .
Notice a l s o t h a t t h e algebra of i n v e r t i n g t h i s type of f u n c t i o n
i s word-for-word t h e same a s i n t h e r e a l c a s e s i n c e t h e s t r u c t u r a l
r u l e s a r e t h e same. Namely, i f w = c 1z + c 2 ( c l # 0) t h e n
etc.
I n o t h e r words, t h e a l g e b r a remains t h e same, b u t t h e g e o m e t r i c
i n t e r p r e t a t i o n i s e l e v a t e d by a dimension o f s o p h i s t i c a t i o n
(so t o speak)
a.
.
Here w e have f ( z ) = clz
+
c 2 where cl =
Jz andc2=i.
By t h e r e s u l t of t h e p r e v i o u s e x e r c i s e clz r o t a t e s z t h r o u g h
a n a n g l e e q u a l t o t h e argument of cl and m u l t i p l e s t h e magnitude
of
by Icll
I n o u r c a s e , lcll = 1 [ i . e . ,
.
w h i l e t h e a r g u e n t o f c1 i s 45O.
Y -
I
Jz
Hence, clz i s a 45" r o t a t i o n of t h e z - p l a n e .
Then s i n c e
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
Complex Functions of a Complex V a r i a b l e
Unit 4:
1.4.5 continued
c 1z
+
(i.e.,
i " t r a n s l a t e s " clz
an amount e q u i v a l e n t t o t h e v e c t o r
t
J
adding i r a i s e d t h e p o i n t by 1 u n i t ; we s e e t h a t
i s e q u i v a l e n t t o r o t a t i n g each p o i n t i n t h e p l a n e through 45'
and t h e n r a i s i n g i t 1 u n i t )
.
Pictorially,
R o t a t e P through 45' and t h e n l i f t i t ( i . e . ,
t h e y - a x i s ) one u n i t .
move it p a r a l l e l t o
Solutions
Block P: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4: Complex F u n c t i o n s o f a Complex V a r i a b l e
1.4.5
continued
Hence i n C a r t e s i a n form, t h e mapping i s given by
1.4.6
z 4 h a s magnitude 16 i f z h a s magnitude 2 , and t h e argument of z 4
i s f o u r t i m e s t h e argument of z.
Hence a s z t r a c e s t h e p o r t i o n
of t h e c i r c l e r = 2 between 8 = O 0 and 0 = 60°, f ( z ) t r a c e s t h e
p o r t i o n of t h e c i r c l e r = 16 between 8 = O 0 and 8 = 240°.
Pictorially
g ( s ) where g ( z ) = z
4
F i n a l l y , adding 3 + 4 i t r a n s l a t e s each p o i n t 5 u n i t s i n t h e
d i r e c t i o n 3 + 4 i.
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4 : Complex unctions of a Complex V a r i a b l e
1.4.6
Adding 3
continued
+
4 i t o each p o i n t on t h e c i r c l e r = 16 t r a n s l a t e s t h e
c i r c l e from c e n t e r a t 0 t o c e n t e r a t 0 ' .
{r = 1 6 , 0 < 8 < 240')
l a t e d by 3: + 4 5 , i . e . ,
$A'
=&3'
= 3;
+
43.
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4: Complex F u n c t i o n s of a Complex V a r i a b l e
1.4.7(L)
Here a g a i n w e s e e how o u r knowledge of v e c t o r c a l c u l u s h e l p s u s
here.
lim
Namely, it i s n a t u r a l , i f only by mimicking, t o d e f i n e
f (z) = L
z+c
which means t h a t , g i v e n
o
a -+ If ( 2 ) -
<Iz
cI<
E >
0 , t h e r e e x i s t s 6 > 0 such t h a t
LI
<
€.
The above d e f i n i t i o n makes s e n s e even though z , c , and L need
n o t be r e a l s i n c e we a r e d e a l i n g only w i t h a b s o l u t e v a l u e s which a r e (non-negative) r e a l numbers.
