e to the i pi

[Lane Straatman]

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <complex.h>
/*
double complex z1, z2, z3;
bool flag;
z1 = .4 + .7I;
z2 = cpow(z1, 2.0);
z3 = z1 * z1;
flag = false;
flag = true;
if (flag)
{
printf("%lf %lf\n", creal(z1), cimag(z1));
printf("%lf %lf\n", creal(z2), cimag(z2));
printf("%lf %lf\n", creal(z3), cimag(z3));
printf("%d\n", N);
}
*/

int main(int argc, char *argv[])
{
double complex z1, z2, z3, z4, z5;
z1=5 +7I;
z2=cpow(z1, 1I);
printf("%lf %lf\n", creal(z1), cimag(z1));
printf("%lf %lf\n", creal(z2), cimag(z2));
z5= 0 + I*(3.14159);

z3=2.54 + 0*I;
z4=cpow(z5,z3);
printf("%lf %lf\n", creal(z4), cimag(z4));

system("PAUSE");
return 0;
}
Why doesn't e^(i *pi) equal what most folks think it does? LS

[Dave Vandervies]

In article <eq***********@pc-news.cogsci.ed.ac.uk>,
Richard Tobin <ri*****@cogsci.ed.ac.ukwrote:

>In article <eL******************************@bt.com>,
Malcolm McLean <re*******@btinternet.comwrote:

>>Can't say this bit about e^(i*x) = cos x + i * sin x is too convincing
though.

Consider the polynomial expansion e^x = 1 + x + x^2/2! + x^3/3! + ...

I think you mean "power series expansion". It's only a polynomial if
it's finite.
dave
(knowing the power series expansions for sin and cos would also be helpful)

--
Dave Vandervies dj******@csclub.uwaterloo.ca

>What would you give if somebody used a silly unit (hogsheads) and gave
a good justification for it? --Joe Zeff and Garrett Wollman

More thought. in the scary devil monastery

[Keith Thompson]

"Lane Straatman" <in*****@invalid.netwrites:

"Richard Tobin" <ri*****@cogsci.ed.ac.ukwrote in message
news:eq***********@pc-news.cogsci.ed.ac.uk...

>In article <ln************@nuthaus.mib.org>,
Keith Thompson <ks***@mib.orgwrote:

>>>Apparently i^i (where i is the imaginary square root of -1 and "^"
denotes exponentiation) is a real number. I'm sure there's a simple
mathematical proof of this, but I'm too lazy to track it down or
reconstruct it.

It's exp(-pi/2). Someone else showed the proof, but you can just type
"i^i" into Google to see the numerical value.

http://www.billfordx.net/screendumps/cstuff_7.htm

Clearly, this latest version is off. Heathfield says I have two specifiers
for one datum. Is it not a datum with a real and a complex part, both of
which need attention in the conversion specifiers?

[...]

This is in reference to your attempt to print a single complex value
using a printf format that expects two floating-point arguments. I
have tried at least twice to explain this to you. I will not waste my
time doing so again.

In fact, I will not waste my time responding to anything you post, or
offering you any sort of help after this. I just read your article in
comp.std.c,
<http://groups.google.com/group/comp.std.c/msg/d91bae8783a38bde>, in
which you replied to a polite, informative, and only mildly critical
article by ku****@wizard.net with:

| You have the manners of ferrets. F*** yourselves collectively and shove the
| standard up your ***. LS

[edited for content]

It's not inconceivable that that article was a forgery. If so, please
say so. (I don't believe it was.)

Until and unless your behavior improves dramatically, I judge you to
be a troll (deliberate or not, I don't care which), and I will ignore
you. I leave it to others to make their own judgements.

Yes, this is off-topic. I decided to post it anyway as an attempt to
improve this newsgroup's signal-to-noise ratio.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Malcolm McLean]

"Dave Vandervies" <dj******@caffeine.csclub.uwaterloo.cawrote in message

In article <eq***********@pc-news.cogsci.eds..ac.uk>,
Richard Tobin <ri*****@cogsci.ed.ac.ukwrote:

>>In article <eL******************************@bt.com>,
Malcolm McLean <re*******@btinternet.comwrote:

>>>Can't say this bit about e^(i*x) = cos x + i * sin x is too convincing
though.

