this post was submitted on 30 Oct 2024
389 points (97.3% liked)

Science Memes

11130 readers
3026 users here now

Welcome to c/science_memes @ Mander.xyz!

A place for majestic STEMLORD peacocking, as well as memes about the realities of working in a lab.



Rules

  1. Don't throw mud. Behave like an intellectual and remember the human.
  2. Keep it rooted (on topic).
  3. No spam.
  4. Infographics welcome, get schooled.

This is a science community. We use the Dawkins definition of meme.



Research Committee

Other Mander Communities

Science and Research

Biology and Life Sciences

Physical Sciences

Humanities and Social Sciences

Practical and Applied Sciences

Memes

Miscellaneous

founded 2 years ago
MODERATORS
 
you are viewing a single comment's thread
view the rest of the comments
[–] ornery_chemist@mander.xyz 34 points 3 weeks ago (2 children)

Isn't the squaring actually multiplication by the complex conjugate when working in the complex plane? i.e., √((1 - 0 i) (1 + 0 i) + (0 - i) (0 + i)) = √(1 + - i^2^) = √(1 + 1) = √2. I could be totally off base here and could be confusing with something else...

[–] diaphanous@feddit.org 16 points 3 weeks ago (1 children)

I think you're thinking of taking the absolute value squared, |z|^2 = z z*

[–] candybrie@lemmy.world 6 points 3 weeks ago

Considering we're trying to find lengths, shouldn't we be doing absolute value squared?

[–] HexesofVexes@lemmy.world 1 points 3 weeks ago* (last edited 3 weeks ago)

Almost:

Lengths are usually reals, and in this case the diagram suggests we can assume that A is the origin wlog (and the sides are badly drawn vectors without a direction)

Next we convert the vectors into lengths using the abs function (root of conjugate multiplication). This gives us lengths of 1 for both.

Finally, we can just use a Euclidean metric to get our other length √2.

Squaring isn't multiplication by complex conjugate, that's just mapping a vector to a scalar (the complex | x | function).