Moreover, from a p i c t o r i a l p o i n t of view ( i . e . ,
i n terms of t h e
Argand diagram) t h e above d e f i n i t i o n i s p r e c i s e l y t h e same a s
o u r l i m i t d e f i n i t i o n when w e d e a l t w i t h v e c t o r f u n c t i o n s of a
vector variable.
R e c a l l i n t h a t c a s e we showsd t h a t t h e d e f i n i t i o n was e q u i v a l e n t
t
+
t
t o s a y i n g t h a t i f f (6) = u ( x , y : ~t v ( x . y ) ] and i f L = L1l + L 2 ) ,
-b
*
c = c1I + c Z j : t h e n
*
*
was e q u i v a l e n t t o
lim
u ( x , y ) = L1
~x,y~+(c1,c2)
lim
v(x,y) = L 2 .
(x,y)+(c,,c,)
T r a n s l a t e d i n t o t h e Argand diagram t h i s s a y s t h a t i f c = c L f c 2 i
then
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4 : Complex Functions of a Complex V a r i a b l e
1 . 4 . 7 ( L ) continued means Re[f(z)l = R ~ ( L ) lirn
( x , y ) + (c1,c2) and Im[f(z)l = Im(L). lirn
( ~ I Y ) + ( ~ ~ , C ~ )
For example, i n t h e given e x e r c i s e
f
(2) =
z3
= (X
+
iy)3
Hence,
The key p o i n t i s t h a t u s i n g t h e Argand diagram model f o r t h e
complex numbers w e need n o t i n v e n t any new i d e a s t o h a n d l e
l i m f ( z ) i f c i s complex and f i s complex-valued.
Z+C
I n p a r t i c u l a r , e v e r y l i m i t theorem t h a t was t r u e i n o u r s t u d y of
v e c t o r f u n c t i o n s of a v e c t o r v a r i a b l e remains t r u e i n o u r s t u d y
of complex f u n c t i o n s of a complex v a r i a b l e . More s p e c i f i c a l l y ,
w e may c o n t i n u e t o use such r e s u l t s a s t h e l i m i t of a sum i s
t h e sum of t h e l i m i t s , t h e l i m i t of a p r o d u c t i s t h e p r o d u c t of
the l i m i t s , etc.
Again, t h e main i d e a i s t h a t once w e view
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4: Complex F u n c t i o n s of a Complex V a r i a b l e
1.4.7 (L) c o n t i n u e d
complex numbers i n t h e Argand diagram we cannot d i s t i n g u i s h
between complex numbers and p l a n a r v e c t o r s s t r u c t u r a l l y .
Thus,
theorems f o r one model remain theorems f o r t h e o t h e r .
1.4.8
( z + h I 2 = z2 t 2zh
Hence,
(z
+
h12
-
= 22
f h2 ( j u s t
z2 = 2zh
+
+
as i n t h e reaL c a s e ) .
h2.
Hence,
h , provided h
# 0.
Hence,
h12
l i m [ (z +
h
h+O -
z2
1 = lim
[2z
+
h1 h+O
= l i m
22+ lim
h+O
h+O
h
N o t i c e t h a t t h i s e x e r c i s e seems t o be t h e complex e q u i v a l e n t
of f i n d i n g f ' ( x ) when f ( x ) = x 2
This i d e a i s t h e t o p i c of t h e
.
next unit.
1.4.9 Our main aim i n t h i s e x e r c i s e i s t o show t h a t t h e s t u d y o f complexv a l u e d f u n c t i o n s o f a s i n g l e r e a l v a r i a b l e was made when we
s t u d i e d t h e p l a n a r problem of v e c t o r f u n c t i o n s of a s c a l a r
variable.
