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Loser66

  • one year ago

Someone explains me, please. Question on attachment.

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  1. Loser66
    • one year ago
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  2. Loser66
    • one year ago
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    I don't get "Equivalently..." why is it?

  3. Loser66
    • one year ago
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    @ganeshie8

  4. ganeshie8
    • one year ago
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    Can the magnitude of a complex number ever be negative ? |dw:1440808727196:dw|

  5. Loser66
    • one year ago
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    No

  6. Loser66
    • one year ago
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    but zw (bar) is not the magnitude.

  7. ganeshie8
    • one year ago
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    well the real part of \(z\overline{w}\) is magnitude of something

  8. ganeshie8
    • one year ago
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    |dw:1440809022356:dw|

  9. Loser66
    • one year ago
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    Question: when a complex number is consider as a negative one?

  10. ganeshie8
    • one year ago
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    complex numbers don't have that sense of binary negative /positive the sense comes from the arguement let me think a bit..

  11. ganeshie8
    • one year ago
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    the magnitude of \(z\overline{w}\) equals the real component of \(z\overline{w}\) can we conclude that the imaginary component of \(z\overline{w}\) is 0 ?

  12. Loser66
    • one year ago
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    Yes, I think so

  13. ganeshie8
    • one year ago
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    since the imaginary component is 0, the complex number \(z\overline{w}\) is just as good as a real number and we can compare it with other reals like 0 etc

  14. ganeshie8
    • one year ago
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    Moreover \(z\overline{w}\) identically equals the magnitude of something, so it has to be nonnegative

  15. Loser66
    • one year ago
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    No, not that, like |dw:1440809469436:dw|

  16. ganeshie8
    • one year ago
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    By definition, ,magnitude of a complex number is the distance from itself to the origin. Length can never be negative

  17. ganeshie8
    • one year ago
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    |dw:1440809618870:dw|

  18. Loser66
    • one year ago
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    I know, I talk about the vector itself, \(z\overline w\), not \(|z\overline w|\)

  19. ganeshie8
    • one year ago
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    \(\large -5+0i = -5\) is on negative side, so it can never be magnitude of somebody.

  20. Loser66
    • one year ago
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    We have \(|z\overline w|=Re (z\overline w)\) I am ok with that, but Equivalently \(z\overline w\), (this guy is not the magnitude or a Real part of anything, and it is \(\geq 0\) That is what I don't know how to argue.

  21. ganeshie8
    • one year ago
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    we have concluded earlier that the imaginary part of \(z\overline{w}\) is 0, yes ?

  22. Loser66
    • one year ago
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    Yes

  23. ganeshie8
    • one year ago
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    therefore \(\Re{(z\overline{w})} = z\overline{w} \)

  24. Loser66
    • one year ago
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    ok, got you. hehehe.. yes. Thanks a lot.

  25. ganeshie8
    • one year ago
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    np

  26. ganeshie8
    • one year ago
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    the solution in docx kinda sucks... its not detailed and you will need to put together the missing pieces

  27. ganeshie8
    • one year ago
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    |dw:1440810132148:dw|

  28. Loser66
    • one year ago
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    oh, for this part, I know how to fix.

  29. ganeshie8
    • one year ago
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    can you explain me that part is that some identity that you're using or do you know why is it true

  30. Loser66
    • one year ago
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    We have |a| = |b| iff |a|^2 =|b|^2 Apply to this , we have: \(||z|-|w||= |z -w|\) iff \( LHS ^2 = RHS^2\)

  31. ganeshie8
    • one year ago
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    got that part

  32. Loser66
    • one year ago
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    \(LHS^2 = (|z|-|w|) \overline{(|z|-|w|)} =(|z|-|w|)(|\overline z| -\overline{|w|})\)

  33. ganeshie8
    • one year ago
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    for LHS aren't you simply using the identity \[(a-b)^2 = a^2-2ab+b^2\] ?

  34. Loser66
    • one year ago
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    \(=|z|^2-|z|\overline{|w|}-\overline{|z|}|w|+|w|^2\)

  35. Loser66
    • one year ago
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    The middle terms only \(\overline{z\overline{w}}= \overline{z}w\)

  36. Loser66
    • one year ago
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    Hence it is \(-(z\overline{w} +\overline{z\overline{w}})= -2Re(z\overline{w})\)

  37. Loser66
    • one year ago
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    That is why it becomes as shown.

  38. Loser66
    • one year ago
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    |dw:1440810939082:dw|

  39. Loser66
    • one year ago
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    I did

  40. ganeshie8
    • one year ago
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    \(=|z|^2-|z|\overline{|w|}-\overline{|z|}|w|+|w|^2\) can you show me how the middle two terms equal to \(2\Re(z\overline{w})\)

  41. ganeshie8
    • one year ago
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    i don't get it, let me try reading it again

  42. Loser66
    • one year ago
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    |dw:1440811335931:dw|

  43. ganeshie8
    • one year ago
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    Got it! thank you :)

  44. Loser66
    • one year ago
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    |dw:1440811409420:dw|

  45. Loser66
    • one year ago
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    hihihi... I have to say thanks to you :)

  46. ganeshie8
    • one year ago
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    looks neat! just writing it again : \[|z-w|^2 = (z-w)(\overline{z-w}) =(z-w)(\overline{z}-\overline{w})\\~\\= |z|^2\color{blue}{-w\overline{z} -z\overline{w}}+|w|^2 = |z|^2\color{blue}{-\overline{\overline{w}z}-z\overline{w}}+|w|^2= |z|^2\color{blue}{-2\Re{(\overline{w}z})}+|w|^2\]

  47. Loser66
    • one year ago
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    Thank you. :)

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