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# Solve for k. With imaginary number.?

(i+1)^k = -(3^k)

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- GusBsAsLv 67 years agoFavorite Answer
(i+1)^k = -(3^k)

=>

((1+i)/3)^k = -1

=>

ln[((1+i)/3)^k] = ln(-1)

=>

k ln((1+i)/3) = ln(-1)

=>

k [ (1/2) ln(x) - ln(3) + (2n+1/4)πi ] = (2m+1)πi

=>

k = (2m+1)πi / [ (1/2) ln(x) - ln(3) + (2n+1/4)πi ]

m,n ∊ ℤ

If the question were "find k in ℕ such that...", the answer wpuld be "it is not possible", because the absolute value of the left side is 2^(k/2) and the abs. v. of the right side is 3^k, and a power of 3 cannot be equal to a power of 2, except if k=0, but in this case the left side is 1 and the right side is -1.

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