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Complex Variable Basic Proof by Induction (I'm lost.)

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data

    Use mathematical induction to show that when n = 2,3,...,

    [tex]\overline{z_1+z_2+\cdots +z_n} = \bar{z_1} + \bar{z_2} + \cdots + \bar{z_n}[/tex]


    2. Relevant equations

    [tex]\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2}[/tex]


    3. The attempt at a solution

    So help me get started on the first step because I am new to induction. The only proofs I've done/seen (not many) were proved for each [tex]n \in N[/tex]. With that said how do I even start by showing that P(1) holds because the n = 2, 3, ... is throwing me off because 1 is not in there. Does it mean show for all numbers starting from 2? So should I be proving P(2) holds in the first step? Thank you.
     
  2. jcsd
  3. Aug 31, 2009 #2
    Yes, Prove that it is true for n=2 as a first step, then assume for n and then using that prove that it is also true for n+1
     
  4. Aug 31, 2009 #3
    So to show P(2) holds:

    [tex]P(2) = \bar{2} = \bar{2}[/tex]

    this seems goofy. So should it be this way instead?

    LHS = [tex]\overline{2+3} = \bar{5}[/tex]

    RHS = [tex]\bar{2}+\bar{3} = \bar{5}[/tex]

    Then [tex]\bar{5} = \bar{5}[/tex]. So P(2) is true.
     
  5. Aug 31, 2009 #4
    In this case the inductive variable n is refering to the indeces of the terms, not the values of the terms. So P(2) is the statement

    [tex]\overline{(z_1+z_2)} = \overline{z_1} + \overline{z_2}[/tex]

    while P(3) is

    [tex]\overline{(z_1+z_2+z_3)}=\overline{z_1} + \overline{z_2} + \overline{z_3}.[/tex]

    (Side comment P(1) is trivially true since z-bar = z-bar is an identity and not useful here.)

    As n increases, the number of terms increases. Assuming it works for n terms, you need to show that the complex conjugate of the sum of n + 1 terms is the sum of the individual conjugates.

    --Elucidus
     
  6. Aug 31, 2009 #5
    Thank you Elucidus for clearing up most of my confusion on P(2). For the induction step 2 I think I figured it out but I'm not sure. Here's what I did let me know what you think:

    Assume [tex]p(n)[/tex] is true for n = 2, 3, ...,

    Show that [tex]p(n+1)[/tex] is true

    LHS = [tex]\overline{z_1 + z_2 + .... + z_n + z_{n+1}} = \overline{(x_n + iy_n) + (x_{n+1} + iy_{n+1})} = (x_n-iy_n) + (x_{n+1} - iy_{n+1}) = \overline{z_n} + \overline{z_{n+1}}[/tex] = RHS
     
  7. Sep 1, 2009 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, don't switch to "x+ iy" now. You know (induction hypothesis) that [itex]\overline{z_1+ z_2+ \cdot\cdot\cdot+ z_n}= \overline{z_1}+ \overline{z_2}+ \cdot\cdot\cdot+ \overline{z_n}[/itex] and now you want to extend that to [itex]z_{n+1}[/itex]. Okay, think of [itex]z_1+ z_2+ \cdot\cdot\cdot+ z_n+ z_{n+1}[/itex] as the sum of two numbers: [itex](z_1+ z_2+ \cdot\cdot\cdot+ z_n)[/itex] and [itex]z_{n+1}[/itex].

    Since you know that for two numbers, [itex]\overline{z_1+ z_2}= \overline{z_1}+ \overline{z_2}[/itex], you know that [itex]\overline{z_1+ z_2+ \cdot\cdot\cdot+ z_n+ z_{n+1}}= \overline{z_1+ z_2+ \cdot\cdot\cdot+ z_n}+ \overline{z_{n+1}}[/itex].
     
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