My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
December 5th, 2010, 05:09 AM   #1
Newbie
 
Joined: Oct 2010

Posts: 27
Thanks: 0

Complex equation

I need a help with this.


Prove that if complex roots and of equation have their absolute values equal ( then is a real number.

Thanks for any advices.
Gustav is offline  
 
December 5th, 2010, 07:41 AM   #2
Senior Member
 
Joined: Nov 2010

Posts: 502
Thanks: 0

Re: Complex equation

The quadratic formula works for all quadratics.
DLowry is offline  
December 5th, 2010, 08:07 AM   #3
Global Moderator
 
Joined: Dec 2006

Posts: 20,926
Thanks: 2205

So?
skipjack is offline  
December 5th, 2010, 12:44 PM   #4
Senior Member
 
Joined: Nov 2010

Posts: 502
Thanks: 0

Re:

Quote:
Originally Posted by skipjack
So?
Are you asking me or him?
DLowry is offline  
December 5th, 2010, 09:23 PM   #5
Global Moderator
 
Joined: Dec 2006

Posts: 20,926
Thanks: 2205

You. There is a reasonably short proof using trigonometry, but I don't see any simple method based on the quadratic formula.
skipjack is offline  
December 5th, 2010, 10:11 PM   #6
Newbie
 
Joined: Oct 2010

Posts: 27
Thanks: 0

Re: Complex equation

And could you please show me both way to do it? Especially trigonometry one.

Common quadratic formula works, but from the result I am unable to work out the proof.
Gustav is offline  
December 5th, 2010, 11:19 PM   #7
Global Moderator
 
Joined: Dec 2006

Posts: 20,926
Thanks: 2205

Let where "cis" means "cos + i sin" and and are real, then
p and q
Now it's fairly easy to show that p/q > 0.
skipjack is offline  
December 6th, 2010, 01:29 AM   #8
Newbie
 
Joined: Oct 2010

Posts: 27
Thanks: 0

Re: Complex equation

But I have to prove that p/q is real number, not that p*p/q*q > 0

Or did I understand badly to it?
Gustav is offline  
December 6th, 2010, 05:35 AM   #9
Senior Member
 
Joined: Nov 2010

Posts: 502
Thanks: 0

Re:

Quote:
Originally Posted by skipjack
Let where "cis" means "cos + i sin" and and are real, then
p and q
Now it's fairly easy to show that p/q > 0.
Ah, this is much shorter than the way I had in mind. I yield for your victory.

To Gustav, consider what we now have. If p^2/q^2 = a > 0, then we also have that p^2/q^2 - a = 0. (I chose this form because I think it's easiest to see).
DLowry is offline  
December 6th, 2010, 08:31 AM   #10
Newbie
 
Joined: Oct 2010

Posts: 27
Thanks: 0

Re: Complex equation

But how can we say that a and b are real?
Gustav is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
complex, equation



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Complex equation zelmac Algebra 2 September 2nd, 2012 01:13 AM
How to solve a complex equation? watanuki Complex Analysis 1 October 24th, 2009 07:37 PM
Complex exponential equation Luigi Castelli Complex Analysis 3 February 23rd, 2009 07:52 AM
roots of a complex equation slade8200 Complex Analysis 1 February 20th, 2009 08:34 AM
Complex numbers equation outlander Complex Analysis 9 February 22nd, 2008 12:57 PM





Copyright © 2019 My Math Forum. All rights reserved.