# Metaphor for imaginary numbers? **URL:** \ **Category:** Factual Questions\ **Created:** [January 20, 2010, 11:08pm UTC](https://boards.straightdope.com/t/metaphor-for-imaginary-numbers/525755 "2010-01-20T23:08:34Z")\ **Posts on this page:** 6\ **Page:** 3
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [January 23, 2010, 11:07pm UTC](https://boards.straightdope.com/t/metaphor-for-imaginary-numbers/525755/41 "2010-01-23T23:07:54Z")
Or, as noted before: since i is a 90 degree turn, the sqrt(i) is something which, when repeated twice, gives a 90 degree turn; namely, a 45 degree turn (or a 235 degree turn). Much simpler this way, eh? 🙂 (But, if one insists on breaking such motions down into their parallel and perpendicular components, yes, “v turned 45 degrees” is the same as “(v scaled by a factor of 1/sqrt(2)) + (v scaled by a factor 1/sqrt(2) rotated 90 degrees)”, in accord with the side ratios of 45-45-90 triangles, as determined by the Pythagorean theorem. As shown above by **Chronos** , this all works out, as it must; indeed, one can go the other way around and use complex numbers to give a simple derivation of the Pythagorean theorem. It all hangs together, as it must.) ---
**Author:** ![Topologist](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/topologist/32/3208_2.png) [@Topologist](https://boards.straightdope.com/u/Topologist)\ **Post date:** [January 24, 2010, 1:52am UTC](https://boards.straightdope.com/t/metaphor-for-imaginary-numbers/525755/42 "2010-01-24T01:52:30Z")
> [@Indistinguishable](#): > > … indeed, one can go the other way around and use complex numbers to give a simple derivation of the Pythagorean theorem. Is this true? Does it depend on the interpretation of the absolute value of a complex number being its distance from the origin? And doesn’t that interpretation depend on the Pythagorean theorem? (I may be off base, but I can’t think of how to do this without it becoming a circular argument.) ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [January 24, 2010, 2:38am UTC](https://boards.straightdope.com/t/metaphor-for-imaginary-numbers/525755/43 "2010-01-24T02:38:26Z")
Like so: Suppose given a right triangle in a plane, with vertices a and b on the hypotenuse and c opposite to the hypotenuse. Let the vectors A, B, and C denote the vectors c - b, a - c, and a - b respectively (thus A + B = C). We must show that ||A||^2 + ||B||^2 = ||C||^2. Let u be a unit length vector pointed in A’s direction. Since A is perpendicular to B, there must be a 90 degree rotation which causes u to be pointed in the same direction as B; let this be the rotation denoted by i. There must also be some rotation which causes u to be pointed in the same direction as C; call this rotation theta. Thus, A = ||A|| \* u, B = ||B|| \* i \* u, and C = ||C|| \* theta \* u. But also A + B = C, so we have that (||A|| + ||B|| \* i) \* u = ||C|| \* theta \* u. Since u is a nonzero vector, we can conclude from this last equation that ||A|| + ||B|| \* i = ||C|| \* theta; essentially, we’ve given both “rectangular” and “polar” descriptions of the complex number sending u to C. Now, take the complex conjugate of both sides to obtain that ||A|| - ||B|| \* i = ||C|| / theta. Finally, multiply these two equations together to see that (||A|| + ||B|| \* i)(||A|| - ||B||\*i) = ||C|| \* theta \* ||C|| / theta, which simplifies to ||A||^2 + ||B||^2 = ||C||^2. Of course, we could just have written “90 degree rotation in the chosen direction” each time instead of “i”, and this would be a straightforward geometric proof. But that’s the point: working with complex numbers _is_ doing the geometry of scaling and rotation of vectors; the two are exactly the same thing, whether you write it this way or that. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [January 24, 2010, 2:55am UTC](https://boards.straightdope.com/t/metaphor-for-imaginary-numbers/525755/44 "2010-01-24T02:55:33Z")
Incidentally, in case anyone is wondering what prevents this proof from being applied to non-Euclidean geometries as well (where the Pythagorean theorem fails), it’s the fact that such geometries do not form affine spaces (so one cannot perform vector arithmetic with the differences between points). ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [January 24, 2010, 3:40am UTC](https://boards.straightdope.com/t/metaphor-for-imaginary-numbers/525755/45 "2010-01-24T03:40:28Z")
> [@Chronos](#): > > Eh, it’s impressive enough that introducing complex numbers completely closes the system with respect to polynomials, but I think it’s even more impressive that it also closes it with respect to things like inverse trig functions and logarithms. I imagine that this stems from the fact that trig functions and the like can be expressed in terms of infinite sums, but that still adds another layer of nontriviality. You don’t need to appeal all the way to infinite sums. Logarithms come for free from the ability to take logarithms of positive reals (since every nonzero complex number is a product of scaling by a positive real and rotation by some angle, with rotation being an exponential function of its angle). And once you have logarithms (and square roots), you can get all the inverse trig functions from them [for free](http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms) (just as exponential functions give the trig functions for free, again as a consequence of rotation being itself an exponential function). ---
**Author:** ![Topologist](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/topologist/32/3208_2.png) [@Topologist](https://boards.straightdope.com/u/Topologist)\ **Post date:** [January 24, 2010, 7:12pm UTC](https://boards.straightdope.com/t/metaphor-for-imaginary-numbers/525755/46 "2010-01-24T19:12:33Z")
**Indistinguishable** : Thanks (for the proof of the Pythagorean theorem). I hadn’t seen that one before. Nice. [Previous page](https://boards.straightdope.com/t/metaphor-for-imaginary-numbers/525755.md?page=2)