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

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  • #1
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Homework Statement



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]


Homework Equations



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


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.
 

Answers and Replies

  • #2
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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
 
  • #3
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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.
 
  • #4
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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
 
  • #5
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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
 
  • #6
HallsofIvy
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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
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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