Complex numbers in geometry

"Let \(A,B,C,D\) be pairwise distinct points. Then \(\overline{AB}\perp\overline{CD}\) if and only if \(\frac{d-c}{b-a}\in\mathbb{iR}\); i.e. \( \frac{d-c}{b-a} +\overline{\bigg( \frac{d-c}{b-a}}\bigg)\ = 0.\)"

Well, what goes wrong when I say that dcab+(dcab) =0. \frac{d-c}{a-b} +\overline{\bigg( \frac{d-c}{a-b}}\bigg)\ = 0.?

And could anyone explain the difference between arg(dcba)arg\Big(\frac{d-c}{b-a}\Big) and arg(dcab)arg\Big(\frac{d-c}{a-b}\Big)?

#Geometry

Note by Dhrubajyoti Ghosh
3 years, 8 months ago

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Comments

Do you know what is the difference between 1i \frac{1}{i} and 1i \frac{ 1}{ - i } ?

What about arg1i \arg \frac{1}{i} and arg1i \arg \frac {1}{-i} ?

Calvin Lin Staff - 3 years, 8 months ago

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Well, arg1i=3π2arg\frac{1}{i} = \frac{3\pi}{2} while arg1i=π2arg\frac{1}{-i} = \frac{\pi}{2}. I hope I am correct. But I still don't see why I should be using dcba+(dcba) =0 \frac{d-c}{b-a} +\overline{\bigg( \frac{d-c}{b-a}}\bigg)\ = 0 instead of dcab+(dcab) =0 \frac{d-c}{a-b} +\overline{\bigg( \frac{d-c}{a-b}}\bigg)\ =0

Dhrubajyoti Ghosh - 3 years, 8 months ago

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Right. The point is that those numbers are different.

For example, (1+i)+(1+i)=0 (1 + i) + \overline{( -1 + i ) } = 0 , but (1+i)+(1i)0 ( 1 + i ) + \overline{ (1-i)} \neq 0 .

Calvin Lin Staff - 3 years, 8 months ago
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