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Fools101 Group Title

Imaginary Numbers Simplify i 9

  • 3 years ago
  • 3 years ago

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  1. Fools101 Group Title
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    Please help!

    • 3 years ago
  2. pottersheep Group Title
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    That is simplified isnt it?

    • 3 years ago
  3. Fools101 Group Title
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    oh also it i^9

    • 3 years ago
  4. pottersheep Group Title
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    ah. ok then so 1^2 = -1 right

    • 3 years ago
  5. Blank123 Group Title
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    simplified version has to be i

    • 3 years ago
  6. Fools101 Group Title
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    Aww idk this kind of confusing me

    • 3 years ago
  7. pottersheep Group Title
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    (i^2)^4 . i (-1)^4 . i 1i

    • 3 years ago
  8. pottersheep Group Title
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    (i^2)^4 . i (-1)^4 . i 1i

    • 3 years ago
  9. Blank123 Group Title
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    any variable equals 1 automatically

    • 3 years ago
  10. pottersheep Group Title
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    Do you get it?

    • 3 years ago
  11. pottersheep Group Title
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    Basically take out an i^2 first so that you work with -1. I.e i^4 = (i^2)^2 so now you have (-1)^2 = 1. Do you understand up to there?

    • 3 years ago
  12. Fools101 Group Title
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    KIND of so it 1? a little ...

    • 3 years ago
  13. Blank123 Group Title
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    if there is no value to a variable it becomes a 1.

    • 3 years ago
  14. Fools101 Group Title
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    then the i Crossout ?

    • 3 years ago
  15. myininaya Group Title
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    \[i^9=i^{4 \cdot 2 +1}=i^{4 \cdot 2} \cdot i^1=(i^4)^2 \cdot i=(1)^2 \cdot i\] \[=1 \cdot i=i\]

    • 3 years ago
  16. satellite73 Group Title
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    pattern is \[i^0=1\] \[i^1=i\] \[i^2=-1\] \[i^3=-i\] \[i^4=1\] and so on. so your real job is to take the integer remainder when you divide the power by 4. for example \[i^{103}=i^3=-i\]

    • 3 years ago
  17. Fools101 Group Title
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    ohhhh i see !!! kk thank u that make sense !!!

    • 3 years ago
  18. satellite73 Group Title
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    yw

    • 3 years ago
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