MHB Complex numbers and area of octagon

[Saitama]

Problem:
Let $\dfrac{1}{a_1-2i},\dfrac{1}{a_2-2i},\dfrac{1}{a_3-2i},\dfrac{1}{a_4-2i},\dfrac{1}{a_5-2i}, \dfrac{1}{a_6-2i},\dfrac{1}{a_7-2i},\dfrac{1}{a_8-2i}$ be the vertices of regular octagon. Find the area of octagon (where $a_j \in R$ for $j=1,2,3,4,5,6,7,8$ and $i=\sqrt{-1}$).

Attempt:
The problem looks too difficult to me. I don't see how to even start when I am given 8 variables! All I can do is find the centre of octagon and the side length but I am not sure if that would even help.

I need a few hints to begin with this problem.

Any help is appreciated. Thanks!

 

[Dustinsfl]

Problem:
Let $\dfrac{1}{a_1-2i},\dfrac{1}{a_2-2i},\dfrac{1}{a_3-2i},\dfrac{1}{a_4-2i},\dfrac{1}{a_5-2i}, \dfrac{1}{a_6-2i},\dfrac{1}{a_7-2i},\dfrac{1}{a_8-2i}$ be the vertices of regular octagon. Find the area of octagon (where $a_j \in R$ for $j=1,2,3,4,5,6,7,8$ and $i=\sqrt{-1}$).

Attempt:
The problem looks too difficult to me. I don't see how to even start when I am given 8 variables! All I can do is find the centre of octagon and the side length but I am not sure if that would even help.

I need a few hints to begin with this problem.

Any help is appreciated. Thanks!

I would multiple by the conjugates to get the vertices in the a form a can work with. I would then plot it on the complex plane. Then what is the formula for the area of an octagon? I don't recall but that will be needed as well.

For the first part, take
\[
\frac{a_1 + 2i}{(a_1 - 2i)(a_1 + 2i)} = \frac{a_1}{a_1^2 + 4} + \frac{2i}{a_1^2 + 4}
\]
where \(a_1^2 + 4\in\mathbb{R}\).
Once you do this, you can plot the points but you must realize that you have a regular octagon so all sides and angles are equal. That should help provide some insight.

 

[Saitama]

For the first part, take
\[
\frac{a_1 + 2i}{(a_1 - 2i)(a_1 + 2i)} = \frac{a_1}{a_1^2 + 4} + \frac{2i}{a_1^2 + 4}
\]
where \(a_1^2 + 4\in\mathbb{R}\).

I did try that but chose not to post it because it doesn't look very useful to me. I know I can find the sides in terms of these variable but I fear that won't give me anything useful. Do you have any other ideas? These 8 variables are throwing me off.

 

[Dustinsfl]

I did try that but chose not to post it because it doesn't look very useful to me. I know I can find the sides in terms of these variable but I fear that won't give me anything useful. Do you have any other ideas? These 8 variables are throwing me off.

The modulus of each complex number should be the same. I just looked up the area of a regular octagon. It is
\[
A = 2\cdot a\cot\Big(\frac{\pi}{8}\Big) = 2(1 + \sqrt{2})a^2
\]
where a is the length of the side (we can use the distance formula).

Or you can use the
\[
A = 2\sqrt{2}R^2
\]
where \(R\) is the radius of the circle that inscribes the octagon. That is the ray from the origin to the vertex of the octagon is the radius. To find that, just take the modulus of \(z_i\) where \(z_i\) is one your vertices. Since all modulus are the same, you will have your R.

 

[Saitama]

The modulus of each complex number should be the same. I just looked up the area of a regular octagon.

Sorry but I don't agree with this. The distance between the adjacent vertices is same i.e
$$\left| \frac{1}{a_1-2i}-\frac{1}{a_2-2i}\right|=\left|\frac{1}{a_2-2i}-\frac{1}{a_3-2i}\right|$$

 

[Dustinsfl]

Sorry but I don't agree with this. The distance between the adjacent vertices is same i.e
$$\left| \frac{1}{a_1-2i}-\frac{1}{a_2-2i}\right|=\left|\frac{1}{a_2-2i}-\frac{1}{a_3-2i}\right|$$

By definition, a regular octagon is a closed figure with sides of the same length and internal angles of the same size.

 

[Saitama]

By definition, a regular octagon is a closed figure with sides of the same length and internal angles of the same size.

Yes, I agree with that but your previous post implies that:
$$\left|\frac{1}{a_1-2i}\right|=\left|\frac{1}{a_2-2i}\right|$$
and this is definitely incorrect.

 

[Dustinsfl]

Yes, I agree with that but your previous post implies that:
$$\left|\frac{1}{a_1-2i}\right|=\left|\frac{1}{a_2-2i}\right|$$
and this is definitely incorrect.

What you are saying is the modulus between the origin and two vertices are the same if the octagon is centered at the origin. I would say that is correct. You would have to make the adjustment for the center shift and then it would be correct.

Suppose our octagon is at the origin (0,0) which would could easily do with a change of coordinates. Wouldn't it be true that the distance from the origin to each vertex is that same? Your statement above is saying the length from the origin to vertx one is the same as the length from the origin to vertex two. If they are not the same, we don't have a regular octagon.

 

[Saitama]

Suppose our octagon is at the origin (0,0) which would could easily do with a change of coordinates. Wouldn't it be true that the distance from the origin to each vertex is that same?

But the problem is, it isn't at origin (I guess you mean the centre of octagon is at origin).

