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03453660 Group Title

why i^2 = -1, where ' i ' is (iota).

  • 2 years ago
  • 2 years ago

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  1. surdawi Group Title
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    \[i = \sqrt{-1}\]

    • 2 years ago
  2. 03453660 Group Title
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    yes then??

    • 2 years ago
  3. Yahoo! Group Title
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    \[i=\sqrt{-1}\] \[i^2 = (\sqrt{-1})^2 = -1\]

    • 2 years ago
  4. shubhamsrg Group Title
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    its the definition of iota.. that i = sqrt(-1) .. it'd be like asking why 2 is called two ? its the definition,,no theoritical reason..

    • 2 years ago
  5. 03453660 Group Title
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    yes but what about this one

    • 2 years ago
  6. 03453660 Group Title
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    |dw:1350311105247:dw|

    • 2 years ago
  7. 03453660 Group Title
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    |dw:1350311152497:dw|

    • 2 years ago
  8. zugzwang Group Title
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    I believe \[\sqrt{ab}= \sqrt{a}\sqrt{b}\] only applies if a and b are both nonnegative

    • 2 years ago
  9. shubhamsrg Group Title
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    true,,i was about to write that only..

    • 2 years ago
  10. zugzwang Group Title
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    I actually looked it up to be sure :P

    • 2 years ago
  11. ZakaullahUET Group Title
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    hmm |dw:1350312680888:dw|

    • 2 years ago
  12. shubhamsrg Group Title
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    sqrt(1) is only +1 and not -1 @zakaullahUET

    • 2 years ago
  13. shubhamsrg Group Title
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    sqrt is a +ve func..

    • 2 years ago
  14. ZakaullahUET Group Title
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    not necessary a function @shubhamsrg ? can b just a relation a*a=(-a)*(-a)

    • 2 years ago
  15. shubhamsrg Group Title
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    when a no. comes out of sqrt, its in mod.. i.e. sqrt(a^2) = |a| see sqrt(x) graph,,it gives out only +ve values of y..

    • 2 years ago
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