Expert: Alon Mandes - 11/8/2009

Question
QUESTION: Using parameterization perform:

∮〖(e^z)dz〗along c:|z|=1

ANSWER: (1/2πi)∫ e^z dz = 0 Because the complex function f(z)=e^z is Analytic is the whole complex
    |z|=1
plane z . This is according to Cauchy's integral theorem .

Alon.




---------- FOLLOW-UP ----------

QUESTION: The question says using parameterization not (C.I.th).
Integrate using parameterization:
1.∮e^3z dz  , along the circular arc c:|z-1|=1 from z1 = 0 and z2 =1+i.
2.∫e^z/z dz , along c:|z|=1

Answer
a) f(z)=e^(3z). Our parametrization will be : z(t)=(1+i)t , where 0≤t≤1 . Therefore :
  dz=(1+i)dt . Thus,

∫e^(3z) dz =

1
∫(1+i)e^[3(1+i)t] dt =
0


[(1+i)]/[3(1+i)]*e^[3(1+i)t] { from t=0 to t=1} =
(1/3)e^[3(1+i)] - (1/3) =
(1/3)[(e^3)*e^(3i) - 1)] =
(1/3)[(e^3)*( cos(3)+isin(3) ) - 1)] =
(1/3)[20.05+1.05i -1] =
(1/3)[19.5+1.05i] =
6.5+0.35i


b)   ∫e^z/z dz = ?
  |z=1|

Our parametrization will be : z=e^(it) --> dz=ie^(it)dt . Thus,
     ∫e^z/z dz =
  |z=1|

2π
∫ [e^(e^it)]*[e^(it)]/[e^(it) idt =
0   

2π
∫ e^(e^it) idt = This integration is not performable, thus we need to find another approach.
0

We will use Taylor series for complex variable :
e^z=1+z+z²/2+z³/3+...   Hence :
∫(e^z)/z dz = ∫(1+z+z²/2+z³/3+...)/z dz = ∫(1/z)+1+z/2+z²/3+...) dz =
∫(1/z) dz + +∫dz +∫z/2 dz + ∫z²/3 dz +... =

2π
∫ [1/(e^it)]e^(it) idt + ∫ie^(it) dt + (1/2)∫ie^(it)e^(it) dt + (1/3))∫ie^(it)e^(2it) dt +.. =  
0

2π
∫ idt + i∫e^(it) dt + (i/2)∫e^(2it) dt + (i/3))∫e^(3it) dt +... =  
0

2πi + [e^(2πi)-e^0] + (1/4)[e^(4πi)-e^0] + (1/6)[e^(6πi)-e^0] +...=
2πi + 0 + 0 + 0 +0 + ... = 2πi

Alon.