EXERCISE(Complex Variable)
1. Show that the following limits does not exit.
Re z 2
 1

(a) lim 2 (b) lim 
+ ix 2 
1
y
z →0
z →0 1 − e


z
2. Examine the continuity of the function
 Img z 3
, z≠0

2
f ( z) =  z

, z=0
 0
3. Show that for the function
 x 3 − 3 xy 2  y 3 − 3 x 2 y 
+i 2
, ( x, y ) ≠ 0

2 
f ( z) =  x2 + y 2
+
x
y



0
, ( x, y ) = 0

Cauchy-Riemann Equations are satisfied at the origin.
Does f ′(0) exist ?
4. Find an analytic function f ( z )= u + iv such that Img f ′( z ) = 1 + y and f (1 + i ) =
0.
5. Find the values of the constants a, b, c & d such that the function
f ( z ) = ( y 2 + axy + bx 2 ) + i (cy 2 + dxy + x 2 ) is analytic.
6. Let f ( z )= u + iv and g ( z )= v + iu be analytic functions for all z . Show that u and
f ( z)
.
v are constant. If f (0) = 2i and g (0) = 5 , then find the value of h( z ) =
g ( z)
7. If f ′( z ) = − f ( z ) for all z , then find f ( z ) .
8. Show that if f ( z )= u + iv and g ( z )= A + iB are differentiable at a point z ,
u x u y Ax Ay
2
Then f ′( z ).g ′( z ) =
vx v y Bx By
9. If f ( z )= u + iv is an analytic function, show that f ( z ) is not Harmonic.
10. If f ( z )= u + iv is an analytic function, show that log f ( z ) is a Harmonic.
11. Prove that f ( z ) = z is differentiable but not analyticat z = 0 .
4
12. If C is the curve y = x 4 + x 2 − x + 1 joining points (0,1) and (1, 2) , then find the
value of ∫ ( z 2 − iz ) dz .
C
sin z + cos z
where C is the circle z = 4 .
π
C z−

( z −π )
2

cos z
4.
if C is the ellipse z − 1 + z + 1 =
14. Evaluate ∫
πi 
C z −πi  z −
(
)

2

1
e zt
15. Evaluate
∫ 2 2 dz if t > 0 and C is the circle z = 3 .
2π i 
C ( z + 4)
13. Evaluate ∫
16. Find Laurent series about the indicated singularity for each of the following
functions. Name the singularity in each case and give region of convergence of each
series.
1
(a ) ( z − 3) cos
, z=
−4
z+4
1
(b)
, z = −3
( z + 3) 2 z 2
17. Expand the function
1
in the Laurent series valid for
f ( z) =
( z + 2)( z + 4)
(a ) 2 < z < 4 (b) z > 4
(c ) 0 < z + 2 < 2 ( d ) z < 2
18. Determine the poles and Residues at the poles for f ( z ) = sec h z .
19. Evaluate
2 cos π2 z
∫ z ( z − 2 )
2
dz .
C
∞
20. Evaluate
∫ (x
0
2
dx
+ 1) ( x 2 + 4) 2
Hint & Answers
Ex -1. (a) Use paths x → 0 , y → 0 and y → 0, x → 0
(b) Use paths x → 0, y → 0 + and x → 0, y → 0 −
2. Continuous. Use r , θ definitions
∂u
∂x
∂u
∂y
∂v
∂x
∂v
∂y
3. At (0,0) we have= 1,= 0,= 0,= 1 , but lim
z →0
f ( z ) − f (0)
does not exist (e.g.
z −0
choose path y=mx)
4. Ans. f ( =
z)
1 2
( z + 2iz + 2 − 4i )
2
5. a =
−2, b =
−1, c =
−1, d =
−2 by showing partial derivatives are continuousand then calculate by
assuming C − R are satisfied
6. h( z ) =
2i
5
7. f ( z ) = ke − z
Use f ′( =
z)
8. Use f ′( =
z)
∂u ∂v
where f ( z )= u + iv then apply C − R equations
+i
∂x
∂x
∂u ∂v
+i
∂x
∂x
9.=
w | f (=
z) |
u 2 + v 2 does not satisfy Laplace equation.
10. Use Laplace equation
for w
=
1
log(u 2 + v 2 )
2
f ( z ) − f (0)
using x r=
cos θ , y r sin θ But C.R. are not satisfied at
2r 3 0=
≤ lim
=
z →0
r →∞
z
(0,0) for f ( z ) =
| z |4 =
( x 2 + y 2 )2 , z ≠ 0
11.=
f ′(0) lim
1 5i
) z 2 − iz is analytic, hence Integral is Independent of path. So choose
+ . As f ( z=
6 2
path c= c1 ∪ c2 when c1 : straight line joining (0, 1) to (1, 1) & c2 : straight line joining (1, 1) (1, 2)
12. Ans.=
∫ (z
c1
2
1 i
1
− iz )dz =+ while ∫ ( z 2 − iz )dz =−
2i
3 2
6
c2
13. Ans. −8i
−π
& π which lie inside |z|=4. Apply Cauchy Integral formula after breaking the
2
Singular points are


2  sin z + cos z
sin z + cos z 
dz − ∫
dz 
Integral into  ∫
π
π | z| = 4 z − π

z−
2


14. Ans. −4 cosh
π
2
Singular points z = π i (lies outside) while
π
2
i (lies inside). Break up the integral into two parts &
apply Cauchy Theorem and Cauchy Integral Formula.
15. Ans.
1
[ −t cos 2t + sin 2t ]
8
Singularities are z = ±2i , both are of order 2 lying inside |z|=3.
Residue
z = +2 i
(z
e zt
2
+ 4)
2
=
e 2it
e zt
e −2it
&
−
16
t
−
8
i
Residue
=
−16t + 8i ]
[
] z = −2i 2 2
4 [
44
4
4
z
+
(
)
Apply Residue Theorem
16. Ans. (a) z − 3 −
1
7
.............; z =−4 is essential singularity. Put z + 4 =
+
4 & use
2( z + 4) 2( z + 4) 2
the expansion of cos
1
around 0. The series converges for all values of z ≠ −4
u
1
 2

2
1 + 3 ( z + 4) + 3 ( z + 3) + ...... series converges 0 <| z + 3 |< 3, z =−3 is double
9 ( z + 3)
1
(b)
2
pole
17. Ans. (a) ...... −
4 2 1
1 1 z
z2
+
−
+
−
+
−
+ ......
z 2 z 3 z 2 2 z 8 32 128
(b)
1 6
− + .....
z 2 z3
(c)
1
1 1
− + ( z + 2) + ......
2( z + 2) 4 8
1
8
(d) −
3
z + .......
32
−1
Hint use
1
1 2
1
1 1
1 
=
=
−
1 +  expand

 . If |z|>2,
( z + 2) z  z 
( z + 2)( z + 4) 2  z + 2 z + 4 
Assuming
18. Ans.
z
=
2
< 1 similarly others
|z|
1
(2n + 1)π i for n= 0, ±1,..... poles; Residue = (−1) k +1 i , k=0,1,....
π
2
=z (2 n +1)
2
19. Ans. 2π i , Apply Residue theorem. Poles are z=0(order 1), z = 2 (order 2).
5π
.Use
20. Ans .
288
dz
∫c ( z 2 + 1)( z 2 + 4) where C:
Poles z = i of order 1& z = 2i of order 2 lies inside c
∫ ( z
c
2
1
11
5π
dz
)=
.
= 2π i ( −
2
+ 1)( z + 4)
18i 288i
288
C
-R O R