How Do You Solve This Complex Integral with a Curved Path?

[malawi_glenn]

Homework Statement

Evaluate:

$$\int _{c} \dfrac{1- Log z}{z^{2}} dz$$

where C is the curve:

$$C : z(t) = 2 + e^{it} ; - \pi / 2 \leq t \leq \pi / 2$$

Homework Equations

I know the independence of path in a domain where f(z) is analytical, but I tried the standard parametrization just to beging with someting.

The Attempt at a Solution

$$z^{2} = 4 + 4e^{it} + e^{2it}$$

$$Log(2 + e^{it} ) = \frac{1}{2} \ln (5 + \cos t) +it$$

$$dz = ie^{it} dt$$

$$i \int _{- \pi / 2} ^{\pi / 2} \dfrac{1 -\frac{1}{2} \ln (5 + \cos t) -it }{4 + 4e^{it} + e^{2it}}e^{it} dt$$

lol iam lost

 

[malawi_glenn]

do you think I shall use independence of path?

 

[malawi_glenn]

never mind, I solved it.