x is not 0! (viral maths meme)

x=0 doesn't even make sense anyway.

Nobody said the x is 0! Everybody was saying that x is 0

Conclussion: i = 1

The vector reasoning is incompatible with other right triangles. Why can't I look at a 3-4-5 triangle and say, well, 3 and 4 are vectors that go in the same direction, so the remaining side is just 1? You're assuming the triangle is drawn to scale, and supposing that the triangle actually does point in the imaginary direction. I'm not saying you're wrong with √2 but there's absolutely more nuance to what you did.

Missing the joke sooooo much 💀 i is the magnitude of the vector. "Vectors can't have imaginary magnitude" yes that's the joke. It doesn't make sense to say you're i meters away from something.

I'm not sure if I agree with this. The diagram shows "i" as the value for the length of the side, not as a vector itself. If you treated the side like a vector, then the vector should have imaginary magnitude (i.e. magnitude = i), not a magnitude equalling 1 as you described in your explanation. I'm not sure if imagninary magnitude is a well-defined concept but if we were to extend mathematics to allow such a concept, it doesn't seem unreasonable that we could extend Pythagoras' theorem to allow us to claim 1^2 + i^2 = 0^2 as an answer to the problem.

The hypotenuse being zero is legal on a pseudo-Riemannian manifold. It shows up all the time in special relativity with light cones.

this isn't correct because the length of the triangles side is i, it isn't a vector pointing to i

The real issue here is everyone assuming without it being specified that the triangle is right angled and that pythagoras is applicable

You miss the point, this is the space-time geometry. A photon has energy 1, momentum i and mass √(1²+i²)=0

magnitudes can be imaginary in the split complex numbers (a+bj, j^2 = 1) as abs(a+bj) = sqrt(a^2-b^2), for example abs(j) = i you can imagine this diagram as the split complex number 1+j, which indeed has a magnitude of 0.

X is 0 What you forgot in the video is that the second side is perpendicular to the first, so you have to multiply it by i (think of a triangle of sides 1 and 1, and how it would be represented in the complex plane; which also explains the case 1, -1, sqrt2). So you have a side being 1+0i, and other being i*(0+1i)=-1+0i. Adding both vectors get you to the origin with modulus 0, as expected.

No, complex numbers are not vectors, strictly speaking. They are something different. They are a basic example of multivectors (a subgroup of a Clifford Algebra). So, it is incorrect to say that they are vectors, even if this is somehow a common mistake, used in applied math!

As far as I know, that is only assuming that the triangle is in the complex plane and not an actual triangle. Though I guess that isn't too far fetched as an actual triangle with a side length of i would be impossible.

conclusion: x = √(2)*e^(iπ/4)

I think that it boils down to how one interprets a mathematical diagram. When one first looks at the diagram, he or she is probably inclined to think that he or she is looking at a triangle and will assume that it is a right one and therefore use the Pythagoras theorem to try to find x. If there were axis in the background, he or she might think about complex numbers and vectors if he or she had studied those things before. The other problem is that one can't assume that that triangle is a right one since there is no indicator that any of the triangle's angles are right angles. The indicator that is usually used is a small square inside the right angle. Therefore, x could be any of many values.

Its not ordinary geometry. Its hyperbolic one like in Minkowski diagram of special relativity where Pitagoras theorem is a2-b2=c2. If a=b then c=0.

Ridiculous question, i is not a length so we cannot have a triangle with a side of i. What you have worked out is based on a side of |i| instead which is quite a different thing.

I remember hearing this idea that complex numbers are vectors, but I couldn't buy it. It's like saying that C = R^2. Learning about the history of vector analysis, the vectors we're familiar with came after when C and H were defined, and were constructed specifically to avoid the isomorphism with the special unitary group. If anything, complex numbers and quaternions are probably best thought of as an object that encodes in multiplication, what vectors encode in both dots and crosses. But just because you can ultimately do the same work with them, doesn't make them the same thing.

It's very interesting point of view... Your way for me looks very close to möbius strip topology. Triangle is the basis with "i"-number to build möbius strip... Is It posible to visualise "your way"in 3-d form?.. with spherical diagonal line as "x"π and ortogonal oriented triangle sides "i" and "1" ...