Consider the polynomial expansion e^x = 1 + x + x^2/2! + x^3/3! + ...

I think you mean "power series expansion". It's only a polynomial if
it's finite.
dave
(knowing the power series expansions for sin and cos would also be
helpful)

You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

[Malcolm McLean]

"Lane Straatman" <in*****@invalid.netwrote in message

>The bigger question is why Malcolm doesn't use a free development
system like MinGW. Combined with the Platform SDK, you can do nearly
everything with it, that you'd do with MSVC++.

I've had a very good experience this last week with Devcpp. Up and
running immediately. I'll post the above link in an ng with MS in the
name and see what comes. LS

I bought Visual Studio when I was still a games programmer. It is handy to
have the same compiler you use at work.
Now I've started a PhD in computer modelling of peptides, and I use a Linux
system for scientific work. However I kept the Visual Studio for doing
little hobby programs, like BASICdraw. I do want people to run my stuff, but
the virus problem means that it is hard to distribute executables. I might
be better off on Linux - sometimes it is better to be a little fish in a
little pond than a little fish in a big one.

So I did actually have a choice, for the first time in many years. I just
assumed that everything would work the same on the new Windows. Now I am not
sure that redm,ond won't remotely delete a dual boot.

http://www.personal.leeds.ac.uk/~bgy1mm

[Keith Thompson]

"Malcolm McLean" <re*******@btinternet.comwrites:

"Dave Vandervies" <dj******@caffeine.csclub.uwaterloo.cawrote in message

>In article <eq***********@pc-news.cogsci.eds..ac.uk>,
Richard Tobin <ri*****@cogsci.ed.ac.ukwrote:

>>>In article <eL******************************@bt.com>,
Malcolm McLean <re*******@btinternet.comwrote:

Can't say this bit about e^(i*x) = cos x + i * sin x is too convincing
though.

Consider the polynomial expansion e^x = 1 + x + x^2/2! + x^3/3! + ...

I think you mean "power series expansion". It's only a polynomial if
it's finite.
dave
(knowing the power series expansions for sin and cos would also be
helpful)

You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

Because when the lengths of the sides of your right triangle and/or
the angles of its vertices are complex numbers, geometric intuition
breaks down. (At least mine does; if you're able to visualize such
things, I'm impressed.)

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[CBFalconer]

Lane Straatman wrote:

>

.... snip ...

>
It looks slick as all get out. Looks don't count for much though.
I can't find a run button or pull-down and don't want to have to
step outside the IDE to run the executable:

On Windoze, ALT-tab can bring you to the terminal window (dosbox),
on which you compile and run the application just edited. Easier
and quicker than fooling with a brain-dead IDE. You can apply the
same technique, unaltered, to Linux etc.

--
<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>

"A man who is right every time is not likely to do very much."
-- Francis Crick, co-discover of DNA
"There is nothing more amazing than stupidity in action."
-- Thomas Matthews

[robertwessel2]

On Feb 7, 11:06 am, "Lane Straatman" <[e-mail]:

"Walter Roberson" <[e-mail] in message

news:eq**********@canopus.cc.umanitoba.ca...In article <[e-mail]>,

Lane Straatman <[e-mail]:

>How many decimal places can we squeeze from a long double? LS

C99 defines minimums for that, but not maximums -- an implementation
could in theory have a long double that was a kilobyte long (or more.)

Right, what is the minimum maximum? LS

The required minimums for long doubles are exactly the same as for
doubles. The minimums for both DBL_DIG and LDBL_DIG are both 10.

