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December 16th, 2012, 12:42 AM  #1 
Newbie Joined: Dec 2012 Posts: 28 Thanks: 0  Complex numbers;
Help on this, if z=1+i

December 16th, 2012, 01:07 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Complex numbers;
If , then what is ?

December 16th, 2012, 01:11 AM  #3 
Newbie Joined: Dec 2012 Posts: 28 Thanks: 0  Re: Complex numbers;
Not listed I think that it changes the  to + and + to 

December 16th, 2012, 01:12 AM  #4 
Newbie Joined: Dec 2012 Posts: 28 Thanks: 0  Re: Complex numbers;
Conjugation

December 16th, 2012, 01:14 AM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Complex numbers;
Yes, the bar over a complex value represents the conjugate, in this case . So, make the substitutions, then simplify. What do you get?

December 16th, 2012, 01:21 AM  #6 
Newbie Joined: Dec 2012 Posts: 28 Thanks: 0  Re: Complex numbers;
Are : (1+i)^2 = 11= 0? Or: (1i)*(1+i)= 1+iii^2= 1+1=2? 
December 16th, 2012, 02:16 AM  #7 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Complex numbers; You won't need the second product. 
December 16th, 2012, 02:20 AM  #8 
Newbie Joined: Dec 2012 Posts: 28 Thanks: 0  Re: Complex numbers;
When it is (1i)^2, what should I do?

December 16th, 2012, 02:24 AM  #9 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Complex numbers;
That won't show up either, but you use the square of a binomial as I did above.

December 16th, 2012, 02:32 AM  #10 
Newbie Joined: Dec 2012 Posts: 28 Thanks: 0  Re: Complex numbers;
I get this: 2i(1+i1+i)i+22i = 2i(2i)i+22i = 2i^2 i +22i = 2(1)i+22i = 2i+22i = 24i Correct?? 

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