What do you think of imaginary triangles, Veeky Forums?

Violating the triangle inequality just means it isn't a constructible triangle, it doesn't mean we can't have imaginary meme triangles.

No the thing that's interesting to me regarding the right triangle is that drawing a straight line from [math]1i[/math] to [math]1[/math] results in a line of length 0

Does that mean

[math]sin^2(\frac{i}{0})+cos^2(\frac{1}{0}) = 1[/math]

is true?

wolfram says it's [math]-\infty[/math] but what the fuck would wolfram know about memes of this magnitude

That's a shame, it would have been comfy if it equaled 1.

drawing a line from i to 1 in the usual sense does not produce a line length of zero. The diagram is invalid. If it were valid and it were somehow possible to make a right triangle with legs i and 1, the other angles (x and y) would have to solve the equations tan(x) = i and tan(y)= -i. No such x and y exist. Therefore whatever object you're referring to as "the right triangle with sides i, 1 and 0" isn't a right triangle.

Hold on a second: wouldn't a side of length zero mean that the i-side is overlapping with the 1-side? Doesn't that mean this triangle is collapsing the complex number line onto the real number line in the complex plane?

Useful for phasor analysis I guess, I don't know what else you would do with them.

Did I solve it, OP?

It does, it means this set in affine space is empty

they are not real what the fuck ks there to think about u virgins

Here's your answer

Degenerate and useless.

And non-existent

>negative side length
really perturbs my pecans

bump

inventive..
unrealistic

So the area of this 'square' is 2i?

Why are you assuming that the triangle is perpendicular?

So we end up with [math]0 = \sqrt{1-1}[/math] which is correct, but also [math]i = \sqrt{0-1}[/math] and [math]1 = \sqrt{0-(-1)}[/math]

>drawing a line from i to 1 in the usual sense does not produce a line length of zero

Alright then, how long is this?

Multi-dimensional surfaces do not have arc-lengths.

its obviously sqrt(2)

Agreed, as I'm aware triangle equality is necessary for a metric space. But I haven't studied complex spaces, would there be an analog?

considering |i|=1, the line ought to be sqrt(2) units long

Brainlets detected.

All three of those are true though?

be more careful when you ask questions on topics you don't understand at all, people might take it the wrong way

complex vector spaces are metric spaces

I think that that was the point of the post

Intradasting
I think ur onto something op

three phase power involves imaginary triangles

They are used to represent power triangles in power systems.

Real Power(Watts) is on the X
Reactive Power(Vars) on the Y
The hypotenuse is Apparent Power (VA)

The [math]\sqrt{2}[/math] is derived from Pythagoras' theorem of a triangle with legs of 1

One of the legs of this triangle is obviously not 1 so the answer is obviously not [math]\sqrt{2}[/math] else [math]i = 1[/math] which invalidates the entire imaginary axis

how the fuck do you have a polygon with a side of length 0

degenerate

So basically the distance between the real axis and imaginary axis is actually zero.

The distance of i from 0 is 1 you brainlet

>degenerate

underrated

We have a triangle A,B,C. One side A=1 is purely spatial side, the side B=i is purely time, ande side C has norm equal to the Minkowski norm root(1^2+i^2)=0.

what would be an example of a metric for complex vectors?

Interesting, so basically the origin of [math]i[/math] can simply be described as the third side of a right triangle with 0 and 1 sides, which is way simpler than trying to shit it out through quadratic equations