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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)?
 one year ago
 one year 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)?
 one year ago
 one year ago

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

Studentc14Best ResponseYou've already chosen the best response.0
i 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)
 one year ago

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

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

Studentc14Best ResponseYou've already chosen the best response.0
yeah 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)
 one year ago

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

Studentc14Best ResponseYou've already chosen the best response.0
i 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:
 one year ago

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

asnaseerBest ResponseYou've already chosen the best response.3
ok, 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
 one year ago

Studentc14Best ResponseYou've already chosen the best response.0
for step 3, would the complex conjugate of z^2 be a^2b^22abi?
 one year ago

Studentc14Best ResponseYou've already chosen the best response.0
then 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
 one year ago

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

Studentc14Best ResponseYou've already chosen the best response.0
not 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?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
yes  that is what I believe the question would have stated
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
question is correct, they are not same
 one year ago

Studentc14Best ResponseYou've already chosen the best response.0
hartnn how are they not the same?
 one year ago

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

hartnnBest ResponseYou've already chosen the best response.0
{z(bar) }^2 = left side = (abi)(abi)
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
right side =conjugate of (a+bi)(a+bi)
 one year ago

asnaseerBest ResponseYou've already chosen the best response.3
correct  and they both work out to be the same value
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
oh dear, sorry, i worked out right side incorrectly
 one year ago

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

Studentc14Best ResponseYou've already chosen the best response.0
yes i beleive that was the question. okay, thanks for all your help!
 one year ago
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