Complex- area enclosed by a polygon

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• Apr 30th 2009, 08:00 PM
kathrynmath
Complex- area enclosed by a polygon
Recall that the area enclosed by the polygon with vertices z1,z2,z3,...,znis1/2I(z1conguatez2+z2congugatez3+...+zncongugatez1)Sh ow that the area enclosed =1/2Isummation[zkcongugate(zsub(k+1)-zk)].Interpret this sum as part of the approximating sum in the definition ofthe line integral about C of z congugatedz. I don't even know where to start. I understand where the 1/2I(z.....) comesfrom, but once the summation comes in, I'm lost.
• Apr 30th 2009, 09:59 PM
ThePerfectHacker
Quote:

Originally Posted by kathrynmath
Recall that the area enclosed by the polygon with vertices z1,z2,z3,...,znis1/2I(z1conguatez2+z2congugatez3+...+zncongugatez1)Sh ow that the area enclosed =1/2Isummation[zkcongugate(zsub(k+1)-zk)].Interpret this sum as part of the approximating sum in the definition ofthe line integral about C of z congugatedz. I don't even know where to start. I understand where the 1/2I(z.....) comesfrom, but once the summation comes in, I'm lost.

Let $\Gamma$ be polygon with its ordered vertices (the vertices you take in order as you travel around it) be $z_1,z_2,...,z_n$.
Also say they are taken in a counterclockwise fashion.

Let $\Gamma_j = [z_j,z_{j+1}]$ be the line segment from $z_j$ to $z_{j+1}$ for $j=1,2...,n-1$. But for $j=n$ define $\Gamma_n = [z_n,z_1]$.

Define $f: \mathbb{C}\to \mathbb{C}$ by $f(x+iy) = (-y+ix)$.
Let $R$ denote the interior of the polygon, then by Green's theorem:
$\frac{1}{2}\oint_{\Gamma} f = \frac{1}{2}\iint_R \frac{\partial (x)}{\partial x} - \frac{\partial (-y)}{\partial y} = \iint_R 1 = \text{area}(R)$.

Thus, the area is equal to: $\frac{1}{2}\oint_{\Gamma}f = \frac{1}{2}\sum_{j=1}^n \int_{\Gamma_j}f = \frac{1}{2}\sum_{j=1}^{n-1} \left( \int_{[z_j,z_{j+1}]} f \right) + \frac{1}{2}\int_{[z_n,z_1]} f$

For $j=1,2,...,n-1$ define $\gamma_j(t) = z_j + (z_{j+1}-z_j) t$ for $0\leq t\leq 1$. Notice that $\gamma_j(t)$ parametrizes $[z_j,z_{j+1}]$.
Thus, $\int_{[z_j,z_{j+1}]} f = \int_0^1 f(\gamma_j(t))\dot {\gamma}_j(t) dt$.

Now, $f(x+iy) = -y + ix = i(x-iy)$, so $f(z) = i\bar z$, while $\dot{\gamma}_j(t) = z_{j+1}-z_j$.

Therefore, $\int_0^1 f(\gamma_j(t))\dot {\gamma}_j(t) dt = i(z_{j+1}-z_j)\int_0^1 \overline{\gamma}_j(t) dt = i(z_{j+1}-z_j)\bar z_j + \tfrac{1}{2}i|z_{j+1}-z_j|^2$

So if you add these up including $[z_n,z_1]$ you get:
$\tfrac{1}{2}i( (z_2 - z_1)\bar z_1 + ... + (z_1 - z_n)\bar z_n) + \tfrac{1}{4}i(|z_1-z_2|+...+|z_n-z_1|)$

Work on simplying that expression now. ;)
• May 1st 2009, 11:00 AM
kathrynmath
Quote:

Originally Posted by ThePerfectHacker
Let $\Gamma$ be polygon with its ordered vertices (the vertices you take in order as you travel around it) be $z_1,z_2,...,z_n$.
Also say they are taken in a counterclockwise fashion.

Let $\Gamma_j = [z_j,z_{j+1}]$ be the line segment from $z_j$ to $z_{j+1}$ for $j=1,2...,n-1$. But for $j=n$ define $\Gamma_n = [z_n,z_1]$.

