# Pythagorean Theorem for imaginary numbers

If we let one leg be real-valued and the other leg equal $bi$ then the Pythagorean Theorem changes to $a^2-b^2=c^2$ which results in some kooky numbers.

For what reason does this not make sense? Does the Theorem only work on real numbers? Why not imaginary?

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The length of the $bi$ leg is $|bi| = \sqrt{0^2 + b^2} = b,$ not $(ib)^2.$ –  user2468 Jul 11 '12 at 22:05
Ideas like this can be useful. For example, hyperbolic geometry can be thought of as geometry on the sphere of imaginary radius. And the identity $\cos^2 x+\sin^2 x=1$ can be though of as a version of the Pythagorean Theorem. In the world of hyperbolic functions, this has the analogue $\cosh^2 x-\sinh^2 x=1$. –  André Nicolas Jul 11 '12 at 22:25

The Pythagorean Theorem is specific to right triangles in Euclidean space, which have nonnegative real lengths. It is properly generalized by considering inner product spaces, which are vector spaces $V$ equipped with a function $\langle \cdot,\cdot \rangle:V\to \mathbb R$ (or $\mathbb C$) satisfying certain axioms. This allows us to define a right triangle as a triple of points $(\vec a,\vec b,\vec c)$ such that $\langle \vec b-\vec a,\vec c-\vec a\rangle=0$, which can be seen as saying the sides $\vec b-\vec a$ and $\vec c-\vec a$ are orthogonal. It also gives us a notion of length, with the length of a vector $\vec v$ being $\sqrt{\langle \vec v,\vec v\rangle}$. The generalization then states that $$\langle \vec b-\vec a,\vec b-\vec a\rangle+\langle \vec c-\vec a,\vec c-\vec a\rangle=\langle \vec b-\vec c,\vec b-\vec c\rangle$$ assuming $\vec b-\vec c$ is the longest side, i.e. that the length of the side $\vec b-\vec a$ squared plus the length of the side $\vec c-\vec a$ square equals the length of the side $\vec b-\vec c$ squared. This is very different from the generalization you gave.

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Clearing comments relating to rendering issues. See meta.math.stackexchange.com/q/4629/1543 instead. –  Willie Wong Jul 12 '12 at 8:27

It does not make sense because lengths are nonnegative real numbers. It does not really make sense for negative real numbers, either.

The theorem does not really have much to do with numbers at all, it's just the analytic statement of a geometric fact, which can be restated as:

For any right triangle, the area of a square whose side is the hypoteneuse is equal to the sum of areas of squares whose sides equal the legs.

which removes the reliance on numbers completely. You can return them by attaching a numerical value to areas, but you don't need real (or complex!) numbers to have a meaningful notion of area of a polygon, like you don't need them to compare lengths.

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And you can prove it (as Euclid did) by purely geometric means, without any numbers at all. All that is needed is that the area of any triangle with a given base and altitude is half the area of a rectangle with the same base and altitude. –  marty cohen Jul 11 '12 at 22:49
@martycohen In fact, you don't even need that much - you can show it by dissection, somewhat implicitly using the fact that the area of a right triangle of given base and altitude is half the area of that rectangle (which is even more trivial than the general statement, since two congruent right triangles obviously fit together along their diagonals to form the rectangle). –  Steven Stadnicki Jul 12 '12 at 0:02

In the vein of @AndréNicolas' comment that "ideas like this can be useful" ...

Heron's formula for the area, $W$, of a (non-obtuse) triangle in terms of the lengths of its sides is equivalent to the Pythagorean Theorem for Right-Corner Tetrahedra:

$$W^2 = X^2 + Y^2 + Z^2$$

where $W$ is the "hypotenuse-face" opposite three mutually-perpendicular edges that form right triangles with areas $X$, $Y$, $Z$.

Specifically, if we position the right corner at the origin in 3-space, choose the other vertices at $(x,0,0)$, $(0,y,0)$, $(0,0,z)$, and write

\begin{align} a^2 = y^2 + z^2 \qquad b^2 &= z^2 + x^2 \qquad c^2 = x^2 + y^2 \\ X = \frac{1}{2} y z \qquad Y &= \frac{1}{2} z x \qquad Z = \frac{1}{2} x y \end{align} then $$x^2 = \frac{1}{2}\left(-a^2 + b^2 + c^2\right) \qquad y^2 = \frac{1}{2}\left(a^2-b^2+c^2\right) \qquad z^2 = \frac{1}{2}\left(a^2+b^2-c^2\right)$$ whence \begin{align} W^2 = X^2 + Y^2 + Z^2 &= \frac{1}{4}\left(y^2 z^2+z^2x^2+x^2y^2\right)\\ &=\frac{1}{16}\left(-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2\right) \\ &=\frac{1}{16}\left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right) \end{align} which is Heron's Formula for $W$ in terms of side-lengths $a$, $b$, $c$.

Note that I wrote the equivalence is for "non-obtuse" triangles. This is because "hypotenuse-face" $W$ cannot have an obtuse angle; to see this, write $\theta$ for the angle opposite side $a$, so that

$$\cos \theta =\frac{-a^2+b^2+c^2}{2b c} = \frac{x^2}{bc}$$

This value is non-negative (and, hence, $\theta$ is non-obtuse) for any edge-length $x$ ... well, any real edge-length $x$. However, if we allow $x$ to be an imaginary length, then $\theta$ can be obtuse, and then Heron's Formula can be seen to apply to all triangles.

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