# Math Help - Complex- area enclosed by a polygon

1. ## 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.

2. 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.

3. 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?

4. 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.

5. 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.

6. 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 $[z_j,z_{j+1}]$.

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.

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.

7. 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?

8. Ok now after I use Green's Theorem I'm trying to understand the summation stuff. I'm not following.

9. 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.

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, $\iint_R 2 = 2\text{area}(R)$.

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.

10. 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?

11. 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.

12. 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

13. 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.

14. 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.

15. 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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