Define $f: \mathbb{C}\to \mathbb{C}$ by $f(x+iy) = (-y+ix)$.
Let $R$ denote the interior of the polygon, then by Green's theorem:
$\frac{1}{2}\oint_{\Gamma} f = \frac{1}{2}\iint_R \frac{\partial (x)}{\partial x} - \frac{\partial (-y)}{\partial y} = \iint_R 1 = \text{area}(R)$.

Thus, the area is equal to: $\frac{1}{2}\oint_{\Gamma}f = \frac{1}{2}\sum_{j=1}^n \int_{\Gamma_j}f = \frac{1}{2}\sum_{j=1}^{n-1} \left( \int_{[z_j,z_{j+1}]} f \right) + \frac{1}{2}\int_{[z_n,z_1]} f$

For $j=1,2,...,n-1$ define $\gamma_j(t) = z_j + (z_{j+1}-z_j) t$ for $0\leq t\leq 1$. Notice that $\gamma_j(t)$ parametrizes $[z_j,z_{j+1}]$.
Thus, $\int_{[z_j,z_{j+1}]} f = \int_0^1 f(\gamma_j(t))\dot {\gamma}_j(t) dt$.

Now, $f(x+iy) = -y + ix = i(x-iy)$, so $f(z) = i\bar z$, while $\dot{\gamma}_j(t) = z_{j+1}-z_j$.

Therefore, $\int_0^1 f(\gamma_j(t))\dot {\gamma}_j(t) dt = i(z_{j+1}-z_j)\int_0^1 \overline{\gamma}_j(t) dt = i(z_{j+1}-z_j)\bar z_j + \tfrac{1}{2}i|z_{j+1}-z_j|^2$

So if you add these up including $[z_n,z_1]$ you get:
$\tfrac{1}{2}i( (z_2 - z_1)\bar z_1 + ... + (z_1 - z_n)\bar z_n) + \tfrac{1}{4}i(|z_1-z_2|+...+|z_n-z_1|)$

Work on simplying that expression now. ;)

Ok, I follow this pretty well until the modulus comes in. Why exactly do we have a modulus?
• May 1st 2009, 01:05 PM
ThePerfectHacker
Quote:

Originally Posted by kathrynmath
Ok, I follow this pretty well until the modulus comes in. Why exactly do we have a modulus?

For $z\in \mathbb{C}$ we have that $z\bar z = |z|^2$.
The modulus comes in because we multiply complex numbers by their conjugates.
• May 1st 2009, 02:17 PM
kathrynmath
Ok, I don't understand this as well as I thought. Why exactly are we using a segment from zj to zsub(j+1)? Is it because we can only evaluate the polygon on each side?

Next, I don't understand the f(x+iy)=-y+ix

The next part I understan up to a certain point. I see how Green's Theorem works, but I don't see where the 1/2 comes from.
The next line following makes me all lost with the summations. I'm thinking it comes somewhat from the definition of a line integral, but I'm a bit confused.
• May 1st 2009, 03:14 PM
ThePerfectHacker
Quote:

Originally Posted by kathrynmath
Ok, I don't understand this as well as I thought. Why exactly are we using a segment from zj to zsub(j+1)? Is it because we can only evaluate the polygon on each side?

Yes, $\Gamma$ is the entire polygon. We have shown that $\frac{1}{2}\oint_{\Gamma} f$ is the area of the polygon by Green's theorem. Now to evaluate $\oint_{\Gamma} f$ we break up the polygon into each side and evaluate every side along those lines $[z_j,z_{j+1}]$.

Quote:

Next, I don't understand the f(x+iy)=-y+ix
I defined $f: \mathbb{C}\to \mathbb{C}$ to be $f(x+iy) = -y+ix$. The reason why I defined it that way is because $\tfrac{\partial f}{\partial x} = 1 \text{ and }\tfrac{\partial f}{\partial y} = -1$ so $\tfrac{\partial f}{\partial x} - \tfrac{\partial f}{\partial y} = 2$. This differencial term is what appeas in Green's theorem.

