How do you evaluate #\sqrt { - 328}#?
1 Answer
Jan 25, 2018
Explanation:
Note that
#328 = 2^3 * 41#
Hence we find:
#sqrt(-328) = sqrt(2^2*82)i = 2sqrt(82)i#
is an irrational multiple of
Since
#sqrt(82) = [9; bar(18)] = 9+1/(18+1/(18+1/(18+1/(18+1/(18+...)))))#
So we could write:
#sqrt(-328) = 2i(9+1/(18+1/(18+1/(18+1/(18+1/(18+...))))))#