In practice, it's often considerably larger. For example, IEEE
doubles will get you 15 decimal digits, 80-bit Intel/x86 style
extended reals (assuming your compiler supports them) will get you 18
digits. Other implementations of extended reals exists - the double-
double format mentioned by Walter Robinson is fairly common in
software implementations, and gets you about 31 decimal digits, the
128 bit extended real on zSeries has a 112 bit mantissa and nets about
33 decimal digits. There is not really a standard for IEEE extended
doubles, but the standard does specify some minimums, which is
approximately what the 80-bit x86 format implements.

[Malcolm McLean]

"Keith Thompson" <ks***@mib.orgwrote in message

>You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

Because when the lengths of the sides of your right triangle and/or
the angles of its vertices are complex numbers, geometric intuition
breaks down. (At least mine does; if you're able to visualize such
things, I'm impressed.)

Say we've got three points in three dimensions. Put a line between a, b, and
c. It is just a normal angle. Three points always lie in a plane.
So if our vertices are complex, that makes 4 dimensions, x, xi, y, yi for
each point. But we still have three points.
Do they still lie in a plane, or is the angle between them complex?

[Keith Thompson]

"Malcolm McLean" <re*******@btinternet.comwrites:

"Keith Thompson" <ks***@mib.orgwrote in message

>>You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

Because when the lengths of the sides of your right triangle and/or
the angles of its vertices are complex numbers, geometric intuition
breaks down. (At least mine does; if you're able to visualize such
things, I'm impressed.)

Say we've got three points in three dimensions. Put a line between a, b, and
c. It is just a normal angle. Three points always lie in a plane.
So if our vertices are complex, that makes 4 dimensions, x, xi, y, yi for
each point. But we still have three points.
Do they still lie in a plane, or is the angle between them complex?

I think that when you're dealing with trig functions applied to
complex numbers, you're no longer dealing with triangles. A triangle
in any number of dimensions lies in a plane. I'm not sure there's a
geometric interpretation for a triangle with complex sides and/or
angles.

Something like (sin x)^2 + (cos x)^2 = 1 makes sense geometically.
e^(i*x) = cos x + i * sin x, as far as I know, doesn't -- but it does
make sense algebraically. There are infinite series for each of the
exp(), sin(), and cos() functions, derived using their properties for
real numbers. Applying the same series to complex numbers gives you
mathematically useful results -- including cool whacko stuff like
e^(i*pi) + 1 = 0.

I imagine that quaternions would yield some interesting results -- but
standard C doesn't support them, so that's off-topic.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Dik T. Winter]

In article <ln************@nuthaus.mib.orgKeith Thompson <ks***@mib.orgwrites:

"Malcolm McLean" <re*******@btinternet.comwrites:

"Keith Thompson" <ks***@mib.orgwrote in message

>You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

Because when the lengths of the sides of your right triangle and/or
the angles of its vertices are complex numbers, geometric intuition
breaks down. (At least mine does; if you're able to visualize such
things, I'm impressed.)

Say we've got three points in three dimensions. Put a line between a, b,
and c. It is just a normal angle. Three points always lie in a plane.
So if our vertices are complex, that makes 4 dimensions, x, xi, y, yi for
each point. But we still have three points.
Do they still lie in a plane, or is the angle between them complex?

I think that when you're dealing with trig functions applied to
complex numbers, you're no longer dealing with triangles.

Indeed. cos(i.ln(10)) = 5.2.

Something like (sin x)^2 + (cos x)^2 = 1 makes sense geometically.
e^(i*x) = cos x + i * sin x, as far as I know, doesn't -- but it does
make sense algebraically.

Also geometrically. Take the complex plane with real and imaginary
axis. Draw a circle with radius 1 around the origin. exp(i * x)
circles around this circle when x increases.

I imagine that quaternions would yield some interesting results

Not really.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, [phone]
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

[Richard Tobin]

In article <JD********@cwi.nl>, Dik T. Winter <Di********@cwi.nlwrote:

Something like (sin x)^2 + (cos x)^2 = 1 makes sense geometically.
e^(i*x) = cos x + i * sin x, as far as I know, doesn't -- but it does
make sense algebraically.

