sqrt(-i) = exp(-PI/4) or exp(3PI/4). Not "purely imaginary" at all--the
real part is +/- sqrt(2)/2. It's as plain as a point on the unit circle.
You could double-check that if a = sqrt(2)/2 - sqrt(2)/2*i, or b=-a,
then a^2=b^2=-i. Simple trig and De Moivre's Theorem...

I must be doing something wrong:
( sqrt(2)/2 - sqrt(2)/2*i ) **2

=(sqrt(2)/2)**2 -*((sqrt(2)/2*sqrt(2)/2*i) - (sqrt(2)/2*i)**2
= 2/4 -2*(2/4*i) - 2/4*(i*i)



Robert, Nice Try! The line above should be 2/4 -2*(2/4*i) +2/4*(i*i)

= 2/4 -4/4i -2/4 = -i, as advertised. You had me worried for a moment.



= 2/4 - 4/4*i + 2/4
= 1-i

I'm tempted to make a facitous remark about an 'off by one' error.


Admittedly, I haven't played with this stuff for 30+ years, but I've got
a vague recollection of 'j' (the 'hyper-imaginary'??) as sqrt(-i).



Of course, on my first exposure to the imaginary exponential, I
immediately
and rather vehemently questioned "exp(2*pi*i) = 1". As follows:

exp(0) = 1 by definition
exp(2*pi*i) = exp(0) 2 thins equal same thing, equal to each other
(2*pi*i) = 0 if bases (non-zero, and non-multiplicative-
identity) equal, exponents equal.

Yes, you assumed that the function e^x is 1-to-1. And it is definitely not.
It wraps the imaginary axis around the unit circle infinitely often.
Indeed,
it maps every horizontal strip of height 2PI in the complex plane onto the
whole complex plane minus the point 0. I use this property of e^x as a
building block to help construct other mappings which are infinity-to-one.
Of course, there is nothing special abought e^x in this context, any
exponential
function a^x (a0, a!=1) may be written in the form e^(kx) for some real
number
k so similar properties hold for a^x. What is special about e^x is that
d/dx (e^x) = e^x, and only constant multiples of it have this property (of
being equal to their derivative on a suitable domain).



This rather upset the 8th grade math teacher. He *knew* my reasoning
was incorrect, but pointing out _where_ the flaw was was not obvious.




...Not obvious to an 8th grade math teacher. They have their hands full
trying to convince students that (a+b)^2 is not a^2 + b^2!!! Tough job!

Actually, my 8th grade math teacher was one of my favorite teachers, K-12.




Simple reasoning on complex issues can lead one to into trouble. grin

Once it was clarified that '2*pi' was angular measure, things clarified
It is true that 2*pi == 0 (plus 1 revolution), even though neither of
the multiplicands is zero.



It is of course Not true that 2*pi =0. What is true is that e^(2pi*i) =
e^0=1.

What is also true is that, for real values x, e^(ix) = cos(x) + i sin (x),
which I see is what you were referring to by "plus 1 revolution" above.


Best,
Bill