Integrating x-y Along Contour: Step-by-Step Guide

[Polamaluisraw]

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]

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

<br /> \cint_{\gamma} f(z) dz = \int_{t_a}^T_b f(z(t)) z&#039;(t) dt

 

[Polamaluisraw]

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

 

[lanedance]

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

 

[Polamaluisraw]

I don't understand how to set up a substitution so I can get it into a form that I know how to work with

 

[HallsofIvy]

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?

 

[Polamaluisraw]

Okay this is what I have so far:

∫_{\gamma}(x-y)dz where \gamma has parametrization: z(t)=e^{it} for \pi/2\leqt\leq3\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)?

 

[lanedance]

you may want to check your "y" value, remember z= x+ iy,

 

[Polamaluisraw]

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