# Contour Integration

## Homework Statement

∫$_{\gamma}$(x-y)dz where $\gamma$ has parametrization: z(t) = e$^{it}$ for $\pi$/2 $\leq$ t $\leq$ 3$\pi$/2

## Homework Equations

the integral of the sum is the sum of the integral

## The Attempt at a Solution

I tried to break it up and see if I could evaluate it as I normally would but it started to get really messy and I think I was going about it wrong.

z=exp(it) and dz=iexp(it)
∫(x)dz - i∫(y)dz

can someone please push me into the right direction? thank you

lanedance
Homework Helper
as you mean to evaluate directly you need to use a substitute for the parameterisation variable

$$\cint_{\gamma} f(z) dz = \int_{t_a}^T_b f(z(t)) z'(t) dt$$

could you elaborate just a little more? I really appreciate your help

lanedance
Homework Helper
what don't you understand? I won't do the problem for you, but am happy to help

I dont understand how to set up a substitution so I can get it into a form that I know how to work with

HallsofIvy
Homework Helper
You intend that z= x+ iy, right? So if $z= e^{it}= cos(t)+ i sin(t)$ what are x and y in terms of t?

Okay this is what I have so far:

∫$_{\gamma}$(x-y)dz where $\gamma$ has parametrization: z(t)=e$^{it}$ for $\pi$/2$\leq$t$\leq$3$\pi$/2

∫$_{\gamma}$(x-y)dz = ∫$_{\gamma}$(cos(t)-isin(t))dz

which we can break up like,

∫$_{\gamma}$(cos(t))dz - i∫$_{\gamma}$(sin(t))dz

since z(t)=e$^{it}$ then dz=ie$^{it}$

so we have

∫$_{\gamma}$(cos(t))(ie$^{it}$) - i∫$_{\gamma}$(sin(t))(ie$^{it}$ )

is this correct so far?

or would my dz be -sin(t)+icos(t)?

Last edited:
lanedance
Homework Helper
you may want to check your "y" value, remember z= x+ iy,

so instead of isin(t) just simply sin(t)

∫$_{\gamma}$(x-y)dz = ∫$_{\gamma}$(cos(t)-sin(t))dz

∫cos(t)dz-∫sin(t)dz

every time I try evaluating the above expression I never can seem to get the correct answer. I'm using ie^(it) for my dz

AHHH im getting so frustrated. I feel like I am missing an important concept, this problem should not take me this much time