Namely, i f we view z a s x
f
i y , then t h e f a c t t h a t z i s a f u n c t i o n
of t h e s c a l a r ( r e a l ) v a r i a b l e t means t h a t we may w r i t e
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4 : Complex Functions of a Complex V a r i a b l e
1.4.9
continued
The c r i t i c a l p o i n t i s t h a t i f we e l e c t t o u s e t h e Argand diagram
a s a g e o m e t r i c model, w e see a t once t h a t e q u a t i o n (1) i s
s t r u c t u r a l l y equivalent t o the vector equation:
I n summary, t h e curve i n t h e xy-plane d e f i n e d by e q u a t i o n ( 2 ) i s
t h e "graph" of t h e complex numbers d e f i n e d by e q u a t i o n (1). I n
o t h e r words, one way of v i s u a l i z i n g a ( c o n t i n u o u s ) complex
f u n c t i o n of a r e a l v a r i a b l e i s a s a c u r v e i n t h e z-plane.
The main p o i n t i s t h a t s i n c e w e may i d e n t i f y a complex f u n c t i o n
of a r e a l v a r i a b l e w i t h a v e c t o r f u n c t i o n of a s c a l a r v a r i a b l e ,
w e may a l s o assume t h a t t h e c a l c u l u s s t r u c t u r e of v e c t o r f u n c t i o n s
o f s c a l a r v a r i a b l e s i s i n h e r i t e d by complex f u n c t i o n s o f r e a l
v a r i a b l e s ; and both p a r t s ( a ) and ( b ) of t h i s e x e r c i s e a r e
designed t o i l l u s t r a t e t h i s .
a . W e assume h e r e t h a t f ' ( t ) h a s t h e u s u a l meaning, e x c e p t t h a t f
i s now a v e c t o r f u n c t i o n r a t h e r t h a n a s c a l a r f u n c t i o n .
p o i n t i s t h a t had we been g i v e n t h e problem
The
we would have been a b l e t o conclude t h a t
3'
( t )=
-+
1
+ 2t3.
S i n c e e q u a t i o n ( 3 ) t r a n s l a t e s , i n t h e Argand diagram, i n t o
i t f o l l o w s t h a t f l ( t ) must be t h e a n a l o g of e q u a t i o n ( 4 ) , namely,
Solutions Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e U n i t 4: Complex F u n c t i o n s o f a Complex V a r i a b l e 1.4.9 c o n t i n u e d
More g e n e r a l l y , t h e n , i n terms o f t h e Argand diagram i f z = f ( t )
where f i s a d i f f e r e n t i a b l e complex f u n c t i o n of a r e a l v a r i a b l e ,
w e may view z = f ( t ) a s t h e curve z = g ( t ) + h ( t ) i where g i s
t h e r e a l p a r t of f and h i s t h e imaginary p a r t of f . I n t h i s
e v e n t , f l ( t ) i s a v e c t o r t a n g e n t t o t h i s curve w i t h magnitude e q u a l
to
The key p o i n t i s t h a t t h e c a l c u l u s h e r e i s a "carbon copy" of t h e
c a l c u l u s o f v e c t o r f u n c t i o n s of a s c a l a r f u n c t i o n .
b.
If R
1 ( t )= t 2i
+
e3tj,
t h e n we a l r e a d y know t h a t
T r a n s l a t i n g t h e r e s u l t ( 6 ) i n t o t h e language of complex numbers
2
e3ti, then
w e have t h a t , i f f ' ( t ) = t
+
f ( t ) = 1 t 3
+
e3ti
+
c , where c i s an a r b i t r a r y complex c o n s t a n t .
(7)
I f w e now u s e t h e f a c t t h a t f (0) = 1 + i, e q u a t i o n ( 7 ) becomes
1 i + c s o t h a t c = 1 + 2 i. P u t t i n g t h i s r e s u l t i n t o (71,
1+ i = 3
2 e3ti
1 T i ,o r w e have t h a t f ( t ) =
t3
+
+
+
+
I n summary, w e a l r e a d y know how t o d i f f e r e n t i a t e and i n t e g r a t e
complex f u n c t i o n s o f a r e a l v a r i a b l e because o u r p r e v i o u s knowledge
o f v e c t o r f u n c t i o n s of s c a l a r v a r i a b l e s .
