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anonymous
 3 years ago
How to find prove that the complex conjugate of Z(with a line over it)^2 is not the same as Z^2(with a line over it)?
anonymous
 3 years ago
How to find prove that the complex conjugate of Z(with a line over it)^2 is not the same as Z^2(with a line over it)?

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asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3If you write the complex number z as:\[z=a+ib\]Then do you know what \(\bar{z}\) equals in terms a and b?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know that z=a+bi, and z(line over)=abi, but when i squared the first one i got (a^2b^2)2abi and i got the same thing when i solve (a+bi)^2(all with a line over it)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3\[(a+ib)^2=(a+ib)(a+ib)=a^2+2abi+i^2b^2=a^2+2abib^2=a^2b^2+2abi\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3That last term is:\[=a^2b^2+2abi\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah i got that part... that's the z^2 part right? because after that the problem wants z^2(with a line over z^2)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3yes, so now find \(\bar{z}^2\) using the same technique, i.e.:\[\bar{z}^2=(aib)^2=(aib)(aib)=?\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i did that and i got a^2b^22abi... so this is the first part of the problem that i have to compare to z^2(with a line over the entire thing including the square... look at the attatched picture, im not sure if I'm typing it right, but this is the question at the top there:

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3So is this the question  prove that:\[(\bar{z})^2\ne\bar{(z^2)}\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3ok, then steps are: 1. set z = a + bi 2. find \(z^2\) using this 3. take the complex conjugate of this expression 4. set \(\bar{z}\)=a=bi 5. find \(\bar{z}^2\) using this 6. compare the two results

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for step 3, would the complex conjugate of z^2 be a^2b^22abi?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then isn't it the same as the other result? i got a^2b^22abi as the complex conjugate squared [z^2(with a line over it)] as well

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3yes  so do I. Are you 100% sure you have the correct question?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0not 100%... it was from a test that we started today and are finishing tomorrow. I remember the z terms correctly but im not exactly sure what it was asking... i guess maybe the question was to prove that they ARe the same?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3yes  that is what I believe the question would have stated

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0question is correct, they are not same

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hartnn how are they not the same?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3@hartnn  the /revised/ question is: \[(\bar{z})^2\ne\bar{(z^2)}\]

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0{z(bar) }^2 = left side = (abi)(abi)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0right side =conjugate of (a+bi)(a+bi)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3correct  and they both work out to be the same value

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0oh dear, sorry, i worked out right side incorrectly

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.3so I believe the question should have been: Prove that \((\bar{z})^2=\bar{(z^2)}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes i beleive that was the question. okay, thanks for all your help!
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