If you shift the origin, you need to shift it to the centre of octagon. Since the figure is regular, this would mean that the distance of $\dfrac{1}{8}\sum_{r=1}^8 \dfrac{1}{a_i-2i}$ is same from all vertices and I don't really think it would be a good idea to equate these distances as it would result in messy expressions.

Your statement above is saying the length from the origin to vertx one is the same as the length from the origin to vertex two. If they are not the same, we don't have a regular octagon.

No, it is not same from origin. I mean $\left|\dfrac{1}{a_1-2i}\right|\neq \left|\dfrac{1}{a_2-2i}\right|$ It isn't explicitly stated in the problem that octagon is centred at origin.

Instead, the distances are same from the centre of octagon as I stated above.

 

[Opalg]

Problem:
Let $\dfrac{1}{a_1-2i},\dfrac{1}{a_2-2i},\dfrac{1}{a_3-2i},\dfrac{1}{a_4-2i},\dfrac{1}{a_5-2i}, \dfrac{1}{a_6-2i},\dfrac{1}{a_7-2i},\dfrac{1}{a_8-2i}$ be the vertices of regular octagon. Find the area of octagon (where $a_j \in R$ for $j=1,2,3,4,5,6,7,8$ and $i=\sqrt{-1}$).

Attempt:
The problem looks too difficult to me. I don't see how to even start when I am given 8 variables! All I can do is find the centre of octagon and the side length but I am not sure if that would even help.

I need a few hints to begin with this problem.

Any help is appreciated. Thanks!

Let $z_j = \dfrac1{a_j-2i}\ (1\leqslant j\leqslant 8)$. The eight points $z_j$, being the vertices of a regular octagon, all lie on a circle. Under the map $w=1/z$, those eight points all get taken to points on the line $\text{Im}\,w = -2$. So the circumcircle of the octagon is the image of the line $\text{Im}\,w = -2$ under the map $z=1/w.$ That gives you all the information you need to find the area of the octagon.

 

[johng1]

Hi,
Here's an addendum to OpalG's excellent suggestion. The only difference is to use the conjugate of the given map.

 

[Saitama]

This "map" thing in complex numbers is new to me, I am having some trouble understanding both of your posts.

Referring to johng's post, I don't understand the statement: "Notice geometrically this map is inversion with respect to unit circle. " What does this mean?

 

[johng1]

Hi Pranav,
In this context "map" is a synonym for function. So OpalG's map is just the function on the complex numbers:
$$f(z)={1\over z}={x\over x^2+y^2}+i\,{-y\over x^2+y^2}$$ where $$z=x+iy$$
So when we find the conjugate of f(z) and view this as a mapping of the (extended) Euclidean plane we get:
$$(x,y)\rightarrow({x\over x^2+y^2},{y\over x^2+y^2})$$
Assuming you know about the geometric construction of inversion with respect to a circle, this is precisely inversion w.r.t. the unit circle. The only reason I brought this up was to "easily" find the circumcircle of the octagon. Of course the original function is defined only for $$z\neq0$$. However, if you're familiar with circle inversion, you know the image of any extended line in the plane is a circle passing through the center of the inverting circle. So I thought the easiest way to get the circumcircle was to view it as a circle inversion. Maybe I should have left well enough alone, and just let OpalG's post be the good answer.

 

[Opalg]

This "map" thing in complex numbers is new to me, I am having some trouble understanding both of your posts.

Referring to johng's post, I don't understand the statement: "Notice geometrically this map is inversion with respect to unit circle. " What does this mean?

If you don't know about inversion or fractional linear transformations, then the easiest way to look at it is this. Let $w = \dfrac1z$, where $z=x+iy$ and $w = a-2i$. Then $$z = \frac1w = \frac1{a-2i} = \frac{a+2i}{a^2+4} = \frac a{a^2+4} + i\frac 2{a^2+4}.$$ It follows that $x = \dfrac a{a^2+4}$ and $y = \dfrac2{a^2+4}$. Therefore $x^2+y^2 = \dfrac{a^2+4}{(a^2+4)^2} = \dfrac1{a^2+4} = \dfrac y2.$ So $z = x+iy$ lies on the circle $x^2+y^2 = \frac12y$, or $x^2 + \bigl(y - \frac14\bigr)^2 = \bigl(\frac14\bigr)^2$, with centre at $\bigl(0,\frac14\bigr)$ and radius $\frac14.$

 

[Saitama]

If you don't know about inversion or fractional linear transformations, then the easiest way to look at it is this. Let $w = \dfrac1z$, where $z=x+iy$ and $w = a-2i$. Then $$z = \frac1w = \frac1{a-2i} = \frac{a+2i}{a^2+4} = \frac a{a^2+4} + i\frac 2{a^2+4}.$$ It follows that $x = \dfrac a{a^2+4}$ and $y = \dfrac2{a^2+4}$. Therefore $x^2+y^2 = \dfrac{a^2+4}{(a^2+4)^2} = \dfrac1{a^2+4} = \dfrac y2.$ So $z = x+iy$ lies on the circle $x^2+y^2 = \frac12y$, or $x^2 + \bigl(y - \frac14\bigr)^2 = \bigl(\frac14\bigr)^2$, with centre at $\bigl(0,\frac14\bigr)$ and radius $\frac14.$

Yes, I am unaware of inversion or fractional linear transformations.

The alternative approach you have shown is great, thanks a lot Opalg! :)