EXERCISE SET #2
Definition: Let (x 0 , y0 ) be an interior point of the domain of definition D of a function
u  u(x, y) . The limit of u  u(x, y) as (x, y) approach (x 0 , y0 ) is said to be L,
lim(x,y)(x0 ,y0 ) u(x, y)  L ,
if for every number   0 there is a corresponding number   0 such that,
if (x, y)  D and
(x  x 0 )2  (y  y0 ) 2   , then u(x, y)  L   .
Definition: A function u  u(x, y) is called continuous at (x 0 , y0 ) , if
lim(x,y)(x0 ,y0 ) u(x, y)  u(x 0 , y0 ) .
Here the limit value needs to be independent of the way in which (x, y) approaches (x 0 , y0 ) .
Definition: A function u  u(x, y) is differentible at (x 0 , y0 ) , if
u  u(x 0  x, y0  y)  u(x 0 , y0 ) can be expressed in the form
u  u x (x 0 , y0 )x  u y (x 0 , y0 )y  1x  2y
where 1 and 2  0 as (x, y)  (0,0) .
Theorem: If the partial derivatives u x and u y exist near (x 0 , y0 ) and are continuous at
(x 0 , y0 ) , then u is differentible at (x 0 , y0 ) .
Definition: The derivative of f at z 0 is defined as
f (z0 )  limzz0
f (z) f (z0 )
z  z0
provided that the limit exists and z 0 is an interior point of the domain of definition of f . Here
the limit value needs to be independent of the way in which z approaches z 0 .
Definition: f (z)  u(x, y)  iv(x, y) is said to satisfy the Cauchy-Riemann equations if
u x  v y and u y  v x .
Differentibility: Let f (z)  u(x, y)  iv(x, y) be defined throughout some neighborhood of a
point z0  x 0  iy0 . For f to be differentible at z 0 ,
(i) it is necessary that the Cauchy-Riemann equations be satisfied at x 0 , y0 ;
(ii) it is sufficient that the Cauchy-Riemann equations be satisfied at z 0 , and that u and v be
continuously differentible in some neighborhood of z 0 .
If f is differentible, then f  is given by any of these four equivalent expressions:
f   u x  ivx  u y  iv y  u x  iu y  v y  iv x
Definition: Suppose f (z) is differentible at z 0 and throughout some neighborhood of z 0 , then
it is analytic at z 0 . If it is not analytic at z 0 , it is singular there. If it is analytic at each point of
a region D, then it is analytic in D.
Definition: If f (z)  u(x, y)  iv(x, y) is analytic in a domain D, then u and v are harmonic in
D, i.e. they are twice continuously differentible and they satisfy the Laplace equation
2 u  u xx  u yy  0,
2 v  vxx  v yy  0
in D. Note that If f (z)  u(x, y)  iv(x, y) is analytic, then the harmonic functions u and v are
a related pair by Cauchy-Riemann equations. They are refereed to as conjugate harmonic
functions..
Cauchy’s Theorem: If f(z) is analytic in a simply connected domain D, then

C
f (z) dz  0 for
every piecewise smooth simple closed curve C in D.
Fundamental Theorem of the Complex Integral Calculus: Let f(z) be analytic in a simply
connected domain D, and let z 0 be any fixed point in D. Then
z
(i) G(z)   f () d is analytic in D and G(z)  f (z) .
z0
(ii) If F(z) ia any primitive of f(z), i.e. F(z)  f (z) , then

z
z0
f () d  F(z)  F(z 0 ) .
Cauchy Integral Formula: Let f (z) be analytic in a simply connected domain D, let C be a
piecewise smooth simple closed curve in D oriented anticlockwise, and let z 0 be any point
within C. Then

f (z)
z  z0
dz  2if (z 0 ) .
C
Furthermore,

f (z)
(z  z0 )n 1
dz 
2 i
n!
f (n) (z 0 ) .
C
1. Find the limit, if it exists, or show that the limit does not exist.
(a) lim(x,y)(1,2)  5x3  x 2 y2  , (b) lim(x,y)(1,1) e xy cos(x  y) , (c) lim(x,y)(2,1)
(d) lim(x,y)(1,0) ln
(g) lim(x,y)(0,0)
  , (e) lim
1 y2
x 2  xy
y2 sin 2 x
x y
4
4
x 2 ye y
(k) lim(x,y)(0,0)
x 4  4y2
(n) lim(x,y)(0,0)
x  y8
x 4  4y2
(x,y) (0,0) x 2  2y2
, (h) lim(x,y)(1,0)
, (l) lim(x,y)(0,0)
xy  y
(x 1)  y
2
x 2 sin 2 y
x 2  2y2
2
, (f) lim(x,y)(0,0)
, (i) lim(x,y)(0,0)
, (m) lim(x,y)(0,0)
5y4 cos2 x
x 4  y4
xy
x y
2
2
4 xy
x 2 3y2
,
,
, (j) lim(x,y)(0,0)
x 2  y2
x 2  y2 11
,
xy4
2
Ans.: (a) 1 , (b) e , (c) 2 7 , (d) 0, (e) DNE, (f) DNE, (g) DNE, (h) DNE, (i) 0, (j) 0, (k) DNE, (l) 0, (m) 2, (n) DNE.
x 4  y4
x 2  y2
,
2. Determine the set of points at which the function is continuous.
(a)
xy
1 e
xy
1 x 2  y 2
, (b) cos 1  x  y , (c)
x y

