What is negative i raised to the i power where i is the imaginary unit equal to square root of -1?

Answer 1

-√(-1)√(-1) can be shortened to -ii, where i = √(-1).

The answer to this question is not trivial. To answer it, we must invoke Euler's formula, which is eix = cos(x) + isin(x).

Substituting in π/2 for x gives us eiπ/2 = cos(π/2) + isin(π/2).

Well, the cos(π/2) is 0, and the sin(π/2) is 1, so the above becomes:

eiπ/2 = i.

So, now we raise both sides to the power of i:

ei*iπ/2 = ii.

Using the basic identity for i:

i*i = i2 = -1.

So, now we have e-π/2 = ii, or -e-π/2 = -ii.

Well, -e-π/2 is a real number, and its value is -0.20788.

For the similar case of (-i)i, we substitute -π/2 into Euler's formula.

This gives us e-iπ/2 = cos(-π/2) + isin(-π/2) = 0 - i = -i.

Once again, raising both sides to the power of i and using the same identity, we get eπ/2 = (-i)i.

Again, the result is a real number, eπ/2, whose value is 4.81047.

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NOTE.

I'd like to add that (-i)ior ii are not (well-)defined, so there is no correct/definite answer to this question.

Indeed, Euler's formula is valid, but the formula is inclomplete; in fact it is

ei.(x+2.k.π) = cos(x) + i.sin(x) for any integer k.

Therefore,

ei.π.(2.k+1/2) = i for any integer k,

and so

e-π.(2.k+1/2) = ii for any integer k,

proving that it is not well-defined.

Other possible values for ii are e-π.(-2+1/2) =eπ.(3/2) = 111.31777... and

e-π.(2+1/2) =e-π.(5/2) = 3.88203203... x 10-4. The reason for this is, that in order to compute ii we like to write ii=ei.log(i), but to compute log(z) for a complex number z, one has to decide on what branch of log(.), the inverse of the periodic exp(.), z is located. Note that this aspect was ignored in the previous answer. Therefore, there is no correct answer for this question: any branche of log(.) can be chosen to produce an answer that may suit the context. The question does not give this context.