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, 2016, 06:40 PM   #1
Newbie
 
Joined: Dec 2016
From: Boston

Posts: 1
Thanks: 0

Prove f(z) =f(z ̄) if f(z) = 0

Prove that f of z where z is a complex number is equal to f of z ̄ (which is the conjugate of z) if f(z)=0. f(x) =x^2+8x+16 Edit: f(z ̄)= z^2+8+16 with a bar over everything.
SpiesWithin is offline  
 
December 5th, 2016, 07:24 PM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 18,048
Thanks: 1395

Is there a typing error?
skipjack is offline  
January 4th, 2017, 03:03 AM   #3
Newbie
 
fungarwai's Avatar
 
Joined: Jun 2016
From: Hong Kong

Posts: 20
Thanks: 2

It holds when $\displaystyle f(z)=\sum_{k=0}^n a_k z^k,a_k\in R$

$\displaystyle \overline{f(z)}=\overline{\sum_{k=0}^n a_k z^k}=\sum_{k=0}^n \overline{a_k z^k}=\sum_{k=0}^n a_k \overline{z}^k=f(\overline{z})$
fungarwai is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
prove



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
How to prove this? jiasyuen Algebra 1 December 21st, 2013 08:35 AM
Prove that for a<x<b ... sivela Calculus 0 January 21st, 2011 10:29 AM
Goldbach's conjecture (to prove or not to prove) octaveous Number Theory 13 September 23rd, 2010 04:36 AM
prove prove prove. currently dont know where to post qweiop90 Algebra 1 July 31st, 2008 06:27 AM
prove prove prove. currently dont know where to post qweiop90 New Users 1 December 31st, 1969 04:00 PM





Copyright © 2017 My Math Forum. All rights reserved.