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I'm reading Beardon's Algebra and Geometry.

Let $T$ be a triangle in $\mathbb{C}$ with vertices at $0$, $w_1$, $w_2$. By applying the mapping $z\mapsto \bar{w_2}z$, show that the area of $T$ is $\frac{1}{2}|Im[w_1\bar{w_2}]|$.

I am lost in this question mainly in two points:

  • The reason for the existence of the mapping $z\mapsto \bar{w_2}z$;

    I believe I know the the reason for the existence of this mapping - it is there to ensure we have 3 points that aren't positioned in, say: $(0,0),(0,2),(0,3)$ because it wouldn't constitute a triangle. Is this correct?

  • I am aware that $\frac{1}{2}|Im[w_1\bar{w_2}]|$ is a form of $\frac{bh}{2}$.

    But I can't understand why it constitute the area of the triangle, I've evaluated this expression and it yields $$\frac{1}{2} \sqrt{\left(a_1 b-a b_1\right){}^2}\tag{1}$$ for $w_1=a+bi$ and $w_2=a_1+b_1i$.

In a trial of understanding the problem, I started thinking on how I would do it myself, I would first get the module of $0$ to $w_1$, then the module from $w_1$ to $w_2$ and then use $\frac{bh}{2}$:

$$\frac{|a+bi|\cdot|a_1+b_1i|}{2}$$

which yields

$$\frac{1}{2} \sqrt{a^2+b^2} \sqrt{a_1^2+b_1^2}$$

And is different of $(1)$, from here I don't know what else to think.

Red Banana
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  • Any luck with the answer there? – Did May 15 '13 at 18:54
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    Too busy to think about what people took pains to explain to you, but not to ask brand new questions on the site? Odd. – Did May 15 '13 at 18:57
  • @Did Whatever. Yes, I am too dumb - just didn't want to look so. I'm still trying to figure out what you answered there (the context I was studying did not involve vector nor scalar products). I also believe that I shouldn't stop to understand it (I have to cover some contents in Cambridge and Oxford Syllabi) and that someday, the answer will come naturally. (The definition for naturally is: "If I look for the question one time per day, I may be able to figure out in a week or two). – Red Banana May 15 '13 at 19:27
  • @Did As an example to the idea of not stoping, you can look this answer in one question I have in the Mathematica comunnity. When I read that I was like: "Riemann Zeta What?!" If I were to try to figure out what that guy told me in that answer, I would take like 10 years and at the end of these 10 years, the only thing I would have in mind would be an explanation to that question - problem is that I must learn a lot of math basics due to the historic note I provided in my profile – Red Banana May 15 '13 at 19:30
  • @Did I understand you took pains to explain it to me, but you need to understand that I take pains to understand it too. I'm not being dishonest nor ungrateful for your answer - I'm just unable to understand it now. Perhaps in the near future when I start reading about vectors, I'll finally be able to understand it. – Red Banana May 15 '13 at 19:35
  • These are strange considerations per se, and anyway quite irrelevant to the case at hand. To invoke dumbness is an avoidance tactics. If something evades you in an answer, just say so, or warn people at the onset that you won't follow up their answers, so that they can decide to not answer your questions. (Say, will you delete these three comments like you did with the previous one?) – Did May 15 '13 at 19:39
  • let us continue this discussion in chat Are you kidding? – Did May 15 '13 at 19:41
  • @Did Sorry - I clicked by accident. – Red Banana May 15 '13 at 19:41
  • Another comment deleted by you. What are you hiding from? – Did May 15 '13 at 19:42
  • @Did I don't want to erase the previous comments. Should I do it? The comment I deleted was the "Let's move this discussion to the chat" comment. – Red Banana May 15 '13 at 19:42
  • False, you deleted another one which was between my own two first. – Did May 15 '13 at 19:43
  • Well, I remember of having deleted only the "move to chat" comment. What comment are you refeering to? – Red Banana May 15 '13 at 19:44
  • Oh, I remember. Sorry, I misread - are you refeering to the comments of "I'm not too dumb, just busy"? I deleted it because I felt I was not being completely honest - then I decided to tell what's happening. – Red Banana May 15 '13 at 19:48
  • @Did If something evades you in an answer, just say so, or warn people at the onset that you won't follow up their answers The problem is that sometimes I have not idea even of what to ask or I don't want to bother people with my lack of understanding. For example, look this question I'm still deeply confused about it, but I know that if I start to tell them that I do not follow their answers completely, I'll make them lose their time trying to explaing it to me. In my oppinion, they've already done more than enough for me. – Red Banana May 15 '13 at 19:51
  • There are even older questions of mine that have an accepted answer, I've been able to figure out some of them - others are pending. I do not think I should bother those users beyond of what I've already bothered. – Red Banana May 15 '13 at 19:53
  • Keeping silent is the worst solution. And now I will follow the recommendation to "Please avoid extended discussions in comments" and let you decide how you want to interact with others on the site. – Did May 15 '13 at 20:07
  • @Did I understand. I just do it because I believe that this stuff I'm trying to understand is simple and I can figure it out if I read a little more. – Red Banana May 15 '13 at 20:20
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    This comment thread has gone on quite long, and so far, none of it seems to relate to this question. Please consider moving to chat (though from the comments, this may not be an option), finishing up, or talking about this question. – robjohn May 15 '13 at 20:38
  • You have the height wrong in your calculation. The height is not defined to be the length of an edge of a triangle (this only happens when the triangle is a right triangle). Rather, the height is the length of the altitude dropped down to whatever edge is chosen as the "base." – anon May 16 '13 at 05:43

