0

Previously, I found $a \bmod b = a - b \left\lfloor \frac{a}{b} \right\rfloor$ and wondered if this could be extended to the complex plane. I did this, and it seems to yield interesting results, such as the fact that when it seems to form a dihedral group under addition. Is this already an area of study within mathematics? I tried looking it up, but whenever I do I get the modulus of a complex number instead.

William Ryman
  • 548
  • 3
    The tag [tag:group-theory] feels misplaced here. Anyway, this begs the question: How do you define the floor function on the complex numbers? As you seem to know about Gaussian integers, I can guess :-). But, if you round both real and imaginary parts either "down" or "towards zero" you run into situations where the remainder is larger (in terms of absolute value) than the divisor $z_2$. A trick (well known in elementary number theory) is that you can avoid this by replacing the floor function with "Round". In other words, you round the quotient $z_1/z_2$ to the nearest Gaussian integer. – Jyrki Lahtonen Oct 31 '21 at 04:16
  • 1
    (cont'd) So sometimes you go up sometimes down, and the real and imaginary parts may be rounded to different directions. The upshot is that doing it this way has nice consequences. You can run Euclid's algorithm in the ring of Gaussian integers! This is kinda important because it does not work that easily with most other similar rings. – Jyrki Lahtonen Oct 31 '21 at 04:18
  • 1
    Start for example here. I tend to think of this question as a duplicate of that one. There is room for different opinions, because somebody else may be able to think of a different focus. Therefore I refrain from using my dupehammer. At least for the time being. – Jyrki Lahtonen Oct 31 '21 at 04:21
  • 1
    I added an upvote and pedal back a bit as I see that you mentioned the dihedral group. That alone implies that you can understand, why rounding to the nearest Gaussian integer will reduce the size of the remainder. The same thing happens in $\Bbb{Z}$ actually. If we seek to minimize the absolute value of the remainder instead of insisting that it be positive, we gain a bit: $|r|\le |b|/2$ instead $0<r<|b|$. – Jyrki Lahtonen Oct 31 '21 at 04:27
  • 1
    How would you define floor for complex numbers – TravorLZH Nov 01 '21 at 02:09
  • @TravorLZH I defined the floor function as $\left\lfloor z \right\rfloor = \left\lfloor \Re(z) \right\rfloor + \left\lfloor \Im(z) \right\rfloor$ – William Ryman Nov 01 '21 at 02:12

0 Answers0