- #1
nomadreid
Gold Member
- 1,672
- 206
I am sure I am overlooking something elementary, but playing around with exponentiation (this is not an assignment), I seem to be making a mistake somewhere. Please don't send me a link for a more compact way of getting the correct result; I wish to know what my particular mistake is.
Suppose r,s∈ℂ, r= exp(a+bi) = ea+bi, and s=c+di, so that
rs =exp((a+bi)(c+di)) = exp((ac-bd)+(ad+bc)i).
(I am not using "r" as length.)
So far, very straightforward, but the problem comes in the interpretation that this result has a length exp(ac-bd) and angle (ad+bc) Radians.
What bothers me about this is that since r = (exp(a +(b+n2π)i) for n∈ℤ, the above result would come out with a length of
exp(ac-(b+n2π)d) = exp(ac-bd- 2ndπ)
and an angle of (ad+(b+n2π)c)=(ad+ bc+2ncπ):
this means that (with a few exceptions) the length will not be the same as the first result, and the angle will not be equivalent to the first result. But I would imagine that rs should have a unique (modulo an angle of 2nπ) result, no?
Suppose r,s∈ℂ, r= exp(a+bi) = ea+bi, and s=c+di, so that
rs =exp((a+bi)(c+di)) = exp((ac-bd)+(ad+bc)i).
(I am not using "r" as length.)
So far, very straightforward, but the problem comes in the interpretation that this result has a length exp(ac-bd) and angle (ad+bc) Radians.
What bothers me about this is that since r = (exp(a +(b+n2π)i) for n∈ℤ, the above result would come out with a length of
exp(ac-(b+n2π)d) = exp(ac-bd- 2ndπ)
and an angle of (ad+(b+n2π)c)=(ad+ bc+2ncπ):
this means that (with a few exceptions) the length will not be the same as the first result, and the angle will not be equivalent to the first result. But I would imagine that rs should have a unique (modulo an angle of 2nπ) result, no?