1 Answers1

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I am not exactly sure answering this question rather than the former is the best course of action, but this question is the more fleshed-out one and is more amenable to the answer I gave you in chat, which I am expanding upon here.

No, the transformation $z\mapsto \bar{w}_2z$ has nothing to do with degenerate triangles. In fact, if you start with a degenerate triangle (where the three endpoints are collinear), applying the map to the triangle will just yield another degenerate triangle - it doesn't change anything. This should be more apparent once you see what multiplication does to the complex plane. But it's not like degenerate triangles pose any issues in the first place!

The idea is that we want to use the formula Area = Base x Height / 2, however it isn't entirely obvious what the "height" of the triangle will be for any choice of edge as "base." This can be fixed, however, if we rotate the triangle (around the origin) so that one of its edges is on the real axis - in this case, the base is the length of that edge, and the height is the imaginary part of the third point.

Let's review some basic facts of the complex plane and its geometry. Every nonzero number $\alpha\in\bf C$ can be written in polar form, $\alpha=re^{i\theta}$. Here, $r=|\alpha|$ is the modulus, or magnitude - the distance between $\alpha$ and $0$. And $\theta$ is the angle made between the line segment from $0$ to $\alpha$ and the positive real axis. Note that $e^{i\theta}$ is the rescaling of $\alpha$ that is on the unit circle in the same direction.

So if $\alpha=re^{i\theta}$ and $\beta=se^{i\phi}$, then $\alpha\beta=rse^{i(\theta+\phi)}$. In words, this means that the effect of multiplying anything by $\alpha$ is to (a) expand/shrink by a factor of $r=|\alpha|$, and (b) rotate about the origin by an angle of $\theta$. Rotations and dilations are achieved by multiplication.

Now, suppose we want to rotate $T$ so it has an edge on the real axis, which will necessitate multiplying by something. We already know that $u\bar{u}=|u|^2\in\bf R$ for any $u\in\bf C$, so we can achieve our goal of rotating the triangle by multiplying the vertices by $\bar{w}_2$. Denote the rotated triangle $T'$, so that the vertices of $T'$ are $\bar{w}_2\cdot0=0$, $\bar{w}_2w_2=|w_2|^2$, and $\bar{w}_2w_1$. The base of this triangle has length $|w_2|^2$, and the height of this triangle (the vertical distance of the third vertex from the real axis) is ${\rm Im}(\bar{w}_2w_1)$ Hence ${\rm Area}(T')=\frac{1}{2}|w_2|^2{\rm Im}(\bar{w}_2w_1)$.

The issue at this point is that multiplying by $\bar{w}_2$ didn't just rotate the triangle $T$, it also stretched it. Luckily, multiplication stretches the complex plane uniformly - all lengths are scaled by a given factor. In particular, since lengths are stretched by a factor of $|w_2|$, areas must be stretched by a factor of $|w_2|^2$. Hence ${\rm Area}(T')=|w_2|^2{\rm Area}(T)$. Solving for ${\rm Area}(T)$ yields the conclusion.

anon
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