1. Expand the following functions in the given region (first two terms should suffice for (b)).

(a) 1
z2 − 1
     |z + 1| > 2,       (b) 1
1 − cos z
     0 < |z| < 2 π.


2. Is the following logic correct ?
Since i2 = −1,
i =
√
 
−1
 
(take the square roots on the both sides). Also since (−i)2 = −1, it follows
−i =
√
 
−1
 
 .
Therefore, i = −i from which it follows 2 i = 0, i.e. i = 0 !!!


3. Evaluate
⌠
(⎜)
⌡


C 
 ez
z2 + 1
 dz
where C is |z − i| = 1 ccw.




4.

    (a) What is the fundamental theorem of algebra ? Why is it so important ?
    (b) If the real part of an analytical function is given as
    u = x2 − x − y2
    find the imaginary part.


5. Evaluate the following integrals.
(a) ⌠
⌡ 		∞

−∞ 
 cos x
x2 − 2 x + 2
 dx
(b)
⌠
⌡ 		2π

0 
 dθ
4 + cos θ
6. Evaluate
⌠
⌡ 		∞

0 
 √x
x2 + a2
 dx
by considering
⌠
(⎜)
⌡
 √z
(z2 + a2)
 dz
along the contour shown below:
7.

    (a) What is the streamline ? How can one find the streamline ?
    (b) What is the residue of
    cos z
    z5
    at z = 0 ?


SOLUTION


1. (a) Note that
1
z2 − 1
 = 1
z − 1
 1
z + 1
 .
Since the region (|z + 1| > 2) contains both singular points (1 and -1), the Laurent series is required.
1/(z + 1) is already in such a format so leave it as it is.

1
z − 1
=
1
(z + 1) − 2
(1)
=
1
(z + 1) (1 − 2
z + 1
 )
(2)
=
1
z + 1
 ⎛
⎝ 		1 + ( 2
z + 1
 ) + ( 2
z + 1
 )2 +( 2
z + 1
 )3 + …⎞
⎠
(3)
so

1
z2−1
 = 1
(z + 1)2
 ⎛
⎝ 		1 + ( 2
z + 1
 ) + ( 2
z + 1
 )2 +( 2
z + 1
 )3 + …⎞
⎠ 		.
(4)


(b) As
cosz = 1 − z2
2!
 + z4
4!
 − …,

1 − cosz  ∼  z2
2!
        as     z → 0.
so

1
1−cosz
=
⎛
⎝ 		z2
2
⎞
⎠
1−cosz
 1
⎛
⎝ 		z2
2
⎞
⎠
(5)
=
g(z) 1
⎛
⎝ 		z2
2
⎞
⎠
(6)
=
⎛
⎝ 		g(0)+g′(0)z + g"(0)2
2!
 z2 + …⎞
⎠ 		1
⎛
⎝ 		z2
2
⎞
⎠
(7)
By long division or direct calculation,
g(0)=1,     g′(0)=0,     g"(0)= 1
12
 , …
so
1
1−cosz
 = ⎛
⎝ 		1+ z2
12
 + z4
240
 +…⎞
⎠ 		2
z2
 .


2.

√
 
−1
 
 =
√
 
1 eπi/2 + 2 n πi
 
 = eπi /2 en πi = ±i.


3.
⌠
(⎜)
⌡
 ez
(z−i)(z+i)
 dz
=
⌠
(⎜)
⌡
 g(z)
z−i
 dz
(8)
=
2πg(0)
(9)
=
2 πi ei
2i
(10)
=
πei
(11)
=
π(cos1 + i sin1).
(12)
where
g(z) ≡ ez
z+i
 .


4.
(a) An algebraic equation has at least one zero on the complex plane. It is significant as all the roots are expressed by a complex number.


(b) Note that (x+i y)2 = x2 −y2 + 2 x y i, come up with
u = 2 x y − y
so that
u + i v = x2−y2−x + i (2 xy −y ) = (x+iy)2 − (x+iy) = z2 −z.


5 (a) The zeroes of x2−2x+2=0 are 1±i of which 1+i is in the upper half plane.

I
=
ℜ⎛
⎝ 		⌠
⌡ 		∞

−∞ 
 eix
x2−2x+2
 dx ⎞
⎠
(13)
=
ℜ⎛
⎝ 		2πi Res ( eiz
z2−2z+2
 ;1+i) ⎞
⎠
(14)
=
ℜ⎛
⎝ 		2πi ei (1+i)
2i
⎞
⎠
(15)
=
ℜ(πei−1)
(16)
=
ℜ⎛
⎝ 		π
e
 (cos1 + i sin1)⎞
⎠
(17)
=
π
e
 cos1.
(18)
(b)
I
=
⌠
(⎜)
⌡
dz
i z
4 + (z + 1/z)/2
(19)
=
2
i
 ⌠
(⎜)
⌡
 dz
z2 + 8 z + 1
(20)
=
2
i
 2πi Residue( 1
z2 + 8 z + 1
 ;−4+
√
 
15
 
 )
(21)
=
4 π 1
−8 + 2
√
 
15
 
 + 8
(22)
=
2 π

√
15
 .
(23)


6
⌠
(⎜)
⌡
=
2πi (Res( a eπi /2 )+Res( a e3πi /2 ))
(24)
=
2πi ⎛
⎝ 		√a eπi/4
2 a i
 + √a e3 πi/4
2 a (−i)
⎞
⎠
(25)
=
2 πi √a
2 a i
(eπi /4−e3 πi/4)
(26)
=
  ⎛
√
2
a
 
 π.
(27)
On the other hand,
⌠
⌡

I 
=
I
(28)
⌠
⌡

II 
=
0
(29)
⌠
⌡

III 
=
⌠
⌡ 		0

∞ 
 eπi √x
x2+a2
 dx
(30)
=
− ⌠
⌡ 		∞

0 
 (−1) √x
x2+a2
 dx
(31)
=
⌠
⌡ 		∞

0 
 √x
x2+a2
 dx
(32)
=
I
(33)
⌠
⌡

IV 
=
0,
(34)
so
2I =   ⎛
√
2
a
 
 π
from which
I = 1
2
   ⎛
√
2
a
 
 π.


7. (a) dx : dy = u: v. Ψ = const.
(b)
cosz
z5
=
1− z2/2! + z4/4! − z6/6!+ …
z5
(35)
=
…+ 1
4!
 1
z
 − …
(36)
So the residue is 1/4!.


File translated from TEX by TTH, version 4.03.
On 25 Feb 2023, 15:12.