Simple Cauchy's integral theorem problem

[John O' Meara]

Integrate f(z) counterclockwise around the unit circle indicating whether Cauchy's integral theorem applies, ( show details of your work).
(A) $$z(t) = \cos t + \iota \sin t =\exp{\iota t} \mbox{ for } 0 \leq \ t \ \leq 2\pi \\$$
So that counterclockwise integration corresponds to an increase of t from 0 to 2$$\pi$$
(B) $$\frac{dz(t)}{dt} = \iota \exp{\iota t} \\$$
(C) f[z(t)] = x(t) = cos(t). Therefore
(D) $$\oint_C \Re z dz = \int_0^{2\pi}\cost \iota\exp{\iota t} \\$$
Integrating by parts I get $$\iota \int_0^{2\pi} \cos t \exp{\iota t} = \frac{\exp{2\pi\iota}-1}{2} \\$$. I could be wrong about x(t)=cos(t). Cauchy's integral theorem does not apply as f(z) = Re(z) = cos(t) is not analytic.

 

[HallsofIvy]

Are you saying that you are given that f(z)= Re(z)? If so, then you are correct.

 

[John O' Meara]

Yes I am saying that f(z)=Re(z), and the answer in the 'book is $$\pi\iota$$. Thanks for the reply. In fact $$\exp{2\pi\iota} = 1$$, hence the integral is zero.