Is it possible/meaningful to have a right triangle with sides $0$, $1$, and $i$?

Question

If $i^2 = -1$, does that mean you could have a triangle with side lengths $0$, $1$, and $i$? But if you draw such as triangle on a complex number grid, the hypotenuse length is $\sqrt 2$, not $0$. Does the pythagorean theorem only apply to real numbers?

Answer 1

No.

That wouldn't be possible because distance is inherently a nonnegative real number, usually -- you can see this in the formal definition of a metric, which is how we define distance in a mathematically rigorous and formal sense.

Within that framework, then, it's not possible to have negative distances, much less complex ones. For instance, we can define the distance between the complex numbers $0,i,1$ by the usual "Euclidean distance" or "$2$-metric", the distance you're familiar with. But for complex numbers, this ties back somewhat to definition of magnitude. We define the distance $d(z_1,z_2)$ between complex numbers $z$ (under this metric) to be

$$d(z_1,z_2) = |z_1-z_2|$$

where $| \cdot |$ denotes magnitude. Notice how this is a real number (magnitudes of complex numbers always are). In fact, if $z_i = x_i + i y_i$ for $i=1,2$, then

$$d(z_1,z_2) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$$

What this all says is that it is nonsense to say that a side length can have length "$i$". This is because what you're ultimately, truly meaning is "what is the distance between $0$ and $i$ in the complex plane (under a certain metric)?" Yes, $0+i=i$, but that distance is meant to be measured in terms of nonnegative, real numbers -- otherwise, the distance between $-2$ and $0$ is $-2$, which doesn't make sense.

Or in a different sense, perhaps bear in mind the difference between "distance/length" and "displacement."