1.
I f z = x ( t ) + y ( t ) i , t h e n
dZ
= dX
+
In particular
8
i ; and
+
I f z = x l ( t ) + y ' ( t ) i, t h e n / z d t = x ( t ) + y ( t ) i
c; where
and c i s an a r b i t r a r y complex
d x (t
t ) and y ' ( t ) = dy
dt
x' (t)= d
2. constant.
Thus, w h i l e complex f u n c t i o n s of a r e a l v a r i a b l e a r e i m p o r t a n t i n
o u r s t u d y o f complex v a r i a b l e s ( e . g . , a s mentioned i n E x e r c i s e
1.4.1,
t h e "graph" of a s e t of complex numbers i n t h e Argand
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4: Complex Functions o f a Complex V a r i a b l e
1.4.9 continued
diagram has t h i s f o r m ) , we do n o t d e v o t e much t i m e t o such a
s t u d y s i n c e t h e main r e s u l t s a r e a l r e a d y a v a i l a b l e t o us from o u r
s t u d y of p l a n a r v e c t o r s .
The r e s u l t of t h i s e x e r c i s e j u s t i f i e s why t h e s t u d y of r e a l v a l u e d f u n c t i o n s of a complex v a r i a b l e i s u s u a l l y i g n o r e d from a
c a l c u l u s p o i n t of view*. Namely, assuming t h a t t h e r e s u l t of
t h i s e x e r c i s e h o l d s , w e have t h a t i f y = f ( z ) and i f dy e x i s t s ,
dy = 0 .
This, i n t u r n , implies t h a t f (z) is constant.
then dz
Thus, i f f : c+R such t h a t f 1 e x i s t s , t h e n f ( z ) must be c o n s t a n t .
df
( = 1
fails to
I n o t h e r words, u n l e s s f (z 1 = c o n s t a n t ,
exist.
Thus, t h e s t u d y of d i f f e r e n t i a b l e r e a l f u n c t i o n s of a
complex v a r i a b l e i s " s h o r t and sweet".
Now, t u r n i n g t o t h e s p e c i f i c s of t h i s e x e r c i s e , we must f i r s t
d e f i n e what w e mean by f i n t h e c a s e t h a t f i s a r e a l - v a l u e d
I n terms of o u r u s u a l approach
f u n c t i o n of a complex v a r i a b l e .
i n t e r m s of s t r u c t u r e , w e d e f i n e
f (zo
f' ( z o ) = l i m [
az+o
+
Az)
-
$=
f 'u
(2)
by
f (zo) 1
Z
p r o v i d e d t h a t t h e l i m i t e x i s t s . S i n c e z , a n d hence Az, i s complex,
i t means t h a t t h e r e a r e many p a t h s by which Az may apprcaeh 0.
One such p a t h i s t h e one d e f i n e d by t h e change i n t h e imaginary
p a r t of Az b e i n g 0; and a n o t h e r , by t h e change of t h e r e a l p a r t
o f A Z b e i n g 0.
*We h a s t e n t o s t r e s s " c a l c u l u s " l e s t y o u e r r o n e o u s l y b e l e d t o
b e l i e v e t h a t such functions a r e unimportant i n a l l r e s p e c t s .
F o r example, t h e a b s o l u t e v a l u e of a complex v a r i a b l e i s
e x t r e m e l y i m p o r t a n t and t h i s i s a n e x a m p l e o f a r e a l - v a l u e d f u n c t i o n
o f a complex v a r i a b l e .
T h a t i s , i f z i s c o m p l e x a n d f(z) = l z l
t h e n t h e r a n g e of f i s t h e n o n - n e g a t i v e r e a l numbers.
