Solve for k. With imaginary number.?

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

    Answer by GusBsAs
    (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)i ] = (2m+1)i
    =>
    k = (2m+1)i / [ (1/2) ln(x) - ln(3) + (2n+1/4)i ]
    m,n

    If the question were “find k in 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.

  2. How do I do that? I mean, to me its clear that it converges to 1 (k->infinity), but how do I prove it?

    Answer by ecapS trebliH
    1 — k/(1+k) = (1+k)/(1+k) — k/(1+k) = 1/(1+k)

    Therefore | 1 – k/(1+k) | = 1/(1+k)

    Now use the definition of convergence. Given epsilon > 0, let N be such that

    1/(1+N) 1/epsilon.

    Now suppose k > N. Then 1/(1+k) < 1/(1+N) N then | 1 – k/(1+k)| = 1/(1+k) oo) k/(1+k) = 1

  3. (i+2)^k = -1^k

    1 to anything is just 1 so
    (i+2)^k = -1
    k = log of -1 to the base i+2

    Answer by Captain Matticus, LandPiratesInc
    (i + 2)^k = -1^k
    (i + 2)^k / 1^k = -1Random Casale Media Sponsor AD relevant to bEt365 :
    bEt365

    ((i + 2) / 1)^k = -1
    (i + 2)^k = -1
    k * ln(i + 2) = ln(-1)
    k = ln(-1) / ln(i + 2)
    k = log[i + 2](-1)

    Looks like you got it right enough. However, it should be noted that log(-1) only exists for complex values.

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Ivan S. Martin

He that is master of himself, will soon be master of others.