hi here is the question, by integrating explicitly in the complex plane, evaluate

$\displaystyle \int_C (z-1)^6 dz $

where C is the half-circle centered at z = 1 with radius 1, starting at z_0 = 2 and ending at z_1=0, and traveresed in the upper half-plane. Show that the fundamental theorem of calculus holds for this example.

here's my working out:

parametrise: $\displaystyle z(t) = 1 + e^{it} $

$\displaystyle dz = ie^{it}, 0 <=t<= \pi $

$\displaystyle \int^\pi_0 (e^{it} - 1 + 1)^6 i e^{it} dt $

let $\displaystyle x = e^{it} $ and $\displaystyle dx = ie^{it} dt $

$\displaystyle \frac{x^7}{7} ]^\pi_0 $

$\displaystyle = -2/7 $

my fundamental part doesnt equal, coz when i did it i got -128/7

so can someone please help me where i got it wrong?

coz when i FIRST did it i parametrise this way:

$\displaystyle z(t) = e^{it} $

then

$\displaystyle \int^\pi_0 (e^{it} - 1 )^6 i e^{it} dt $

and i got the same answer as the fundamental part, but since the question says centered at z = 1 dont you have to add the 1 to z(t)?