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Find all the possible values of 2^(-i)?? Any help with it?
I can't figure it out. I know i should use the complex logarithmic function, but i can't work it out. :S
1 Answer
- BrianLv 78 years agoFavorite Answer
Note first that 2 = e^(ln(2)).
So 2^(-i) = (e^(ln(2))^(-i) = e^(-i*ln(2)) = cos(ln(2)) - i*sin(ln(2)),
which equals 0.76924 - i*(0.63896) to 5 decimal places.
Edit: Note that e^(ix) = cos(x) + i*sin(x), so
e^(-i*ln(2)) = cos(-ln(2)) + i*sin(-ln(2)) = cos(ln(2)) - i*sin(ln(2)).
Edit #2: Just saw the "all" in the instructions. These will be
2^(-i) = cos(ln(2) + k*2pi) - i*sin(ln(2) + k*2pi) for any integer k.
Sorry for the initial oversight. :)
