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posted by  jkim8251 on 10/6/2008 1:44:16 AM  |  status: Live  

Rational Functions

Course Textbook Chapter Problem
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Question Details:
a) Show that 2i and 1-i are both solutions of the equation x^2 -(1+i)x +(2+2i)=0 but that their complex conjugates -2i and 1+i are not..
 
b) Explain why the result of part (a) does not violate the Conjugate Zeros Theorem
 
 
Please help me on these problems.. A.S.A.P.
It will be great if you explain both of these problems step by step so i can easily understand..
Thanks~
Joseph Kim

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posted by Whackpro on 10/6/2008 5:31:35 AM  |  status: Live
Asker's Rating: Somewhat Helpful   
Response Details:
 
 
 you can easily prove part a by putting all four values in place of x, getting value of the expression to be zero.
 
  you can do part b this way :
 
  the conjugate zeroes theorem is meant for expressions with all real coefficients.
did my answer help you ? if it was wrong or you didn't quite get it kindly send me a private message giving the details.
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posted by Diego P on 10/6/2008 6:00:03 AM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
For a), we get:

Using the quadratic formula, we then get:




This means that:


Therefore, 2i and 1-i are solutions to the given equation, and the only ones (because of the fundamental theorem of algebra).

For b), you were already told why (the one above me).
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