 2 2
(g)  2x  y

 1
2 3
if
if
1 x 2  y 2
,
(d)
xy

2
 2
, (h)  x  xy  y
(x, y)  (0, 0)

 1
(x, y)  (0, 0)

ex  ey
e xy 1
, (e) ln  x 2  y2  4  , (f) tan 1
if
(x, y)  (0, 0)
if
(x, y)  (0, 0)

1
(x  y) 2

.



(d)
x  e4y ; (3,0) .
Ans.: (a) everywhere, (b) (x, y) y  x  1 , (c) (x, y) x 2  y2  1 , (d) (x, y) x  0, y  0 , (e) (x, y) x 2  y2  4 ,
(f) (x, y) y  x , (g) (x, y) x  0, y  0 , (h) (x, y) x  0, y  0 .
3. Explain why the function is differentiable at the given point.
(a) 1  x ln(xy  5), (2,3) , (b) x 3 y4 , (1,1) , (c)
x
xy
, (2,1) ,
Ans.: The function is differentible for (a) (x, y) xy  5 , (b) everywhere, (c) (x, y) y  x , (d) Both  x and  y are continuous near
(3,0) ,
4. Show that the function is differentiable by finding 1 and  2 that satisfy the definition of
differentiability. (a) x 2  y2 , (b) xy  5y2 .
Ans.: (a) 1  x , 2  y , (b) 1  y , 2  5y .
5. Determine where f (z)  (x  y)2  2i(x  y) is analytic for real constant  .
6. Determine f (z) , where it exists, and state where f is analytic and where it is not.
(a) (1  2z3 )5 , (b) (x  iy) (x 2  y2 ) , (c) z sin z , (d) 1 (z 2  3iz  2) , (e) 1 (z3  1) ,
(f) x  isin y , (g) z100 , (h)
z , (i) 1
z , both defined by the branch cut (, 0]
Ans.: (a) everywhere, (b) nowhere, (c) differentiable at z  n, n  0, 1, 2 ; analytic nowhere, (d) analytic everywhere except
at z  i, 2i , (e) analytic everywhere except at z  1, 12 (1  i 3) , (f) differentiable along the lines y  (2k  1) 2 , k  0,1, 2,... ;
analytic nowhere, (g) everywhere, (h) & (i) everywhere in the cut plane.
7. Let f  u  iv be differentible with continuous partials at a point z  rei , r  0 . Show
that the polar form of the Cauchy-Riemann equations is
v
u 1 v
1 u
and
.


r
r r 
r 
Use the polar form above to verify the analyticity of the function
f (z) 
y2  x 2
2xy
i 2
.
2
2 2
(x  y )
(x  y 2 ) 2
8. Show that the Cauchy-Riemann conditions (equations) are satisfied by
 (x 3  y3 )  i(x 3  y3 )
, z0

f (z)  
x 2  y2

0,
z0

at z  0 , but that f is not differentible at that point. Thus conclude that Cauchy-Riemann
conditions are insufficient.
9. Let C : z  z(t) , a  t  b , be a regular arc and let z be a particular point on C and let
z  z  z(t  t)  C . The directional derivative of a function f, defined at least on C, at z
in the direction of the arc is defined by
f (z  z)  f (z)
z
z zC
f C (z)  lim0,
provided that the limit exists.
Show that fC (z)  f z  f z e2i where z(t)  x(t)  y(t)  z(t) ei . Thus conclude that if
f z  0 then the directional derivative of f at z depends on  where the value f z is called
the deviation from analyticity.
10. Determine whether or not the given function u is harmonic and if so, in what region. If it
is harmonic, find the most general conjugate function v and corresponding analytic
function f (z)  u  iv expressed in terms of z.
(a) ex cos y , (b) e2x sin 2y , (c) x 3  3xy2 , (d) r 3 sin 3 , (e) r 2 cos 2  4 , (f) r ,
(g) x cos 2x cosh 2y  ysin 2x sinh 2y , (h) x 2  y2 .
Ans.: (a) f (z)  ez , (b) f (z)  e2z , (c) f (z)  z3 , (d) f (z)  iz3 .
11. Let C be the unit circular path centered at origin traversed in the anticlockwise direction in
the complex plane. Show that for 0  k  n
n
n!
1 (1  z) n


 
 k1 dz .
 k  k!(n  k)! 2i C z
12. Show that Green’s theorem in the plane
 v
u 
 u dx  vdy    x  y  dx dy

