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

[DEMJ]

Homework Statement

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

\overline{z_1+z_2+\cdots +z_n} = \bar{z_1} + \bar{z_2} + \cdots + \bar{z_n}

Homework Equations

\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2}

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 n \in N. 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.

 

[rootX]

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

 

[DEMJ]

So to show P(2) holds:

P(2) = \bar{2} = \bar{2}

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

LHS = \overline{2+3} = \bar{5}

RHS = \bar{2}+\bar{3} = \bar{5}

Then \bar{5} = \bar{5}. So P(2) is true.

 

[Elucidus]

In this case the inductive variable n is referring to the indeces of the terms, not the values of the terms. So P(2) is the statement

\overline{(z_1+z_2)} = \overline{z_1} + \overline{z_2}

while P(3) is

\overline{(z_1+z_2+z_3)}=\overline{z_1} + \overline{z_2} + \overline{z_3}.

(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

 

[DEMJ]

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 p(n) is true for n = 2, 3, ...,

Show that p(n+1) is true

LHS = \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}} = RHS

 

[HallsofIvy]

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 p(n) is true for n = 2, 3, ...,

Show that p(n+1) is true

LHS = \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}} = RHS

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

Since you know that for two numbers, \overline{z_1+ z_2}= \overline{z_1}+ \overline{z_2}, you know that \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}}.