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Math Help - [SOLVED] inductive proof

  1. #1
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    [SOLVED] inductive proof

    for n in the natural numbers prove that (2+i)^n is never real.

    base case
    (2+i) is not real

    inductive hypothesis
    for k=n assume (2+i)^k is not in the reals
    must prove
    (2+i)^(k+1)

    Ive got to the point where
    2(2+i)^k + i(2+i)^k must not be real. its seems logical that the i terms would not cancel out making this sum not real but i cant figure out a way to prove it.
    I've tried using the binomial theorem. Any HINTS would be greatly appreciated.
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  2. #2
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    If (2+i)^k is not real then you can write it as a + i \cdot b for a, b \in \mathbb{R}. Now, what is (2+i)(2+i)^k ?
    Last edited by Defunkt; May 9th 2010 at 01:20 PM. Reason: Context
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  3. #3
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    (2+i)((2+i)^k) = (2+i)^(k+1)
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  4. #4
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    Correct. Now use my hint to prove the inductive step.
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  5. #5
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    interesting so following your reasoning
    (2+i)(2+i)^k = (a+ib)(2+i)
    so the expansion would be 2a-b + i(2b+a)
    its clear that this cannot be zero unless a and b are zero because for the sum to be zero b=2a and b= -a/2. But how do we prove that b=-a/2, making the imaginary term zero we know nothing about a and b besides that they are reals
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