>Also geometrically. Take the complex plane with real and imaginary
axis. Draw a circle with radius 1 around the origin. exp(i * x)
circles around this circle when x increases.

It certainly *corresponds* to something geometrical - cos x and sin x
are the coordinates of exp(ix), but to *make sense* geometrically
you'd need an intuition as to why exp(ix) should be a circle for real x.

-- Richard

--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.

[Yevgen Muntyan]

Dave Vandervies wrote:

In article <ln************@nuthaus.mib.org>,
Keith Thompson <ks***@mib.orgwrote:

[Snip everything C-related]

>Apparently i^i (where i is the imaginary square root of -1 and "^"
denotes exponentiation) is a real number. I'm sure there's a simple
mathematical proof of this, but I'm too lazy to track it down or
reconstruct it.

Note that e^(i*pi/2) = i (this follows from the theorem that says
e^(i*x)=cos x + i*sin x). This tells us that ln(i) = i*pi/2.

So:
i = e^(ln i)
i^i = (e^(ln i))^i
= e^(ln i * i)

Try this one:

e^(i*5*pi/2) = i =>
i^i = (e^(5*i*pi/2))^i = e^(i*5*i*pi/2) = e^(-5*pi/2)

Yevgen

[Dik T. Winter]

In article <eq**********@pc-news.cogsci.ed.ac.ukri*****@cogsci.ed.ac.uk (Richard Tobin) writes:

In article <JD********@cwi.nl>, Dik T. Winter <Di********@cwi.nlwrote:

Something like (sin x)^2 + (cos x)^2 = 1 makes sense geometically.
e^(i*x) = cos x + i * sin x, as far as I know, doesn't -- but it does
make sense algebraically.

Also geometrically. Take the complex plane with real and imaginary
axis. Draw a circle with radius 1 around the origin. exp(i * x)
circles around this circle when x increases.

It certainly *corresponds* to something geometrical - cos x and sin x
are the coordinates of exp(ix), but to *make sense* geometrically
you'd need an intuition as to why exp(ix) should be a circle for real x.

In mathematics it is easy to be lead astray by intuition.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, [phone]
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

[Dik T. Winter]

In article <ADGyh.5401$384.1094@trnddc05Yevgen Muntyan <mu****************@tamu.eduwrites:

Dave Vandervies wrote:

....

Note that e^(i*pi/2) = i (this follows from the theorem that says
e^(i*x)=cos x + i*sin x). This tells us that ln(i) = i*pi/2.

This is the main branch of ln.

e^(i*5*pi/2) = i =>
i^i = (e^(5*i*pi/2))^i = e^(i*5*i*pi/2) = e^(-5*pi/2)

In the complex numbers it is not necessarily true that (a^b)^c = a^(b*c):
-i = i^3 = ((-1)^(1/2))^3 = (-1)^(3/2) = ((-1)^3)^(1/2) = (-1)^(1/2) = i ?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, [phone]
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

[Yevgen Muntyan]

Dik T. Winter wrote:

In article <ADGyh.5401$384.1094@trnddc05Yevgen Muntyan <mu****************@tamu.eduwrites:

Dave Vandervies wrote:

...

Note that e^(i*pi/2) = i (this follows from the theorem that says
e^(i*x)=cos x + i*sin x). This tells us that ln(i) = i*pi/2.

This is the main branch of ln.

Yep. Note how I didn't use ln() in my "proof".

e^(i*5*pi/2) = i =>
i^i = (e^(5*i*pi/2))^i = e^(i*5*i*pi/2) = e^(-5*pi/2)

And note how I did *exactly* same thing as in the original proof.
While "ln" (with small "l") is well-defined, something^something
isn't. You got to tell what it means (like man page does).
Anyway, point really was: "see how off-topic stuff everybody likes can
be so easily wrong?" ;)
(I hope nobody will get a bright idea to cross-post to nonsense.math
or something).

