MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

15. We shall remember Example 4:
∫
C
dz
z 2 = 0
where C is the unit circle. This result does not follow from Cauchy’s theorem, because f(z) = 1
z 2
not analytic at z = 0. Hence the condition that f be analytic in D is sufficient rather than necessary for
Cauchy’s integral theorem to be true.
Then they can be deformed each other. So integral of f is 0.
16. İf C is the unit circle, ∫ f(z)dz = 3
İf C is |z| = 2, ∫ f(z)dz = 5.
is
f isn’t analytic in the annulus 1 < |z| < 2 because of the principle of deformation of path.
17.
a) C 1 : x : 0 → π and y : 0 → π
y = x =⇒ dy = dx.
İf z = x + iy, dz = dx + idy.
=⇒ ∫ C 1
cos zdz = ∫ cos(x + iy)(dx + idy) = ∫ π
0 cos((1 + i)x)(1 + i)dx = sin((1 + i)x)∣ ∣ π 0 =
sin((1 + i)π) − sin 0 = sin(π + iπ) = − sin(iπ).
b) ∫ C cos zdz = ∫ C 2
+ ∫ C 3
C 2 : x : 0 → π and y = 0 =⇒ dy = 0
İf z = x + iy, dz = dx + idy = dx.
=⇒ ∫ C cos zdz = ∫ π
0 cos(x + iy)dx = ∫ π
0 cos xdx = sin x∣ ∣ π 0 = sin π = 0
C 3 : x = π =⇒ dx = 0 and y : 0 → π
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