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 November 26th, 2009, 04:20 PM #1 Member   Joined: Oct 2009 Posts: 85 Thanks: 0 Very Easy Circle Evaluate $|2\bar{z} + i|= 4$ 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: $(\frac{1}{2})|2\bar{z} + i|= 4(\frac{1}{2})$ $| \bar{z} + \frac{1}{2}i|= 2$ $|\bar{\bar{z} + \frac{1}{2}i}|= \bar{2}$ $z - \frac{1}{2}i= 2$ so then we have a circle centered at $(0, \frac{1}{2})$ 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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