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Imaginary playmate


Well-known member
Feb 2, 2012

I played with imaginary numbers while in college.
I've made some mind-stretching "discoveries" over the years.
I'll post some of them here, a few at a time.
If any of you have found such "facts", I've love to see them.

Consider a line with slope $i.$
. . It would have the form: $y \:=\:ix + b$
Since $i = \text{-}\frac{1}{i}$ ($i$ is its own negative reciprocal),
. . the line $y \:=\: ix + b$ is perpendicular to itself.

Consider the circle $x^2+y^2 \:=\:\text{-}1$.
. . It has center $(0,0)$ and radius $i$.
Its area is: $\pi(i^2) \:=\:\text{-}\pi$, a real number.

Surprisingly, $i^i$ is a real number: .$i^i \:=\:e^{\text{-}\frac{\pi}{2}}$


Well-known member
MHB Math Scholar
Feb 15, 2012
It turns out that when you consider perpendicularity in complex vector spaces, you don't determine it by:

$(x,y) \perp (x',y') \equiv xx' + yy' = 0$

but rather:

$(x,y) \perp (x',y') \equiv x\overline{x'} + y\overline{y'} = 0$.

So a perpendicular line to:

$y = ix + b$

is actually of the form:

$y = -ix + b'$

(for much the same reasons as the product $(a+bi)(a-bi)$ is a SUM of squares, rather than a difference of squares).

However, your charming observation is why the Rastafarians say: "I and I are orthogonal".