
Algebra PreAlgebra and Basic Algebra Math Forum 
 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 Ì„) if f(z) = 0
Prove that f of z where z is a complex number is equal to f of z Ì„ (which is the conjugate of z) if f(z)=0. f(x) =x^2+8x+16 Edit: f(z Ì„)= z^2+8+16 with a bar over everything.

December 5th, 2016, 07:24 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 17,734 Thanks: 1360 
Is there a typing error?

January 4th, 2017, 03:03 AM  #3 
Newbie Joined: Jun 2016 From: Hong Kong Posts: 20 Thanks: 2 
It holds when $\displaystyle f(z)=\sum_{k=0}^n a_k z^k,a_k\in R$ $\displaystyle \overline{f(z)}=\overline{\sum_{k=0}^n a_k z^k}=\sum_{k=0}^n \overline{a_k z^k}=\sum_{k=0}^n a_k \overline{z}^k=f(\overline{z})$ 

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 