MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

∫C f(z)dz = ∫ a
b f(z(t))z′ (t)dt,
z ′ = dz
dt
C : z(t) = cos t + i sin t, 0 ≤ t ≤ 2π
z ′ (t) = ie it
f(z(t)) = z(t) + 1
z(t) = eit + e −it
∫C z + z−1 dz = ∫ 2π
0 (eit + e −it ).ie it
= ∫ 2π
0
i.(e 2it + 1)dt
= i.( 1 2i e2it + t) ∣ ∣ 2π
0
= ( 1 2 e2it + it) ∣ ∣ 2π
0
= 1 2 e4iπ + 2iπ − 1 2 e0 − 0
= 1 2 (cos 4π + i sin 4π) + 2iπ − 1 2
= 1 2 + 2iπ − 1 2
= 2iπ
25. f(z) = cosh 4z is analytic in C. So, we use first method. Let F (z) = 1 4
sinh 4z. F (z) is analytic also
and F ′ (z) = 1 4
.4. cosh 4z = cosh 4z = f(z).
∫C cosh 4zdz = ∫ πi
8
cosh 4z
− πi
8
= F ( πi
8
= 1 4
) − F (−
πi
8 )
4πi
sinh(
8 ) − 1 −4πi
4
sinh(
8
)
= 1 4 (i sin π 2 + i sin π 2 )
= 1 4 .2i
= i 2
26. z(t) = t + it 2 , (−1 ≤ t ≤ 1), z ′ (t) = 1 + 2it, ¯z = t − it 2 , so that
∫ 1
−1
(t − it 2 )(1 + 2it)dt =
∫ 1
−1
(2t 3 + it 2 + t)dt = 1 3 it3 ∣ ∣∣∣
1
33
−1
= 2 3 i.