C
where u and v are two real-valued functions having continuous partial derivatives on a
domain D containing an open set  and its boundary C , can be expressed by
 f
f 
f
(z)dz

i

i

C
  x y  dx dy
in the complex plane for f (z)  u(x, y)  iv(x, y) . Thus, show that
 z dz  2i  Area enclosed by C .
C
13. Evaluate the following line integrals
(a)

C

C
2
z dz ; C is a straight line from 0 to 1  i , (b)
2

C
z dz ; C is the same as in (a), (c)
z dz ; C is a clockwise semicircle from 2 to -2 centered at 0, (d)

C
dz z ; C consists of
piecewise line segments from 1 to 1  i , from 1  i to 1  i , from 1  i to 1 ,
(e)

ez dz ; C consists of piecewise line segments from i to 1  i , from 1  i to 1  2i ,
(f)

Re(z) dz ; C is a clockwise quarter circle from 3i to 3 centered at 0, (g)
C
C

C
Im(z)dz ;
C is a straight line from i to 2  2i .
1
3
Ans.: (a)
(2  2i) , (b) 1 , (c) 4i , (d) i , (e) (ecos2  cos1)  i(sin1  esin 2) , (f)
1
4
(18  9i) , (g)
1
2
(6  3i) .
14. Evaluate the following integrals using Cauchy’s theorem, if applicable.
(a)

Re(z)dz , (b)
(f)

dz z(z  2) , (g)
(k)

z dz , (l)
z dz , (m)

(p)

z dz (z 2  3z  2) , (q)

C1
C1
C1
C4

C3

C1

Im(z)dz , (c)

C3
dz z(z  5) , (h)
C2
C4

C1
C3

C4
dz z3 (z 2  1) , (r)
dz (z 2  3) , (e)

esin z dz , (i)
dz z(z  1) , (n)
C4

Im(z)dz , (d)
C2
C4
C1
sin(cos z)dz , (j)
dz z(z  5) , (o)



C4
z 4dz ,

C3
dz z ,
z dz (z 2  1) ,
z dz (z 3 1) ,
where C1 : z  1 ; anticlockwise, C2 : z  1 ; clockwise, C3 ; the square with vertices at
1  i , 1  i , 1  i , 1  i anticlockwise, C4 : z  3 ; anticlockwise.
Ans.: (a) i , (b)  , (c) 4 , (d) 0 , (e) 0 , (f) i , (g)  52 i , (h) 0 , (i) 0 , (j) 0 , (k) 2i , (l) 8i , (m) 0 , (n) 0 , (o) 2i ,
(p) 2i , (q) 0 , (r) 0 .
15. Use Cauchy Integral Formula to evaluate the following integrals.
(a)

(f)

(k)

C
cos z
z
e2 z
5
C z
dz , (g)
z3
2
C z i

dz , (b)
dz , (l)

C
sin z
z
dz , (c)
sinh 3z
2
2
C (z 1)

dz , (h)
1
C z(z  2)(z  4)
1
2
C z 5z
dz , (d)


dz , (i)


z2
4
C z 1
z 2 1
2
C z 1

ez dz , (e)

2
ez
C zcos(z 2)
dz , (j)
z 1
2
C (z 1)(z  2)
z
2
C (z i)(z 1)
dz ,
dz ,
dz
where C : z  3 ; anticlockwise.
Ans.: (a) 2i , (b) 0 , (c)  52 i , (d) 4isin1 , (e) 0 , (f)
4
3
i , (g) i(sin3  3cos3) , (h) 0 , (i) 2i , (j) 0 , (k) 2 .
16. Cauchy Integral Formula
f (z)  21i

f ( )
C  z
d .
expresses an analytic function f (z)  u  iv in terms of its boundary values, thus it is
expected there to exist a similar integral formula expressing a harmonic function u(x, y)
in terms of its boundary values. For this purpose, let C be the anticlockwise circle   R
or   Rei . Show that the Cauchy integral formula can be re-expressed as
f (z) 
1
2 i

1
 z
 R12
z
 f () d 
C
1
2
 
2
0

 z

 z z f () d ,
where the bracketed quantity is real. Show that it leads to the Poisson integral formula for
the circular disk
u(r, ) 
1
2

2
0
(R 2  r 2 )
R  2Rr cos(  )  r 2
2
u(R, ) d ,
where z  r ei .