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:

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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