Complex Rectangular Triangle Hypotenuse

[SKG_Scintill]

This is something that bothered me in high school. After graduation I thankfully forgot about it. But ever since SlayerApocalypse666 asked me to do some simple math, my mind went mathematical berserk, resulting in me hitting the wall I hit in the past.

The idea is that you have a rectangular triangle, one leg on a real axis, the other on an imaginary axis. The real leg has a length of 5, the imaginary leg has a length of 4i.

a^2 + b^2 = c^2 (olol duh)
25 - 16 = c^2
c^2 = 9
c = 3

The hypotenuse is shorter than the real leg, but it still has a real length. It can be measured, it is there, the point in which it makes an angle with the imaginary leg also exists.
As long as the real leg stays longer than the imaginary leg, the hypotenuse will be a real number.

But what about when they are equal? legs being 5 and 5i respectively...
The hypotenuse is 0.
The maths behind it don't bother me, but the fact that the distance is to be drawn on a sheet of paper.

Now, the same way that a square is the distance of a line drawn behind that same line (staying in a 1-dimensional space obviously) and a cube is the distance of a square drawn behind that square (in a 2-dimension space), you can draw the real leg with length 5, extend a line from an end in that line segment into a theoretical 4-dimensional space and have a measurable hypotenuse of 0.
But when you try to do that with the triangles that have a measurable hypotenuse with a length > 0, the hypotenuse would have to rotate around the real leg (still reasonable theoretically due to the angle being imaginary as well).

tan ∠A = 4i/5
∠A = arctan(4i/5) ≈ 38.67i deg

My question (finally): Following what curve (formula if it's possible) does the hypotenuse extend from the real leg (length of 5) for all lengths < 5i for the imaginary leg?

[cetaka]

I think you need to rephrase your question. "Distance" as such between two points in the complex plane is always real-valued (4, not 4i, in the above case) and so using the Pythagorean theorem in the way that you do and then calling 3 the hypotenuse of that triangle isn't correct.

[SKG_Scintill]

How so? If for instance you have a time-axis in physics, you use the value along with the unit it is measured in, seconds in that case for example. Similarly you can have an imaginary axis, yes its value would be 3 not 3i, but it would still have the unit in which it is measured: i. (or imaginary cm if you prefer)

[cetaka]

well you can do that but then it's a different space so, for instance, the pythagorean theorem as used by you does not hold. for example below using it on the two triangles on the real axis as you say gives their hypotenuses of zero, then using it again on the right triangle formed the two zero-length things gives you something else of length zero, even though it should naively have length two. and the same segment having two different lengths isn't good

http://i.imgur.com/1eAc6.png

[SKG_Scintill]

I'm not sure if that blows my mind even more or proves that my math is wrong :p
But I guess I shouldn't just use Pythagoras' Theorem for complex numbers and assume it works without it being proven. That put my mind to rest thankfully.

Edit: Yeah nvm, I stupidly forgot about the modulus...

[reuben_tate]

Yeah, make sure you know that the domains you are working with are suitable for any functions/formulas/etc that you are working with before trying to do something xP

EDIT:

Although, I think you can make the theorem work if using only imaginary units

Triangle with legs 3i and 4i:

(3i)^2+(4i)^2 = c^2
-9-16= c^2
-25 = c^2
c = sqrt(-25) = 5I

Trying to apply the theorom to both real and imaginary units seems tricky though.

[UserNameGoesHere]

Draw a picture because "rectangular triangle" does not compute. I have to stop there.

At least cetaka included some sort of picture. You should include one too.

[Cavernio]

He meant a right angle triangle. The triangle that has a hypotenuse.

[Cavernio]

It's because you're trying to visualize something that you weren't designed to visualize. Space only has 3 static dimensions. There's probably someone out there who visualizes different dimensions somehow, but I wouldn't be too concerned if you can't, or be too dumbstruck about the whole thing.
It's always bothered me that math is largely grounded in visuals.

[SKG_Scintill]

sadly indeed

[qqwref]

Yeah, I don't think it really makes sense to use the pythagorean theorem for cases like this. The very idea of "imaginary distance" might not make sense.

How so? If for instance you have a time-axis in physics, you use the value along with the unit it is measured in, seconds in that case for example. Similarly you can have an imaginary axis, yes its value would be 3 not 3i, but it would still have the unit in which it is measured: i. (or imaginary cm if you prefer)
You wouldn't use the units when you're doing geometry, though. For instance, suppose you have a right-angled triangle, with one leg of length 3 along the seconds axis and another leg of length 4 along the centimeters axis. You definitely wouldn't want to be taking sqrt(3 s^2 + 4 cm^2) to find the hypotenuse's length - that just doesn't make sense. To be fair, though, if your two legs have different units, it's generally a good sign that the hypotenuse you are looking for is meaningless and that you shouldn't try to compute it.

And even on the complex plane, the distance between (say) 0 and 2i is just 2 - the points on that plane are at real, two-dimensional positions, and simply represent imaginary numbers.

[Renevatia]

Pretty sure you can do this algebraically. You just have to change your plane of reference to where you are viewing both lines orthogonally. The lines will have to be written in algebraic form to adjust for the plane of reference.

Doing it this way, the imaginary i will only be a placeholder to denote the difference in dimension instead of the i^2=-1. Once it has been put into algebraic form you could just use x,y,z as axis (since you are not really using the real z axis in a basic Pythagorean problem anyway)

Edit: didn't see you mentioned modulus lol.

[Reincarnate]

Keep in mind that if you want the "length" of something imaginary, you need the modulus
|z| = sqrt(x^2 + y^2)
so the "length" of side b*i would just be b

Besides, the Pythagorean Theorem (as we use it for triangles) is all about right triangles in Euclidean space, which have real-valued lengths. It's formally equipped with vector spaces (inner product spaces http://en.wikipedia.org/wiki/Inner_product_space ) that are real-valued. It's what allows you to formally declare that two sides are orthogonal to each other when it comes to real-valued triangles.

So I mean just because you have a^2+b^2=c^2 doesn't necessarily mean you can just throw whatever you want into it. Apple^2+Orange^2=Carburetor^2 doesn't make much sense either. It's going to "mean" different things in different frameworks.

Generalized Pythagorean Theorem via inner product spaces (if you want to know more): http://en.wikipedia.org/wiki/Pythagorean_theorem#Inner_product_spaces
Pythagorean Theorem and imaginary numbers: http://en.wikipedia.org/wiki/Pythagorean_theorem#Complex_numbers

More generally, my point is that when you start working with imaginary numbers, you need to 1. Calculate them correctly, and 2. Understand that trying to invoke axioms of the real won't always make sense. It's like how Euler's Identity holds true even though "it doesn't make much real sense" -- but it makes perfect sense as long as you operate within the proper frameworks and their corresponding axioms.

[SKG_Scintill]

Keep in mind that if you want the "length" of something imaginary, you need the modulus
|z| = sqrt(x^2 + y^2)
so the "length" of side b*i would just be b

Yeah nvm, I stupidly forgot about the modulus...

People can stop answering now, I got it like a week ago :P

[Reincarnate]

haha, my bad -- I'm usually too lazy to read the thread

[qqwref]

da smartest man on the forums

not smart enough to read a post date