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Yahoo!

  • 3 years ago

if |z|>= 3 the least value of |z + 1/z| is?

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  1. hartnn
    • 3 years ago
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    i think we need to use |x|+|y|>|x+y| but not sure.

  2. Yahoo!
    • 3 years ago
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    But wat Does that Least Value For?

  3. hartnn
    • 3 years ago
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    i m just trying.... if |z|>=3 what can u say about |1/z| ??

  4. Yahoo!
    • 3 years ago
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    |z -(- 1/z)| > = |z| - |-1/z|

  5. Yahoo!
    • 3 years ago
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    I Did nt Understand....wat is here Least..""

  6. hartnn
    • 3 years ago
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    means |z+1/z| is > = 'some number' u need to find that number.

  7. hartnn
    • 3 years ago
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    but |z+1/z| <= |z|+|1/z| how did u get > sign there ?

  8. Yahoo!
    • 3 years ago
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    u see..i have inserted a -ve there....Since it asks for least value."

  9. Callisto
    • 3 years ago
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    Hmm.. Am I right for this: |z| ≥ 3 i.e. z ≥ 3 or z ≤ -3 ?

  10. hartnn
    • 3 years ago
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    i think z is complex here, isn't it @Yahoo! ??

  11. Yahoo!
    • 3 years ago
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    Yes!

  12. hartnn
    • 3 years ago
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    so |z|>= 3 represents region outside the circle with radius 3.....if i have not mis-interpreted

  13. Callisto
    • 3 years ago
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    Wow! Something I don't know! I'm sorry!!

  14. Yahoo!
    • 3 years ago
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    "OUTSIDE" hw

  15. hartnn
    • 3 years ago
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    because of > sign, if < then inside.

  16. Yahoo!
    • 3 years ago
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    Ok...)

  17. Yahoo!
    • 3 years ago
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    @hartnn Any Idea....

  18. hartnn
    • 3 years ago
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    i got something from yahoo answers , trying to understand it, see if that makes some sense to u..... http://in.answers.yahoo.com/question/index?qid=20100423221137AAAgV9K

  19. Yahoo!
    • 3 years ago
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    i think we have to find the least dist b/w Circles

  20. Yahoo!
    • 3 years ago
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    @hartnn i Did nt Understand.....Can u Explain...

  21. Ishaan94
    • 3 years ago
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    \[|z| \geq 3\implies z\leq-3\;or\;z\geq3\] \(\left|z + \frac1z\right|\) is just a symmetrical hyperbola about Y-axis. If you're well versed with hyperbola you will get the idea of the graph with constraint \(|z|\geq3\).

  22. Yahoo!
    • 3 years ago
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    lol...i Dont knw Hyperbola....

  23. Ishaan94
    • 3 years ago
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    Okay. \[x + \frac1x\] \(x\) is ever increasing, so to find the minima of \(x + \frac1x\) you should find value for which \(x\) is minimum i.e 3 as \(1/x\) is decreasing for \(x>1\).

  24. Ishaan94
    • 3 years ago
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    Important to note \(1/x\) is decreasing not negative.

  25. Ishaan94
    • 3 years ago
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    Negative with respect to \(x\).

  26. Pallavi06
    • 3 years ago
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    express Z= x+ iy.......and its mod value is suar root of x^2 +y^2.....now mod of Z+1 can be expressed as (x+1 )+ iy......and then solve it.....i know its very lengthy but it works when u have no other option....

  27. Yahoo!
    • 3 years ago
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    If r is the modulus of z, then |z+ 1/z| >= r - 1/r. Now the minimum of r - 1/r for r>=3 is clearly 3 - 1/3 = 8/3. The minimum for |z+1/z| is attained if |z| = 3 and if z and 1/z have opposite directions. Take z = 3i so that z = 1/3i = - i / 3, and the minimum 8/3 is attained for that value. Can u Explain this!

  28. Ishaan94
    • 3 years ago
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    I was afk. Complex Numbers?

  29. Yahoo!
    • 3 years ago
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    afk. ===?

  30. Ishaan94
    • 3 years ago
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    away from the keyboard.

  31. Vaidehi09
    • 3 years ago
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    yup. complex nos.

  32. Yahoo!
    • 3 years ago
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    Yes.. it is Complex Numbers

  33. Pallavi06
    • 3 years ago
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    let Z= x + iy and try to get value of mod 1/z

  34. Pallavi06
    • 3 years ago
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    can u do it?

  35. Pallavi06
    • 3 years ago
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    if mod z is r then ull get mod i/z as 1/r.....

  36. Ishaan94
    • 3 years ago
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    I will be back, my trackpad isn't working somehow.

  37. Vaidehi09
    • 3 years ago
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    shouldn't it be this way? or is something wrong with this? |dw:1347108277591:dw|

  38. hartnn
    • 3 years ago
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    if its <= then its max value , not least value.

  39. Yahoo!
    • 3 years ago
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    The answer should be 8/3

  40. hartnn
    • 3 years ago
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    and yes,thats the correct way to find max value.

  41. Yahoo!
    • 3 years ago
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    lol.....Can we Do like this |z -(- 1/z)| > = |z| - |-1/z|

  42. Yahoo!
    • 3 years ago
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    = 3 - 1/3 = 8/3

  43. hartnn
    • 3 years ago
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    is this true ? |x-y| >= |x|-|y| if yes, then u can do it like that.

  44. hartnn
    • 3 years ago
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    and i guess its true.

  45. Yahoo!
    • 3 years ago
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    Yes.....

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