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex V a r i a b l e
U n i t 4: Complex F u n c t i o n s of a Complex V a r i a b l e
1.4.10 continued
I n t e r m s o f t h e Argand diagram we have, Y
p,Re[Az]+
0 a l o n g l i n e R e ( z ) = Re(z0)
1,
Zo
Im(Az)+O a l o n g l i n e Im(z) = lm(zo)
= (x0,yo)
X
A l g e b r a i c a l l y s p e a k i n g , W e &re saying t h a t I f z = x
Az = Ax
+
+
fy t h e n
iAy; and w e a r e l o o k i n g a t Az i n one c a s e w i t h Ay = 0
and i n t h e o t h e r w i t h Ax = 0.
The key p o i n t i s t h a t numerator i n t h e b r a c k e t e d e x p r e s s i o n i n
e q u a t i o n (1) must b e r e a l s i n c e f i s g i v e n t o be r e a l v a l u e d .
Thus, w i t h Ay = 0 ,
i s equal. t o
i n which case, f ' , i f it e x i s t s must be given by f (xo+
f'(zo) = l i m [
Ax+O
AX,
yo)
Ax
-
f ( X ~ , Y ~ )
] =
a-f
-ax
I
(x0,yo)
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s o f a Complex V a r i a b l e
U n i t 4: Complex Functions of a Complex V a r i a b l e
1.4.10 c o n t i n u e d
S i m i l a r l y w i t h Ax = 0, e q u a t i o n (1) becomes
-
f ( z o + iAy)
f t ( z 1 = lirn [
0
f
(xo'~,
I
+
Ay)
-
f
(x0,yo)
I
iAy
Ay+O
- 1lim
(zo) iAy
Ay+O
= lim [
f
f
[
(xory0
+
Ay)
-
f
(xo,y0)
I
S i n c e t h e e x i s t e n c e of t h e l i m i t i n (1) means t h a t t h e v a l u e of
f l ( z o ) must b e independent of t h e d i r e c t i o n i n which z+0, w e may
e q u a t e t h e v a l u e s of f 1 (zo) found i n ( 2 ) and ( 3 ) t o conclude
Solutions
Block 1: An I n t r o d u c t i o n t o F u n c t i o n s of a Complex v a r i a b l e
U n i t 4 : Complex F u n c t i o n s o f a Complex V a r i a b l e
1.4.10
continued
E q u a t i n g t h e r e a l and imaginary p a r t s i n t h e e q u a l i t y g i v e n by ( 4 1 ,
we conclude t h a t
f x (xoty0 ) = 0 and f Y (xoly0)
= 0.
F i n a l l y , s i n c e ( x0 ,yo) = zo was an a r b i t r a r y p o i n t (number) i n
t h e domain of f we may conclude from e q u a t i o n (5) t h a t
and from o u r knowledge o f r e a l - v a l u e d f u n c t i o n s of s e v e r a l ( t w o )
r e a l v a r i a b l e s * , w e may conclude t h a t
f (x,y) = constant.
Then s i n c e f ( x , y ) i s simply t h e geometric e q u i v a l e n t o f f ( z ) ,
w e conclude t h a t f
(2)
= ( r e a l ) constant.
*
N o t i c e t h a t we h a v e i d e n t i f i e d f ( z ) w i t h f ( x , y ) b y v i e w i n g z
a s t h e p o i n t ( x , y ) i n t h e Argand d i a g r a m .
Since f is real-valued
i t f o l l o w s t h a t f ( x , y ) i s a r e a l f u n c t i o n of t h e r e a l v a r i a b l e s
x and y.
Consequently t h e statement given i n (6) is independent
of o u r knowing a n y t h i n g a b o u t complex numbers ( a l t h o u g h t h e
d e r i v a t i o n o f ( 6 ) came f r o m o u r t r e a t m e n t o f t h e c o m p l e x n u m b e r s ) .
Odx
Ody
A c c o r d i n g l y ( 7 ) is m e r e l y a r e a f f i r m a t i o n t h a t i f df
then f(x,y) i s constant.
-
+
MIT OpenCourseWare
http://ocw.mit.edu
Resource: Calculus Revisited: Complex Variables, Differential Equations, and Linear Algebra
Prof. Herbert Gross
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