Yevgen

[=?utf-8?B?SGFyYWxkIHZhbiBExLNr?=]

Dik T. Winter wrote:

In article <ADGyh.5401$384.1094@trnddc05Yevgen Muntyan <mu****************@tamu.eduwrites:

Dave Vandervies wrote:

...

Note that e^(i*pi/2) = i (this follows from the theorem that says
e^(i*x)=cos x + i*sin x). This tells us that ln(i) = i*pi/2.

This is the main branch of ln.

e^(i*5*pi/2) = i =>
i^i = (e^(5*i*pi/2))^i = e^(i*5*i*pi/2) = e^(-5*pi/2)

In the complex numbers it is not necessarily true that (a^b)^c = a^(b*c):
-i = i^3 = ((-1)^(1/2))^3 = (-1)^(3/2) = ((-1)^3)^(1/2) = (-1)^(1/2) = i ?

That's not specific to the complex numbers, is it? ((-1)^2)^(1/2) !=
(-1)^(2*(1/2)).

[Tak-Shing Chan]

On Thu, 8 Feb 2007, Richard Tobin wrote:

In article <JD********@cwi.nl>, Dik T. Winter <Di********@cwi.nlwrote:

>>Something like (sin x)^2 + (cos x)^2 = 1 makes sense geometically.
e^(i*x) = cos x + i * sin x, as far as I know, doesn't -- but it does
make sense algebraically.

>Also geometrically. Take the complex plane with real and imaginary
axis. Draw a circle with radius 1 around the origin. exp(i * x)
circles around this circle when x increases.

It certainly *corresponds* to something geometrical - cos x and sin x
are the coordinates of exp(ix), but to *make sense* geometrically
you'd need an intuition as to why exp(ix) should be a circle for real x.

-- Richard

Do matrices look geometric enough for you?

[ 1 0] [ 0 1]
Define I = [ ] and J = [ ]. It can be proven that
[ 0 1] [-1 0]

aI + bJ is isomorphic to the complex number a + bi. It can
further be proven that

[ 0 x] [ cos x sin x]
exp [ ] = [ ]
[-x 0] [-sin x cos x]

which corresponds to exp (ix) = cos x + i sin x.

Tak-Shing

[¬a\\/b]

On 8 Feb 2007 08:06:24 -0800, "Harald van D?k" <tr*****@gmail.com>
wrote:

>Dik T. Winter wrote:

>In article <ADGyh.5401$384.1094@trnddc05Yevgen Muntyan <mu****************@tamu.eduwrites:

> Dave Vandervies wrote:

...

> Note that e^(i*pi/2) = i (this follows from the theorem that says
e^(i*x)=cos x + i*sin x). This tells us that ln(i) = i*pi/2.

don't know if it is right:
i=0+i*1=cos(pi/2)+i*sin(pi/2)=e^(i*pi/2) it seems ok

i^i=(e^(i*pi/2))^i=e^(-pi/2)=0,20787957635076190854695561
i have seen that number in this thread...

|#include <stdio.h>
|#include <complex.h>
|int main(void)
|{
| const long double complex z1 = cpowl(I, I);
| printf("%Lf %Lf\n", creall(z1), cimagl(z1));
| return 0;
|}

|and its output:

|0.207880 0.000000

>This is the main branch of ln.

> e^(i*5*pi/2) = i =>

are you sure of that?
if the answer is yes i^i has more than 1 result

> i^i = (e^(5*i*pi/2))^i = e^(i*5*i*pi/2) = e^(-5*pi/2)

In the complex numbers it is not necessarily true that (a^b)^c = a^(b*c):
-i = i^3 = ((-1)^(1/2))^3 = (-1)^(3/2) = ((-1)^3)^(1/2) = (-1)^(1/2) = i ?

That's not specific to the complex numbers, is it? ((-1)^2)^(1/2) !=
(-1)^(2*(1/2)).