Quote:

The next part I understan up to a certain point. I see how Green's Theorem works, but I don't see where the 1/2 comes from.
Because we get (without the 1/2): $\iint_R 2 = 2\cdot \text{area}(R)$.
Therefore, $\oint_{\Gamma} f$ is twice the area, we need to divide by $2$ to get the actual area.
• May 1st 2009, 03:28 PM
kathrynmath
ok I think I get all that, but is there a reason we choose a certain f or would this f(x+iy) be the case every time?
Also, you say 1/2 of the integral f is the area of the polygon by Green's Theorem. Does this come from somewhere?
• May 1st 2009, 03:37 PM
kathrynmath
Ok now after I use Green's Theorem I'm trying to understand the summation stuff. I'm not following.
• May 1st 2009, 06:58 PM
ThePerfectHacker
Quote:

Originally Posted by kathrynmath
ok I think I get all that, but is there a reason we choose a certain f or would this f(x+iy) be the case every time?

I choose $f(x+iy) = -y+ix$ because it works. There are other functions that can give $\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y} = 1$. This is a typical function to use for finding area by Green's theorem.

Quote:

Also, you say 1/2 of the integral f is the area of the polygon by Green's Theorem. Does this come from somewhere?
Remember that $\iint_R 1 = \text{area}(R)$.
Therefore, $\iint_R 2 = 2\text{area}(R)$.

Quote:

Originally Posted by kathrynmath
Ok now after I use Green's Theorem I'm trying to understand the summation stuff. I'm not following.

What part do you not understand? Which part?
The first thing I did was break up $\Gamma$ into line segments.
• May 2nd 2009, 05:45 PM
kathrynmath
Quote:

Originally Posted by ThePerfectHacker
I choose $f(x+iy) = -y+ix$ because it works. There are other functions that can give $\frac{\partial f}{\partial x} - \frac{\partial f}{\partial y} = 1$. This is a typical function to use for finding area by Green's theorem.

Remember that $\iint_R 1 = \text{area}(R)$.
Therefore, $\iint_R 2 = 2\text{area}(R)$.

What part do you not understand? Which part?
The first thing I did was break up $\Gamma$ into line segments.

Did you break up into line segments of length say z1 to z2, z2 to z3 and so on?
• May 3rd 2009, 12:10 PM
ThePerfectHacker
Quote:

Originally Posted by kathrynmath
Did you break up into line segments of length say z1 to z2, z2 to z3 and so on?

When I write [z1,z2] I mean the line segment from z1 to z2. Now $\Gamma$ consists of the line segments [z1,z2],[z2,z3],...,[zn,z1]. And that is all I did, I just integrate that function along each line segment.
• May 3rd 2009, 02:40 PM
kathrynmath
Quote:

Originally Posted by ThePerfectHacker
When I write [z1,z2] I mean the line segment from z1 to z2. Now $\Gamma$ consists of the line segments [z1,z2],[z2,z3],...,[zn,z1]. And that is all I did, I just integrate that function along each line segment.

Ok that makes sense
• May 4th 2009, 07:14 AM
shawsend
Hi. Nice problem! I worked it using $f(z)=-y-ix=-i\overline{z}$ and obtained:

$A=\sum_{j=1}^{N}(z_{n+1}-z_{n-1})\overline{z_n};\quad z_0=z_N, z_{N+1}=z_1$

However, I think a nice addition is to code it in software: Write a routine to generate a random set of complex numbers, say 10. Then find the convex hull, then calculate the area of the polygon using the build-in command for example Mathematica has build-in code to do this, then use the formula you get with the analysis above then compare the results. First do it with a square. Fun I think.(Rock)
• May 4th 2009, 09:45 AM
kathrynmath
I have a question. Do I absolutely have to use Green's Theorem or is there a way to work around this? I'm just asking because my book never really went over it. I have no problem using it, though.
• May 6th 2009, 10:19 AM
kathrynmath
Ok I figure it all out except for the last part of the line integral of z congugate. I know that the integral of f(z)=lim summation(f(p_k)(x_k-x_(k-1)).
My problem is just pitting this together since it starts at z_1 and not at z_0. My intial thought for f(p_k) is to be zcongugate_k. Any thoughts?
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