Complex Conjugates/Proof by Induction

  • Context: MHB 
  • Thread starter Thread starter gucci1
  • Start date Start date
  • Tags Tags
    Complex Induction
Click For Summary
SUMMARY

The discussion focuses on proving by induction that the conjugate of the product of complex numbers \( g_1, g_2, \ldots, g_m \) equals the product of their conjugates. The proof is established for \( m = 2 \) and then extended to \( m = n + 1 \) using the inductive hypothesis. The key steps involve showing that if the statement holds for \( m = n \), it also holds for \( m = n + 1 \) by defining \( z = g_1g_2\cdots g_n \) and applying the properties of complex conjugates.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with mathematical induction
  • Knowledge of complex conjugates
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of complex conjugates in detail
  • Learn more about mathematical induction techniques
  • Explore examples of proofs involving complex numbers
  • Practice problems on proving statements by induction
USEFUL FOR

Students studying complex analysis, mathematicians interested in proof techniques, and anyone looking to strengthen their understanding of mathematical induction and complex number properties.

gucci1
Messages
12
Reaction score
0
So I am having a bit of trouble with a proof by induction that I need to write. The problem is to prove that the conjugate of the product g1 * ... * gm equals the product of the conjugates of g1 ... gm. This is for g1 ... gm complex numbers.

I have proven this for m = 2, by simple calculation of the conjugates. Now I need to prove that if this is true for m = n then it is true for m = n + 1.

I appreciate any help you can offer, thank you.
 
Physics news on Phys.org
gucci said:
So I am having a bit of trouble with a proof by induction that I need to write. The problem is to prove that the conjugate of the product g1 * ... * gm equals the product of the conjugates of g1 ... gm. This is for g1 ... gm complex numbers.

I have proven this for m = 2, by simple calculation of the conjugates. Now I need to prove that if this is true for m = n then it is true for m = n + 1.

I appreciate any help you can offer, thank you.

If you showed that $\overline{g_1g_2}=\overline{g_1}\cdot \overline{g_2}$, and then assume that $\overline{g_1g_2\cdots g_n} = \overline{g_1}\cdot\overline{g_2}\cdots \overline{g_n}$, then it follows that
\[\overline{g_1g_2\cdots g_ng_{n+1}} = \overline{(g_1g_2\cdots g_n) g_{n+1}}\]
Let's define $g_1g_2\cdots g_n=z$. Then we're left with $\overline{z g_{n+1}}$, and we know from the $m=2$ case that this is the same as $\overline{z}\cdot\overline{g_{n+1}}$. Now rewrite this as $\overline{g_1g_2\cdots g_n}\cdot \overline{g_{n+1}}$ and apply the inductive hypothesis to finish the problem.

Does this make sense?
 
Thanks so much man, that really cleared it up for me!
 

Similar threads

Replies
8
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K