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

## 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.

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

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.

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

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

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

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
Homework Helper
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 $\overline{z_1+ z_2+ \cdot\cdot\cdot+ z_n+ z_{n+1}}= \overline{z_1+ z_2+ \cdot\cdot\cdot+ z_n}+ \overline{z_{n+1}}$.