anonymous
  • anonymous
why i^2 = -1, where ' i ' is (iota).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[i = \sqrt{-1}\]
anonymous
  • anonymous
yes then??
anonymous
  • anonymous
\[i=\sqrt{-1}\] \[i^2 = (\sqrt{-1})^2 = -1\]

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shubhamsrg
  • shubhamsrg
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..
anonymous
  • anonymous
yes but what about this one
anonymous
  • anonymous
|dw:1350311105247:dw|
anonymous
  • anonymous
|dw:1350311152497:dw|
zugzwang
  • zugzwang
I believe \[\sqrt{ab}= \sqrt{a}\sqrt{b}\] only applies if a and b are both nonnegative
shubhamsrg
  • shubhamsrg
true,,i was about to write that only..
zugzwang
  • zugzwang
I actually looked it up to be sure :P
anonymous
  • anonymous
hmm |dw:1350312680888:dw|
shubhamsrg
  • shubhamsrg
sqrt(1) is only +1 and not -1 @zakaullahUET
shubhamsrg
  • shubhamsrg
sqrt is a +ve func..
anonymous
  • anonymous
not necessary a function @shubhamsrg ? can b just a relation a*a=(-a)*(-a)
shubhamsrg
  • shubhamsrg
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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