November 26th, 2009, 04:20 PM  #1 
Member Joined: Oct 2009 Posts: 85 Thanks: 0  Very Easy Circle
Evaluate Ok so this problem is so easy yet I got it wrong for some reason on my HW that I turned in and I want to know why. Here is what I did: so then we have a circle centered at with radius of 2. I do not see how this is wrong but maybe I am missing something simple with the center or something. Thanks. 
November 26th, 2009, 07:08 PM  #2 
Member Joined: Oct 2009 Posts: 64 Thanks: 0  Re: Very Easy Circle
The conclusion is correct but your third step isn't particularly logical and in your fourth step you are missing the  on the left side. In the third step it looks like you are saying that if a = b then a* = b*, where * denotes complex conjugation. This is not true. The norm of the complex conjugate of a is the norm of a, without exception. 
November 27th, 2009, 12:16 PM  #3 
Member Joined: Oct 2009 Posts: 85 Thanks: 0  Re: Very Easy Circle
So how should I write out the third step? because I have no clue besides the way I did it.

November 27th, 2009, 03:39 PM  #4 
Member Joined: Oct 2009 Posts: 64 Thanks: 0  Re: Very Easy Circle
"The norm of the complex conjugate of a is the norm of a , without exception." That is, z* + i/2 = (z* + i/2)* = z  i/2 = 2 Conjugating the 2 like you did was spurious, though it didn't change the truth of the equation in step 3. 

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