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 The Powers that Be (Posted on 2003-11-25)
Is it possible for two complex numbers to have a real exponentiation?

In other words, if x and y are complex (each have the form a+bi), show that x^y can have a real value, or prove that it is impossible.

Note: i is the imaginary value defined as the number that yields -1 when squared. a and b are any real numbers, but b is not 0.

 See The Solution Submitted by DJ Rating: 4.4444 (9 votes)

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 Further elaboration | Comment 2 of 19 |
Some bits of notation. Let R\P be polar notation for R(cos(P)+i.sen(P)). Let ^ denote exponentiation.

By definition, X^Y = e^(Y.ln(X)). If the imaginary part of Y.ln(X) is zero, the result is real.

Let X=R\P; then ln(X)=ln(R)+i.P. Also, let Y=R'\P' = R'(cos(P')+i.sen(P')). The imaginary part of Y.ln(X) is then R'(ln(R).sen(P')+P.cos(P'))=0.

If we don't allow R'=0, we could take
P=-1/cos(P'), ln(R)=1/sen(P'), and we would have the sought result.

Did I miss anything? Opinions?
 Posted by Federico Kereki on 2003-11-25 13:23:55

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