Nothing 
My TI says that (-1)^(3/2) = -i .
So does (1/2-sqrt(3)/2).
For (1/2+sqrt(3)/2) ^(3/2) it says +i because it is doing
(1/2+sqrt(3)/2)^3 == -1
and then taking the square root and only considering one of the 2 values
See:
(-i)^(2/3) = exp(i*(3pi/2 - 2Pi) * 2/3)
= exp(i*(pi -4pi/3))
= exp(-i*pi/3)
= 1/2 - sqrt(3)/2
(-i)^(2/3) = exp(i*(3pi/2 + 2Pi) * 2/3)
= exp(i*(pi +4pi/3))
= exp(+i*pi/3)
= 1/2 + sqrt(3)/2
1 has in n n-order roots. You must add their arguments to that of your
(only) result.
Consider the following:
(-i)^(2/3) = ((-i) ^ (1/3)) ^ 2
It turns out that
a^(1/3) = |a|^(1/3) * exp( j * 2PI / 3 * n + j * arg(a)/3 ) with n = 0,1,2
In this case
(-i) ^ (1 / 3) = 1 * exp ( j * -PI/6 + j * 2PI/3*n )
( ) ^2 = exp(j (-pi/3))
exp(j)
exp(j (pi/3))
You can also see it graphically:
(-i) ^(2/3)
means the cube (?) root of -1
If I could draw here, I’d point at the 3 cube (is that the terminology
in English?) roots of the “minus unity”
What should a well-behaved Complex library do about multiple roots? It
even happens for 1!
sqrt(1) ===> 1 taken by convention
-1 silently discarded
PS: eqs.gsub!(/j/, “i”)
My EE background haunts me
···
On Thu, Feb 06, 2003 at 07:20:33AM +0900, Daniel Carrera wrote:
Martin,
I can see that your answer is right. Perhaps you can show me what I’m
doing wrong. My colleagues and I can’t figure it out:
-i = exp(i3pi/2)
2/3
(-i) = exp(i * 3pi/2 *(2/3))
= exp(i * pi)
= -1
What am I doing wrong?
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