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03453660

  • 3 years ago

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

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  1. surdawi
    • 3 years ago
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    \[i = \sqrt{-1}\]

  2. 03453660
    • 3 years ago
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    yes then??

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

  4. shubhamsrg
    • 3 years ago
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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..

  5. 03453660
    • 3 years ago
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    yes but what about this one

  6. 03453660
    • 3 years ago
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    |dw:1350311105247:dw|

  7. 03453660
    • 3 years ago
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    |dw:1350311152497:dw|

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

  9. shubhamsrg
    • 3 years ago
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    true,,i was about to write that only..

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

  11. ZakaullahUET
    • 3 years ago
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    hmm |dw:1350312680888:dw|

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

  13. shubhamsrg
    • 3 years ago
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    sqrt is a +ve func..

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

  15. shubhamsrg
    • 3 years ago
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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..

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