Pythagoras' Theorem

 

The Boy Next Door

The boy next door
I just adore him
So I can't ignore him
The boy next door

If you adore something, then you can’t ignore that something!

In checking on the internet, I am not the first engineering student who could not get his head around the concept of ignoring the imaginary portion of solutions involving AC currents. I was always taught that ignoring something does not make that something go away. I could not bring myself to ignore the imaginary portion, so I ignored electrical engineering. In hindsight, maybe I should have not ignored anything.

The solution for a hypotenuse in Euclidean geometry, Pythagoras’ Theorem, is c=√(a²+b²) which is true because cos²(x)+sin²(x)=1.  But this is only true on a flat surface.  The solution on a spherical surface is cos(c/R)=cos(a/R)*cos(b/R) where R is the radius of that spherical surface, and the solution on a hyperbolic surface is cosh(c)=cosh(a)*cosh(b).  But cosh(i*x)=cos(x)+i*sin(x).  This means that the hypotenuse on a hyperbolic surface can be restated, rotated by 90 degrees, since cosh(x) = cosh(ix) rotated by 90 degrees,  as 

cosh(c/R)-i*sin(i*c/R)=cosh(a/R)*cosh(b/R)
                                           -i*cosh(a/R)*sin(i*b/R)
                                           -i*cosh(b/R)*sin(i*a/R)
                                          +sin(i*a/R)*sin(i*b/R).

The limit as R→∞ because sin(0)= 0, is

Cosh(c)-i*sin(i*0)=cosh(a)*cosh(b)-i*cosh(a)*sin(i*0)-i*cosh(b)*sin(i*0)+sin(i*0)*sin(i*0)

Because cos²(x)+sin²(x)=1, it is also true that sin(x)=√(1-cos²(x)). Thus all three formulae for the different surfaces are the same, and Pythagoras’ Theorem is merely the case of ignoring the imaginary components when R→∞.  This is also why, when a²+b² is negative, the solution to Pythagoras’ Theorem becomes imaginary. They said on the Chiffon Margarine TV commercial when I was growing up, ”It’s not nice to fool Mother Nature!”.  It is apparently not nice to ignore her either!