# Thread: integrating in the complex plane

1. ## integrating in the complex plane

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

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

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: $z(t) = 1 + e^{it}$
$dz = ie^{it}, 0 <=t<= \pi$

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

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

$\frac{x^7}{7} ]^\pi_0$
$= -2/7$

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

coz when i FIRST did it i parametrise this way:
$z(t) = e^{it}$

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

2. You got the first part right:

A parametrisation of $C$ is $C\left(t\right)=1+e^{it}$.

Now, for $f\left(z\right)=\left(z-1\right)^6$ the integral is

$\int_C f=\int_C f\left(z\right)$ $=\int_a^b f\left(C\left(t\right)\right)C^\prime\left(t\right )~dt$ $=\int_0^\pi\left(1+e^{it}-1\right)^6\left(ie^{it}\right)~dt$ $=\int_0^\pi ie^{7it}$ $=\left[\frac{1}{7}e^{7it}\right]_0^\pi=-\frac{2}{7}$.

You didn't show your working for the FTC part, but I believe you got that part wrong.

For $f=\left(z-1\right)^6$, the anti-derivative $F=\frac{1}{7}\left(z-1\right)^7$.

The fundamental theorem of calculus states that

$\int_C f=F\left(C\left(b\right)\right)-F\left(C\left(a\right)\right)$ $=F\left(C\left(\pi\right)\right)-F\left(C\left(0\right)\right)$ $=F\left(0\right)-F\left(2\right)$ $=-\frac{1}{7}-\frac{1}{7}$ $=-\frac{2}{7}$.

As expected, the result derived from the fundamental theorem and the result from explicit integration are the same.

Maths help