Thanks Michael & Peter.

Michael's expansion makes sense.

This is what I expected:

 > a = pi + 0i
 > complex( real = cos(Re(a)), imaginary = sin(Im(a)) )
[1] -1+0i

Not this:
 > exp(a)
[1] 23.14069+0i

Is this not an implementation of Euler's formula:
 > complex( real = cos(2*pi), imaginary = sin(2*pi) )
[1] 1-0i

And that is a result Michael depends on in his
expansion, yet if we pass this argument to exp:
 > exp( (complex( real = 2*pi, imaginary = 2*pi) ) )
[1] 535.4917-0i

That would not work in Michaels expansion, the answer must
be 1 + 0i.

Which seems to suggest that exp( ix ) and cos x + i sin x (as
written above) are different interpretations.


On 01/30/2012 12:47 PM, Peter Langfelder wrote:
> Not sure why you think the formula does not hold... but am guessing
> you think that sin(x) and cos(x) are have values in [-1, 1]? Well that
> only holds for real x. If you have a complex x, sin(x) and cos(x) are
> unbounded - indeed, if you can write x=iy and y is real, you can show
> (up to my own ignorance of possible signs) cos(x) = cosh(y), and
> sin(x) = -sinh(y) simply by expressing (from the formula you wrote)
> cos(x) and sin(x) as
>
> cos(x) = ( exp(ix) + exp(-ix) )/2
> and sin(x) = ( exp(ix) - exp(-ix) )/2
>
> In any case, plug any complex number into
> exp( ix )
> and
> cos x + i sin x
>
> in R and you will get the exact same answers.
>
> HTH,
>
> Peter
>
> On Mon, Jan 30, 2012 at 7:37 AM, Joseph Park<josephpark@ieee.org>  wrote:
>> Hi,
>>
>> Am i doing something silly here in expecting Euler's
>> formula to be handled by exp? exp( ix ) = cos x + i sin x.
>> The first example below follows this, the others not.
>>
>> Thanks for the education!
>>
>>   >  exp( complex(real = 0, imag = 2*pi) )
>> [1] 1-0i
>>   >  exp( complex(real = pi, imag = 2*pi) )
>> [1] 23.14069-0i
>>   >  exp( complex(real = pi/2, imag = 0) )
>> [1] 4.810477+0i
>>
>>
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>>
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