What Happens to sin(i) for Complex i?

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The discussion centers on the definition and understanding of the imaginary unit "i," which is defined as the square root of -1. Participants explore the concept of complex numbers as points in a two-dimensional plane, where "i" is represented as (0, 1) and its multiplication leads to the conclusion that i*i = -1. There is debate over the ambiguity of defining "i" since both +i and -i are valid square roots of -1, raising questions about mathematical definitions and their interpretations. The conversation also touches on the philosophical aspect of existence in mathematics, comparing abstract concepts like "i" to more tangible numbers. Ultimately, the discussion highlights the complexities and nuances in understanding imaginary numbers and their foundational role in mathematics.
  • #51


HallsofIvy said:
But now your distinction between "i" and "-i" depends upon which way is "up" and which way is "down", again an arbitrary choice.

It would be arbitrary if I was the first person to adopt this notation, but it is by now conventional that i is up and –i is down. In Electrical Engineering, j is up and represents inductive reactance and –j is down for capacitive reactance so it seems a convention is already established.
 
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  • #52


schroder said:
It would be arbitrary if I was the first person to adopt this notation, but it is by now conventional that i is up and –i is down. In Electrical Engineering, j is up and represents inductive reactance and –j is down for capacitive reactance so it seems a convention is already established.

but it doesn't matter. whether you call it "j" or "-j", other than the fact that they are negatives of each other, every property of these two imaginary numbers are precisely the same; they both square to be -1. if all math (and physics and electrical engineering) literature was changed and every occurance of i or j were replaced with -i or -j, every statement and every theorem would be just as valid as it was before. but you can't say that about replacing 1 with -1.

check out:

http://en.wikipedia.org/wiki/Imaginary_unit#i_and_.E2.88.92i
 
  • #53


rbj said:
but it doesn't matter. whether you call it "j" or "-j", other than the fact that they are negatives of each other, every property of these two imaginary numbers are precisely the same; they both square to be -1. if all math (and physics and electrical engineering) literature was changed and every occurance of i or j were replaced with -i or -j, every statement and every theorem would be just as valid as it was before. but you can't say that about replacing 1 with -1.

All of that is true. The only point that I was making is that all ambiguity is removed by making a choice, and staying with that as a convention which has already been made in applied science and engineering. Then only in the realm of the pure mathematician does any ambiguity reside but that’s what pure mathematicians thrive on! :smile: I do not reside in that realm. :smile:
 
  • #54


schroder said:
... only in the realm of the pure mathematician does any ambiguity reside but that’s what pure mathematicians thrive on! :smile: I do not reside in that realm. :smile:

Nor do I.

The realm of pure mathematics is a nice place to visit, but I wouldn't want to work there. :biggrin:
 
  • #55
… don't talk to me about reality …

Redbelly98 said:
The realm of pure mathematics is a nice place to visit, but I wouldn't want to work there. :biggrin:

Mathematician to physicist:

:biggrin: Reality is a nice place to visit … but I wouldn't want to live there! :biggrin:
 
  • #56


WE know that sin(x)\leq1 but sin(i) is something else an anyone explain
 
  • #57
hadi amiri 4 said:
WE know that sin(x)\leq1 but sin(i) is something else an anyone explain

Haven't we already gone over this :confused::
tiny-tim said:
Hi hadi amiri 4! :smile:

… and so cosh(ix) = cos(x), sinh(ix) = i sinx,

and, conversely, cos(ix) = cosh(x), sin(ix) = -i sinx.

sin(A + iB) = sinA cosiB + cosA siniB = sinA coshB - i cosA sinhB.

cosh and sinh are known as hyperbolic cosine and sine. :smile:

sin(i) = sin (i.1) = -i sinh(1) = -i(e - 1/e)/2 :smile:
 
  • #58


hadi amiri 4 said:
WE know that sin(x)\leq1 but sin(i) is something else an anyone explain

We know that sin(x)\le 1 for real x. sin(z) for complex z is not bounded- it looks like an exponential along the imaginary axis. In fact, there is a famous theorem that if an entire function is bounded, it is constant.
 
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