Again, in interests of precision:
SKB wrote:
If VN had in mind i-the-imaginary-square-root-of-minus-one (it does occur in general, non-technical writing, e.g., Einsteinšs own popular introductions  to Relativity), then the statement is false. Since i-squared is minus 1, išs-square-roots cannot be +-or-minus-i. If I havenšt lost you, getting the  FOUR square-roots-of-(i-the-square-root-of-minus-one) means solving the
quadratic equation x^4 + 1 = 0; using EULERšs MOST beautiful & mysterious formula e^(i*pi) = -1, we obtain  x = +/- 1/(2^(1/2)(cos pi/4 +/- i*sin pi/4).

JF responded:
Off-topic, but we're all in favor of the precision of poetry, so I'll mention that there are only two square roots of i, namely +/- (1/2^(1/2) + 1/2^(1/2) * i), or if you prefer, +/- (cos pi/4 + i*sin pi/4). The other two numbers you had in mind are square roots of -i.
AS now replies:
I noticed the "error" in SKB's assertion, but assumed that he meant that, since it is arbitrary which of "i" and "-i" we actually call "i", there are therefore four, rather than two, numbers which are a square root of a (i.e., some or other, not a given) square root of -1; i.e., as SKB says, there are four fourth roots of -1.
Anthony Stadlen

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