Complex- area enclosed by a polygon

[kathrynag]

Homework Statement

Recall that the area enclosed by the polygon with vertices z1,z2,z3,...,zn is
1/2I(z1conguatez2+z2congugatez3+...+zncongugatez1)

Show that the area enclosed =1/2I\Sigmazkcongugate(zsub(k+1)-zk).
Interpret this sum as part of the approximating sum in the definition of the line integral about C of z congugatedz.

Homework Equations

The Attempt at a Solution

I don't even know where to start. I understand where the 1/2I(z...) comes from, but once the summation comes in, I'm lost.

 

[Billy Bob]

This really is not as hard as you think, especially since you already believe A=1/2 Im[ bar(z_1) z_2 + bar(z_2) z_3 +...+ bar(z_n) z_1 ]

For the summation 1/2 [ Im( bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n) ] just "remove parentheses" and observe that bar(z) z=|z|^2 so its imaginary part is 0.

As for the line integral, use the fact that \int_C f(z) dz is approximately \lim\sum_{k=1}^n f(z_k^*)(z_k-z_{k-1})

 

[kathrynag]

Do i need to show using green's thm?

I did some work and have a couple questions:
Let x be polygon with its ordered vertices (the vertices you take in order
as you travel around it) be .

Also say they are taken in a counterclockwise fashion.
Let x be the line segment from zsubj,zsub(j+1) for j=1,2,...n-1. But for
j=n define x as[zsubn,zsubn].


Define C-->C by f(x+iy)=-y+ix. I'm iffy on this part because I wasn't
quite sure how to define f(x+iy).
Let R denote the interior of the polygon, then by Green's theorem:
1/2integral(f)=double integral(Green's Thm)=R

Then I use a summation of integrals for each line segment.

For j=1,2,3,,,n-1 define y=zsubj +(zsub(j+1)-zsubj)t for 0<t<1.
Notice that y parametrizes [zj,zj+1].

f(x+iy)==y+ix=i(x-iy) so f(z)=izconjugate while y=zj+1-zj

Then we evealuate about the integral form 0 to 1.
integral[f(y)ydt]=i(zj+1-zj)integral[yconjugatedt]=i(zj+1-zj)zjconjugate+1/2imod(zj+1-zj)^2

So adding all up, we get
1/2i((z2-z1)z1conjugate+...+(z1-zn)znconjgate)+1/4i(mod(z2-z1)+...+mod(zn-z1))

I got this far, but don't really understand what I'm doing. My biggest
problem is using Green's Theorem and understanding my choices for f(x+iy)
and why i parameterized.
I'm looking for some help explaining what I did and where to go from here.

 

[Billy Bob]

I was thinking more geometrically. I didn't use Green's Theorem at all.

A triangle with vertices (0,0), (a,b), and (c,d) has area 1/2 * (ad - bc). This is a fact from analytic geometry. (The cross product of two vectors has a magnitude equal to the area of a certain parallelogram, and this triangle has half that area.)

So now, if you call z1 = (a,b) = a + bi and z2 = (c,d) = c + di, then it is interesting that 1/2 * conjugate(z1) * z2 has imaginary part equal to 1/2 * (ad - bc).

Now generalize slightly, so that the third vertex is not at the origin, but instead is at z3. If you use translation, you can derive a formula for the area of this triangle.

Now suppose you have a convex polygon with vertices z1, z2, ... , zn. Draw segments from z1 to each of the other vertices, dividing the polygon into triangles. Add up the areas.

This is how you can obtain the formula A=1/2 Im[ bar(z_1) z_2 + bar(z_2) z_3 +...+ bar(z_n) z_1 ].

Maybe I'm not understanding your question.

 

[kathrynag]

I don't think i show using geometry.

 

[kathrynag]

I was thinking more geometrically. I didn't use Green's Theorem at all.

A triangle with vertices (0,0), (a,b), and (c,d) has area 1/2 * (ad - bc). This is a fact from analytic geometry. (The cross product of two vectors has a magnitude equal to the area of a certain parallelogram, and this triangle has half that area.)

So now, if you call z1 = (a,b) = a + bi and z2 = (c,d) = c + di, then it is interesting that 1/2 * conjugate(z1) * z2 has imaginary part equal to 1/2 * (ad - bc).

Now generalize slightly, so that the third vertex is not at the origin, but instead is at z3. If you use translation, you can derive a formula for the area of this triangle.

Now suppose you have a convex polygon with vertices z1, z2, ... , zn. Draw segments from z1 to each of the other vertices, dividing the polygon into triangles. Add up the areas.

This is how you can obtain the formula A=1/2 Im[ bar(z_1) z_2 + bar(z_2) z_3 +...+ bar(z_n) z_1 ].

Maybe I'm not understanding your question.

I'm having trouble going from a triangle from a vertice of (0,0) to one without. My probelm is finding the area. Also I'm confused on going from our area formula to the summation. Any hints?

 

[kathrynag]

Like I see how to show with the 3 vertices with one being (0,0) but then when I try to go to the summation, I'm not sure how to show that. Can someone help me show that? I know once I get there I will be able to get the complex line integral.

 

[kathrynag]

Ok, I figured out how to show that the summation area was the same with the (0,0) vertice by plugging in the vertices points. So I figured that out.

