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Studentc14 Group Title

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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  1. asnaseer Group Title
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    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
  2. Studentc14 Group Title
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    i know that z=a+bi, and z(line over)=a-bi, but when i squared the first one i got (a^2-b^2)-2abi and i got the same thing when i solve (a+bi)^2(all with a line over it)

    • one year ago
  3. asnaseer Group Title
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    \[(a+ib)^2=(a+ib)(a+ib)=a^2+2abi+i^2b^2=a^2+2abi-b^2=a^2-b^2+2abi\]

    • one year ago
  4. asnaseer Group Title
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    That last term is:\[=a^2-b^2+2abi\]

    • one year ago
  5. Studentc14 Group Title
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    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
  6. asnaseer Group Title
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    yes, so now find \(\bar{z}^2\) using the same technique, i.e.:\[\bar{z}^2=(a-ib)^2=(a-ib)(a-ib)=?\]

    • one year ago
  7. Studentc14 Group Title
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    i did that and i got a^2-b^2-2abi... 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
  8. asnaseer Group Title
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    So is this the question - prove that:\[(\bar{z})^2\ne\bar{(z^2)}\]

    • one year ago
  9. Studentc14 Group Title
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    yes

    • one year ago
  10. asnaseer Group Title
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    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
  11. Studentc14 Group Title
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    for step 3, would the complex conjugate of z^2 be a^2-b^2-2abi?

    • one year ago
  12. asnaseer Group Title
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    yes

    • one year ago
  13. Studentc14 Group Title
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    then isn't it the same as the other result? i got a^2-b^2-2abi as the complex conjugate squared [z^2(with a line over it)] as well

    • one year ago
  14. asnaseer Group Title
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    yes - so do I. Are you 100% sure you have the correct question?

    • one year ago
  15. Studentc14 Group Title
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    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
  16. asnaseer Group Title
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    yes - that is what I believe the question would have stated

    • one year ago
  17. hartnn Group Title
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    question is correct, they are not same

    • one year ago
  18. Studentc14 Group Title
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    hartnn- how are they not the same?

    • one year ago
  19. asnaseer Group Title
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    @hartnn - the /revised/ question is: \[(\bar{z})^2\ne\bar{(z^2)}\]

    • one year ago
  20. hartnn Group Title
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    {z(bar) }^2 = left side = (a-bi)(a-bi)

    • one year ago
  21. hartnn Group Title
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    right side =conjugate of (a+bi)(a+bi)

    • one year ago
  22. asnaseer Group Title
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    correct - and they both work out to be the same value

    • one year ago
  23. hartnn Group Title
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    oh dear, sorry, i worked out right side incorrectly

    • one year ago
  24. asnaseer Group Title
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    so I believe the question should have been: Prove that \((\bar{z})^2=\bar{(z^2)}\)

    • one year ago
  25. Studentc14 Group Title
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    yes i beleive that was the question. okay, thanks for all your help!

    • one year ago
  26. asnaseer Group Title
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    yw :)

    • one year ago
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