How do I integrate this continuous complex function?

[dla]

1. Homework Statement [/b]
\int _{C} Re z^{2} dz clock wise around the boundary of a square that has vertices of 0, i, 1+i, 1.

Homework Equations

\int_{c} f(z) dz = \int \stackrel{b}{ _{a}} f[z(t)] \stackrel{\cdot}{z(t)}dt

The Attempt at a Solution

Since it is piece-wise continuous I know I need to integrate by the use of the path. So I first need to represent the path in form z(t) from a≤t≤b, find that derivative, and then sub in the values and integrate over it.

But I'm having a really hard time representing the path C in the form of z(t) (a≤t≤b), I was looking at examples but still don't get how to do it. I feel like I need to assume Re z = x and do something to get it in terms of t

Can anyone explain the steps in general any help will be very much appreciated!

 

[SteamKing]

Take the line segment from 0 to i. Can't z(t) be represented by z(t) = (0,t) for 0<=t<=1? Likewise for the other portions of the path. The total integral should be the sum of the integrals on each portion of the path C.

 

[dla]

Hi SteamKing,

Thanks for replying, I see what you did. I tried it for the other curves just to check if my understanding is correct, so from i to 1+i is the representation, z(t)=t+i 0≤t≤1? How did you know t is between 0≤t≤1?

Also how do I take direction into account for example if it is from 1+i to i, would it then be z(t)= -t+i?

 

[SteamKing]

t is just a parameter. You get to decide what range of values you wish to use for t, so I chose to make it convenient for that particular segment of C.

For other segments of C, there may be several different formulations using t. For 1+i to i, you could use
z(t) = ((1-t), i), 0<=t<=1.

 

[dla]

How did you get z(t)=(1-t, i) for i to 1-i? I don't think I have the general idea down, I thought it'd be -t for the real part since it's changing and the imaginary parameter will be constant, thus z(t)=(-t, 1+i)?

 

[Tsunoyukami]

You can parameterize a line connecting $z_{1}$ to $z_{2}$ in C as:

$\gamma (t) = tz_{2} + (1 - t)z_{1}$, where $t \in [0, 1]$

This is easy to see because if t = 0, we are at the point $z_{1}$ and when t = 1 we are at the point $z_{2}$ - and this forms a straight line connecting these points from $z_{1}$ to $z_{2}$.

 

[SteamKing]

It's a good idea to lay out the vertices of C in a regular way.

Remember, complex numbers have two components: a real part and an imaginary part.

From the OP, the vertices of C are 0, i, 1 + i, 1, and back to 0 to close the path.
So, in ordered pair form, the vertices are (0,0), (0,1), (1,1), (1,0), and (0,0), where it is understood that
(x,y) = x + iy

I made a mistake in the parameterization above for the second segment from i to 1+i:
it should be z(t) = (t, 1), where 0<=t<=1. The form of (x(t),y(t)) will depend on your choice of the range for the values of the parameter t.

 

[dla]

Oh I get it now thank you so much Tsunoyukami that's a really useful method! And thanks for the explanations SteamKing I know where I went wrong now, I find that drawing it actually really helps a lot.

 

[Tsunoyukami]

I'm glad to have been of help. If this course is an introduction to complex variables the majority (if not all) of the curves you will be asked to integrate over will be formed by a combination of straight line segments and arcs of circles so you may find it useful to memorize parameterizations for each of these curves.

The parameterizations of a circle is most commonly:

$\gamma(t) = Re^{it}$ where $t \in [0, 2\pi]$