[¬a\\/b]

On Fri, 09 Feb 2007 08:46:05 +0100, "¬a\\/b" <[e-mail]:

>On 8 Feb 2007 08:06:24 -0800, "Harald van D?k" <tr*****@gmail.com>
wrote:

>>Dik T. Winter wrote:

>>In article <ADGyh.5401$384.1094@trnddc05Yevgen Muntyan <mu****************@tamu.eduwrites:
Dave Vandervies wrote:
...
Note that e^(i*pi/2) = i (this follows from the theorem that says
e^(i*x)=cos x + i*sin x). This tells us that ln(i) = i*pi/2.

my book says if a, beR
a+i*b=r(cos(t)+i*sin(t))=r*e^(i*t) (r,t)eRx[0,2*pi)

if i have a+i*b than find
r=sqrt(a^2+b^2)
if r=0 than a+i*b=0*e^0
else resolve
cos(t)=a/r sin(t)=b/r te[0, 2*pi)
(note 1=a^2/r^2+b^2/r^2=cos(t)^2+sin(t)^2)
and write a+i*b=r*e^(i*t)

if i have r*e^(i*t)=r(cos(t)+i*sin(t))=
r*cos(t)+i*r*sin(t)=a+i*b

>don't know if it is right:
i=0+i*1=cos(pi/2)+i*sin(pi/2)=e^(i*pi/2) it seems ok
i^i=(e^(i*pi/2))^i=e^(-pi/2)=0,20787957635076190854695561

it seems right
"e^(i*5*pi/2) = i" seems to me wrong because
not(5*pi/2e[0,2pi))

[¬a\\/b]

On Fri, 09 Feb 2007 10:48:56 +0100, "¬a\\/b" <[e-mail]:

>On Fri, 09 Feb 2007 08:46:05 +0100, "¬a\\/b" <[e-mail]:

>>On 8 Feb 2007 08:06:24 -0800, "Harald van D?k" <tr*****@gmail.com>
wrote:

>>>Dik T. Winter wrote:
In article <ADGyh.5401$384.1094@trnddc05Yevgen Muntyan <mu****************@tamu.eduwrites:
Dave Vandervies wrote:
...
Note that e^(i*pi/2) = i (this follows from the theorem that says
e^(i*x)=cos x + i*sin x). This tells us that ln(i) = i*pi/2.

my book says if a, beR

my book don't say really this but seems right in this way

>a+i*b=r(cos(t)+i*sin(t))=r*e^(i*t) (r,t)eRx[0,2*pi)

a+i*b=r(cos(t)+i*sin(t))=r*e^(i*t) (r,t)eR^(+)x(-pi,+pi]

>if i have a+i*b than find
r=sqrt(a^2+b^2)
if r=0 than a+i*b=0*e^0
else resolve
cos(t)=a/r sin(t)=b/r te[0, 2*pi)

cos(t)=a/r sin(t)=b/r for te(-pi,+pi]

>(note 1=a^2/r^2+b^2/r^2=cos(t)^2+sin(t)^2)
and write a+i*b=r*e^(i*t)

if i have r*e^(i*t)=r(cos(t)+i*sin(t))=
r*cos(t)+i*r*sin(t)=a+i*b

>>don't know if it is right:
i=0+i*1=cos(pi/2)+i*sin(pi/2)=e^(i*pi/2) it seems ok
i^i=(e^(i*pi/2))^i=e^(-pi/2)=0,20787957635076190854695561

>it seems right
"e^(i*5*pi/2) = i" seems to me wrong because
not(5*pi/2e[0,2pi))

not(5*pi/2e(-pi,pi])

[David Wade]

"Malcolm McLean" <re*******@btinternet.comwrote in message
news:H9*********************@bt.com...

>
"Dave Vandervies" <dj******@caffeine.csclub.uwaterloo.cawrote in message

In article <eq***********@pc-news.cogsci.eds..ac.uk>,
Richard Tobin <ri*****@cogsci.ed.ac.ukwrote:

>In article <eL******************************@bt.com>,
Malcolm McLean <re*******@btinternet.comwrote:

Can't say this bit about e^(i*x) = cos x + i * sin x is too convincing
though.

