In article ,
Bill wrote:


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)

ARGH!!! like I said, "I must be doing something wrong' wry grin

I inverted the sign on the last term _and_ left the i^2 in. one or the
other, but not both. Takes more than 2 i's to find it.

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

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.

Well, It _is_, for exponents without an imaginary component. The concept
of multiplying something by itself an imaginary number of times is utterly
lacking in any intuitive foundation.

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.

His own fault! _He_ was the one that sprung it on me, cold, and wanted me
to agree it was true. I'd stuck my head in the classroom, after school
was out, to ask him something, and he threw that at me, from the middle of
a discussion he was having with somebody else -- "e to the two pi eye
equals one, right?" I thought for about half a second and said, firmly,
"no". He asked "why not?" And I wrote the above 'disproof' on the black-
board. Finding -where- to kick a hole in it is difficult. Every line
_does_ follow validly from the prior one. It's merely that the 'meaning'
of the numbers changed.
They have their hands full
trying to convince students that (a+b)^2 is not a^2 + b^2!!! Tough job!

HUH?????

I think my 5th or 6th grade class spent -maybe- one afternoon on that.
Re-write the '^2' as an explicit multiplication, and apply the distributive
property (a total of 3 times) to eliminate the parens, then consolidate the
like terms.

The error is also 'painfully obvious' if you simply work through a couple
of 'concrete' examples.

Admittedly, I was learning this stuff _just_before_ the "new math" teaching
started. (My younger brother, 3 years behind me, got a *very* different math
education!)

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.

Depends on what you mean by '='. grin type='Clintonesque'

They're both the same "direction". "equal" in that sense.

Anything derived from the point-value (ignoring the 'path' that got you
there) is the same for both.

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.

Quoting a line from an old comic strip -- where they got it absolutely
wrong (unintentionally!) in context: "How do you tell a pilot he's 360
degrees off-course?"