Evaluating Integral on Parabolic Curve: Im(z)dz

[squaremeplz]

Homework Statement

evaluate the contour integral

int.cont. Im(z) dz

where cont. is the path from the point z_o = 1 + i to the point z_1 = 3 + 9i

along the parabolic curve y = x^2. sketch the contour

Homework Equations

int.[a to b] z(t)z'(t) dt

The Attempt at a Solution

I parametrized z(t) = t + t^2*i and runs 1 <= t <= 3

z'(t) = 1 + 2ti

Here is where I get confused

I know Im(z(t)) = t^2*i, but do I also have to do Im(z'(t)) = 2ti?

so I am unsure whether to evaluate

int [from 1 to 3] Im(z(t))*Im(z'(t)) dt

or

int [from 1 to 3] Im(z(t))*(z'(t)) dt

Also, since we are integrating the imaginary part only, how does the sketch differ from a contour integral of let's say z(t)dt

thanks

 

[LeonhardEuler]

I think you are being confused by the fact that the function you are integrating takes the imaginary part of its argument. Think about how you would solve the problem generally, look at f(z) instead of Im(z). Im(z) is just a special case.

If you were integrating f(z), would you then also take f(z') (If that makes sense)?

Also, your equation under 'relevant equations' is not quite right. The equation
$$\int_{a}^{b}z(t)z'(t)dt$$
does not take into account that you are looking at the imaginary part. How should this equation look for a general f(z)?

 

[NJunJie]

Hi,

I 've a question here:
sin (z) / (z^15)

How to find singularities; locale etc?
Related to this theory?

 

[squaremeplz]

for general f(z)

isnt it

int [ a to b ] [u(t) + iv(t)] dt

=

int [a to b] u(t) dt + i* int [ a to b] v(t) dt

then I am just solving

i* int [ a to b] v(t) z'(t) dt

?

Thanks

 

[LeonhardEuler]

for general f(z)

isnt it

int [ a to b ] [u(t) + iv(t)] dt

=

int [a to b] u(t) dt + i* int [ a to b] v(t) dt

then I am just solving

i* int [ a to b] v(t) z'(t) dt

?

Thanks

How are you defining u(t) and v(t)? Is it f(z(t)) = u(z(t)) + i*v(z(t)), or z(t) = u(t) + i*v(t) ? Depending on which one you mean, your last equation may be correct. Your first two equations, though, are missing something. How do you convert an integral 'dz' to an integral 'dt'?

 

[squaremeplz]

please ignore the first two equations as they were pertaining to f(t) not f(z(t)). I know u have to add the derivative of z(t).

in the last equation,

i* int [ a to b] v(t) z'(t) dt

u(t) and v(t) are defined in accord to

f(z(t)) = u(z(t)) + i*v(z(t))

my final integral looks like this

int [from 1 to 3] (t^2*i)(1 + 2ti) dt

is this correct? thanks for the help!

 

[LeonhardEuler]

You have the idea exactly right. There is a minor problem with your interpretation of Im(z). The way it is defined is that for z = x + i*y (x and y real), Im(z) = y, not Im(z) = i*y. You are treating it like it is the second case.