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hydroptic ( hydroptic@sopuli.xyz ) deleted by creator to Science Memes@mander.xyzEnglish · 2 years ago

This feels wrong. I love it.

sopuli.xyz

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This feels wrong. I love it.

sopuli.xyz

hydroptic ( hydroptic@sopuli.xyz ) deleted by creator to Science Memes@mander.xyzEnglish · 2 years ago
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  • blackbrook ( blackbrook@mander.xyz ) 
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    2 years ago

    You need to add some disclaimer to this diagram like “not to scale”…

    • hydroptic ( hydroptic@sopuli.xyz ) deleted by creatorOP
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      2 years ago

      It’s to scale.

      Which scale is left as an exercise to the reader.

      • Jerkface (any/all) ( jerkface@lemmy.ca ) 
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        9 months ago

        Removed by mod

        • hydroptic ( hydroptic@sopuli.xyz ) deleted by creatorOP
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          2 years ago

          I may not have been entirely serious

  • puchaczyk ( puchaczyk@lemmy.blahaj.zone ) 
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    2 years ago

    This is why a length of a vector on a complex plane is |z|=√(z*×z). z* is a complex conjugate of z.

    • randy ( randy@lemmy.ca ) 
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      2 years ago

      I’ve noticed that, if an equation calls for a number squared, they usually really mean a number multiplied by its complex conjugate.

    • SystemNeo ( drbluefall@toast.ooo ) 
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      2 years ago

      [ you may want to escape the characters in your comment… ]

  • ornery_chemist ( ornery_chemist@mander.xyz ) 
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    2 years ago

    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 + - i2) = √(1 + 1) = √2. I could be totally off base here and could be confusing with something else…

    • diaphanous ( diaphanous@feddit.org ) 
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      2 years ago

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

  • BorgDrone ( BorgDrone@lemmy.one ) 
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    2 years ago

    Now calculate the angles

    • Rivalarrival ( Rivalarrival@lemmy.today ) 
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      2 years ago

      That’s actually pretty easy. With CB being 0, C and B are the same point. Angle A, then, is 0, and the other two angles are undefined.

      • Drew ( crmsnbleyd@sopuli.xyz ) 
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        2 years ago

        A is clearly a right angle

        • Rivalarrival ( Rivalarrival@lemmy.today ) 
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          2 years ago

          A is drawn in such a way that it resembles a right angle, but it is not labeled as such. The length of the hypotenuse is given as zero. The opposite angle cannot be anything but 0°.

          • Drew ( crmsnbleyd@sopuli.xyz ) 
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            2 years ago

            The pythagoras theorem only holds if A is a right triangle

            • Rivalarrival ( Rivalarrival@lemmy.today ) 
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              2 years ago

              What is depicted here isn’t even a polygon, let alone a triangle, let alone a right triangle. This is just a line segment. Line AB is the same as line AC. There is no line BC. BC is a single point.

              I suppose it could possibly depict a weird cross section of two orthogonal circles in a real and an imaginary plane.

    • hydroptic ( hydroptic@sopuli.xyz ) deleted by creatorOP
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      2 years ago

      No thank you

  • produnis ( produnis@discuss.tchncs.de ) 
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    2 years ago

    Too complexe for me ;)

  • ShinkanTrain ( ShinkanTrain@lemmy.ml ) 
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    2 years ago

  • Jerkface (any/all) ( jerkface@lemmy.ca ) 
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    Removed by mod

    • xor ( xor@lemmy.blahaj.zone ) 
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      2 years ago

      I think BAC is supposed to be defined as a right-angle, so that AB²+AC²=CB²

      => AB+1²=0²

      => AB = √-1

      => AB = i

      • Jerkface (any/all) ( jerkface@lemmy.ca ) 
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        Removed by mod

  • I_am_10_squirrels ( I_am_10_squirrels@beehaw.org ) 
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    2 years ago

    The length would be equal to the absolute value

  • Ace ( AcesFullOfKings@feddit.uk ) 
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    2 years ago

    deleted by creator

  • Boomkop3 ( Boomkop3@reddthat.com ) 
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    2 years ago

    Turn around…

    • hydroptic ( hydroptic@sopuli.xyz ) deleted by creatorOP
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      2 years ago

      https://www.youtube.com/watch?v=lcOxhH8N3Bo

      • Boomkop3 ( Boomkop3@reddthat.com ) 
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        2 years ago

        Exactly what I thought of, but then I was like… nah that’s too cheesy

        • hydroptic ( hydroptic@sopuli.xyz ) deleted by creatorOP
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          Too cheesy?!

    • Rivalarrival ( Rivalarrival@lemmy.today ) 
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      2 years ago

      Every now and then, I get a little bit lonely and you’re never coming 'round

  • Boomkop3 ( Boomkop3@reddthat.com ) 
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    2 years ago

    Turn around…

    • Maiq ( Maiq@lemy.lol ) 
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      2 years ago

      Bright eyes.

      • Zoop ( Zoop@beehaw.org ) 
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        2 years ago

        Every now and then, do ya fall apart?

  • Stomata ( Stomata@buddyverse.one ) deleted by creator
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    deleted by creator

  • mariusafa ( mariusafa@lemmy.sdf.org ) 
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    2 years ago

    What if not a Hilbert space?

  • barsoap ( barsoap@lemm.ee ) 
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    2 years ago

    Looks like a finite state machine or some other graph to me, which just happens to have no directed edges.

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