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December 5th, 2016, 07:40 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, 08:24 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,594 Thanks: 1492 
Is there a typing error?

January 4th, 2017, 04:03 AM  #3 
Newbie Joined: Jun 2016 From: Hong Kong Posts: 20 Thanks: 2 
It holds when $\displaystyle f(z)=\sum_{k=0}^n a_k z^k,a_k\in R$ $\displaystyle \overline{f(z)}=\overline{\sum_{k=0}^n a_k z^k}=\sum_{k=0}^n \overline{a_k z^k}=\sum_{k=0}^n a_k \overline{z}^k=f(\overline{z})$ 

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