 

[kathrynag]

Ok, I'm all set except fo using a line integral about z congugate using the definition of a summation. i know the line integral of f(x) =Lim summation[f(psubk)(xsubk-xsubk-1)]
My problem is seeting this up. My initial assumption is to do (zsubk congugate)z_kcongugate -z-(k-1))? I think if I were to be able to set it up right, i'd be fine on evaluating it.

 

[Billy Bob]

Well, can't you say for a simple closed path C, that \int_C bar(z)\,dz is approximated like this: take points z_1, z_2, \dots, z_n around the path C and on each segment [z_i,z_{i+1}] let \zeta_i = z_i, so that the integral is approximately

bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)


But the imaginary part of this sum is twice the area bounded by the polygon with vertices z_1, z_2, \dots, z_n


Then say something about the limit.

 

[kathrynag]

So take the limit of bar zk(z_k+1-z_k)
which will end up being 1/2[zk+z_k+1](z_k+1-zk}
1/2[zk+1^2-zk^2}
1/2[zn^2-z1^2]
1/2(b^2-a^2)

 

[Billy Bob]

I'm sorry, I don't follow what you are saying when you say "which will end up being 1/2[zk+z_k+1](z_k+1-zk}"

In the "Riemann sum," you don't have to take the midpoint of segment [z_i,z_{i+1}]

You can take any point on the segment. I choose to take z_i

However, I don't know if this comment is related to what you are saying or not.

 

[kathrynag]

I'm sorry, I don't follow what you are saying when you say "which will end up being 1/2[zk+z_k+1](z_k+1-zk}"

In the "Riemann sum," you don't have to take the midpoint of segment [z_i,z_{i+1}]

You can take any point on the segment. I choose to take z_i

However, I don't know if this comment is related to what you are saying or not.

Ok then I'm a bit confused since i want to evaluate using the limit definition and want to show an answer like how if I chose to just do zdz I would get 1/2(b^2-a^2). Do you see what I'm seeing?

 

[Billy Bob]

Maybe I see what you are meaning.

But I don't mean evaluate a limit. The original question was to "Interpret this sum as part of the approximating sum in the definition of the line integral about C of z congugatedz."

So as I understand the problem, you simply make the statement or observation that \int_C bar(z)\,dz is approximately bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)

 

[kathrynag]

I see what you mean. Ok then, what if I wanted to go a little further and evaluate using that limit definition.

 

[kathrynag]

Oh, so it has to be half the sum of the imaginary parts because of the imaginary parts being twice the area?

 

[kathrynag]

And I can say something about when evaluating the limit we use 1/2 of the imaginary part?

 

[Billy Bob]

bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)\approx\int_C bar(z)\,dz

so take imaginary parts of both sides and divide by 2

 

[kathrynag]

bar(z_1) (z_2-z_1) + bar(z_2) (z_3-z_2) +...+ bar(z_n) (z_1- z_n)\approx\int_C bar(z)\,dz

so take imaginary parts of both sides and divide by 2

Ok, I think this makes sense because we are picking a point in the interval and from that we get that, but we get twice the area, so we must divide by 2. Correct me if I'm wrong.

 

[squidsoft]

This is how I analyzed it (hope I don't confuse things):

We can start with the known relation for the area encircled by a closed contour (see Calculus textbook, "Topics in Vector Calculus") :

\frac{1}{2}\mathop\oint\limits_{C} (-ydx+xdy)=\mathop\iint\limits_{R} dA

Note the expression -ydx+xdy is the real part of the contour integral:

\oint f(z)dz=\oint udx-vdy+i\oint vdx+udy

with f(z)=-y-ix=-i\overline{z}

Now, by Green's Theorem:

<br /> \oint vdx+udy=\oint (-xdx-ydy)=\mathop\iint\limits_{R}\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=0<br />

and therefore:

\mathop\iint\limits_R dA=-\frac{i}{2}\oint \overline{z}dz

We can parameterize the straight-line segments (z_n,z_{n+1}) as:
z_n(t)=z_n+(z_{n+1}-z_n)t

and therefore:

<br /> \begin{align*}<br /> \mathop\oint\limits_{z_n(t)} \overline{z}dz&amp;=\int_0^1 \overline{z_n(t)}(z_{n+1}-z_n)dt \\<br /> &amp;=(z_{n+1}-z_n)\int_0^1 \left(\overline{z_n}+\overline{(z_{n+1}-z_n)}t\right)dt\\<br /> &amp;=(z_{n+1}-z_n)\left[1/2\left(\overline{z_n}+\overline{z_{n+1}}\right)\right]<br /> \end{align*}<br />

Then:
<br /> \begin{align*}<br /> \mathop\iint\limits_{R} dA&amp;=-\frac{i}{4}\sum_{j=1}^N (z_{n+1}-z_n)\left(\overline{z_n}+\overline{z_{n+1}}\right);\quad z_{N+1}=z_1 \\<br /> &amp;=-\frac{i}{4}\sum_{j=1}^N\big(z_{n+1}-z_{n-1}\big)\overline{z_n};\quad z_0=z_N,\:z_{N+1}=z_1<br /> \end{align*}<br />