• captainlezbian@lemmy.worldEnglish
    311·
    2 days ago

    It’s just dimensionally shifted. This is not only true, its truth is practical for electrical engineering purposes. Real and imaginary cartesians yay!

    • randy@lemmy.caEnglish
      18·
      3 days ago

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

    • bitcrafter@programming.devEnglish
      71·
      2 days ago

      If you are comfortable with negative numbers, then you are already comfortable with the idea that a number can be tagged with an extra bit of information that represents a rotation. Complex numbers just generalize the choices available to you from 0 degrees and 180 degrees to arbitrary angles.

    • Klear@lemmy.worldEnglish
      19·
      3 days ago

      After delving into quaternions, complex numbers feel simple and intuitive.

    • enkers@sh.itjust.worksEnglish
      29·
      3 days ago

      I never really appreciated them until watching a bunch of 3blue1brown videos. I really wish those had been available when I was still in HS.

      • driving_crooner@lemmy.eco.brEnglish
        22·
        3 days ago

        After watching a lot of Numberphile and 3B1B videos I said to myself, you know what, I’m going back to college to get a maths degree. I switched at last moment to actuarial sciences when applying, because it’s looked like a good professional move and was the best decision on my life.

  • ornery_chemist@mander.xyzEnglish
    35·
    3 days 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…

    • HexesofVexes@lemmy.worldEnglish
      1·
      1 day 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).

      • candybrie@lemmy.worldEnglish
        6·
        3 days ago

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

  • jerkface@lemmy.caEnglish
    14·
    2 days ago

    Doesn’t this also imply that i == 1 because CB has zero length, forcing AC and AB to be coincident? That sounds like a disproving contradiction to me.

    • Rivalarrival@lemmy.todayEnglish
      15·
      3 days 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.

        • Rivalarrival@lemmy.todayEnglish
          3·
          2 days 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°.

            • Rivalarrival@lemmy.todayEnglish
              1·
              2 days 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.

      • L0rdMathias@sh.itjust.worksEnglish
        23·
        3 days ago

        Circles are good at math, but what to do if you not have circle shape? Easy, redefine problem until you have numbers that look like the numbers the circle shape uses. Now we can use circle math on and solve problems about non-circles!

      • owenfromcanada@lemmy.worldEnglish
        311·
        3 days ago

        The short version is: we use some weird abstractions (i.e., ways of representing complex things) to do math and make sense of things.

        The longer version:

        Electromagnetic signals are how we transmit data wirelessly. Everything from radio, to wifi, to xrays, to visible light are all made up of electromagnetic signals.

        Electromagnetic waves are made up of two components: the electrical part, and the magnetic part. We model them mathematically by multiplying one part (the magnetic part, I think) by the constant i, which is defined as sqrt(-1). These are called “complex numbers”, which means there is a “real” part and a “complex” (or “imaginary”) part. They are often modeled as the diagram OP posted, in that they operate at “right angles” to each other, and this makes a lot of the math make sense. In reality, the way the waves propegate through the air doesn’t look like that exactly, but it’s how we do the math.

        It’s a bit like reading a description of a place, rather than seeing a photograph. Both can give you a mental image that approximates the real thing, but the description is more “abstract” in that the words themselves (i.e., squiggles on a page) don’t resemble the real thing.

        • ggtdbz@lemmy.dbzer0.comEnglish
          2·
          2 days ago

          I remember the first time we jumped into the complex domain in an electronics course to calculate something that we couldn’t reach with the equations we had so far.

          … and then popping out the other side with a simple (and experimentally verified) scalar, after performing some calculation in the complex domain, using, bafflingly, real world inputs.

          I suddenly felt like someone from the future barged into my Plato’s cave and proceeded to perform some ritual.

          Like I know what’s happening, I’ve done these calculations before, but seeing them used as an intermediate step in something real in the real world was pretty cool!

          Did not prepare me for all the Laplace et al shenanigans later. Did I test well in those courses? No. Did I have the most fun building the circuits regardless? You bet.

          Oh to be a student again. Why are real world jobs so boring.

    • Bassman1805@lemmy.worldEnglish
      322·
      3 days ago

      The reason it doesn’t work is that 1 is a scalar while i is a vector (with magnitude 1). The Pythagoras theorem works with scalars, not vectors, so you’d get 1^2 +1^2 = 2.

      • someacnt_@lemmy.worldEnglish
        5·
        2 days ago

        I am sorry, but… to be pedantic, pythagorean theorem works on real-valued length. Complex numbers can be scalars, but one does not use it for length for some reason I forgor.

      • hydroptic@sopuli.xyzOPEnglish
        81·
        3 days ago

        Far as I understand it (which is not very far), i is a scalar even if you take it to be the complex number 0 + i. Just by itself i is the imaginary unit that’s defined as i = sqrt(-1) (edit: or, well, the solution to x² + 1 = 0, but same difference), and nothing in that definition says it’s a vector quantity.

        Even though complex numbers do extend real numbers into a 2D plane doesn’t mean they’re automatically vectors, and – again, as far as I’ve understood things – they’re still treated as single entities, ie. scalars. i by itself isn’t a complex number I think, though.

        The joke is that i² = -1 by definition, so i² + 1² = 0²

        Edit: eg. nothing on the imaginary number wiki page implies that the imaginary unit is not a scalar value

        • affiliate@lemmy.worldEnglish
          71·
          3 days ago

          whether or not i is a scalar depends entirely on the context.

          every vector space has an associated field of coefficients. in practice, this field is typically the real numbers. but you can have lots of other kinds of vector spaces as well, and they can be useful for certain things.

          anyways, if you have a vector space over the complex numbers, then i is a scalar, because it is a complex number. if you have a vector space over the real numbers, then i is not a scalar, because it’s not a real number.

          its worth mentioning that you can view the complex numbers as a vector space over itself. this is just a fancy way of saying that you can add complex numbers together, and you can multiply a complex number by a complex number. (one of those numbers is playing the role of scalar, and the other is playing the role of vector.) but you can also view the complex numbers as a vector space over the real numbers. and this is just a fancy way of saying that you can add complex numbers, and you can multiply a complex number by a real number.

          • hydroptic@sopuli.xyzOPEnglish
            2·
            3 days ago

            Right, sort of my vague understanding as well although it’s been 15 years since my university math courses. My point was more that “1 is a scalar while i is a vector” just didn’t seem correct to me, at least on a general level

            • affiliate@lemmy.worldEnglish
              2·
              1 day ago

              ah true. i missed the context of the original comment since it got deleted, but you’re completely right that the “1 is a scalar while i is a vector” statement is not entirely accurate.

    • owenfromcanada@lemmy.worldEnglish
      122·
      3 days ago

      If AB = i and BC = 0, then B would be in the same 2D space as C, but one of them would be “above” the other in 3D space (which doesn’t exist in this context, just as sqrt(-1) doesn’t exist in the traditional sense).

      So this triangle represents a 2D object that is “standing up” on the page.

      • rtxn@lemmy.worldEnglish
        7·
        3 days ago

        It makes sense if you represent complex numbers as (a, b) pairs, where a is the real part and b is the imaginary part (just like the popular a + bi representation that can be expanded to a * (1, 0) + b * (0, 1)). AB’s length is (1, 0), AC’s length is (0, 1), and BC’s length will also be a complex number.

        I think.

        • TowardsTheFutureEnglish
          1·
          3 days ago

          Yes. Also if you think of i as a 90° rotation (with a length of the scalar coefficient infront of i, in this case 1) . Thus one rotates you outwards away from the 2D plane, and two of those gets you back to the 2D plane, just going the other direction.