Consider the polynomial expansion e^x = 1 + x + x^2/2! + x^3/3! + ...

I think you mean "power series expansion". It's only a polynomial if
it's finite.
dave
(knowing the power series expansions for sin and cos would also be
helpful)

You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

For the same reason that the formulas for the roots of quadratic equations
are not self evident. In general, to mathematicians, things only get
interesting when we leave the tangible and touchable behind, and move to a
higher level of abstraction. Trying to relate these back to physical models
in these cases is often unproductive. What you are asking is almost the same
as asking what specific numbers are "x" and "y" in an formula such as
y=x+4...

[Dik T. Winter]

In article <bo******************************@eclipse.net.uk"D avid Wade" <g8***@yahoo.comwrites:

"Malcolm McLean" <re*******@btinternet.comwrote in message
news:H9*********************@bt.com...

"Dave Vandervies" <dj******@caffeine.csclub.uwaterloo.cawrote in message

....

(knowing the power series expansions for sin and cos would also be
helpful)
>

You've got to know them? Why isn't something as simple as opposite /
hypotenuse self evident ?

For the same reason that the formulas for the roots of quadratic equations
are not self evident.

No. It is much more basic. How would you define a trangle with one right
angle and two complex angles? The question being: "what *is* a complex
angle?"
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, [phone]
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

[Malcolm McLean]

"Dik T. Winter" <Di********@cwi.nlwrote in message

No. It is much more basic. How would you define a trangle with one right
angle and two complex angles? The question being: "what *is* a complex
angle?"

Let's imagine a triangle with complex instead of real coordinates.

So instead of x1, y1, x2, y2, x3, y3 we have xr1, xi1, yr1, yi1, xr2, xi2,
yr2, yi2, xr3, xi3, yr3, yi3.

You agree that the sine of such a triangle could be complex, if only we can
decide which is the right angle, because we are dividing one complex number
by another.

To reduce thr strain on our brain, lets declare point one to be at the
origin.

Now we have

0, 0, 0, 0, xr2, xi2, yr2, yi2, xr3, xi3, yr3, yi3.

To reduce the strain further, we can say that point two always alies along
the real x axis.

0, 0, 0, 0, xr2, 0, 0, 0, xr3, xi3, yr3, yi3.

Once we've done that we can rotate about the x axis until one of the other
points goes to zero. Let's choose to lose yi3.

Now we've got

0, 0, 0, 0, xr2, 0, 0, 0, xr3, xi3, yr3, 0.

Now things are really looking up. We've only got one imaginary component
complicating things.

However can can play another trick. Three points in four dimensions are
exactly the same as four points in three dimensions.

So now we;ve got

0, xr2, xr3 - xreal
0, 0, xi3 - x imaginary
0, 0, yr3 - y real
0, 0, 0 - y imaginary.

very nicely, we've got a point at the origin. And four points can be defined
by two lengths, two angles, and one torsion angle - assuming we ignore
translation and rotation. However when we look at our zeroes, we can knock
out the x column, and the two middle values are in a straight line. So we
are actually left with a regular triangle, and we can see if it is
right-angled. Unfortunately we actually have three such "triangles", because
we've introduced more possible connections. Only one of them has no BC edge
and thus isn't a triangle.

[Dik T. Winter]

In article <KM*********************@bt.com"Malcolm McLean" <re*******@btinternet.comwrites:

"Dik T. Winter" <Di********@cwi.nlwrote in message

No. It is much more basic. How would you define a trangle with one right
angle and two complex angles? The question being: "what *is* a complex
angle?"

Let's imagine a triangle with complex instead of real coordinates.

I ask what a complex angle is. Or more basic: I want to compute the
sine of 2.i. You would be a lot more convincing if you could construct
a triangle with angles pi/2, 2.i and ( pi/2 - 2.i). In that case we may
use the definition of quotients of sides. We will not, of course, because
the modulus of such a quotient will alway be less than or equal to 1, but
sin(2.i) = 3.626860408 .i.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, [phone]
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/