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itsjustme_lol Group Title

Operations on Complex Numbers

  • 2 years ago
  • 2 years ago

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  1. itsjustme_lol Group Title
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    simplify the expression

    • 2 years ago
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  2. frx Group Title
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    Ok so start by rewrite \[\sqrt{-10}=i \sqrt{10}\]

    • 2 years ago
  3. itsjustme_lol Group Title
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    ok, im following

    • 2 years ago
  4. frx Group Title
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    Do you get that idea?

    • 2 years ago
  5. itsjustme_lol Group Title
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    yes I understand that i believe

    • 2 years ago
  6. frx Group Title
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    Do the same with sqrt(-5)

    • 2 years ago
  7. frx Group Title
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    Then distribute

    • 2 years ago
  8. itsjustme_lol Group Title
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    ok so its i\[\sqrt{-5}\]

    • 2 years ago
  9. itsjustme_lol Group Title
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    hang on

    • 2 years ago
  10. itsjustme_lol Group Title
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    |dw:1355782789409:dw|

    • 2 years ago
  11. frx Group Title
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    Not really, the reason for why you take out the i is that you want the minus under the squareroot to disappear

    • 2 years ago
  12. frx Group Title
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    \[i \sqrt{5}\]

    • 2 years ago
  13. frx Group Title
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    Do you know the definition of i?

    • 2 years ago
  14. frx Group Title
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    \[i=\sqrt{-1}\] \[i ^{2}=-1\]

    • 2 years ago
  15. itsjustme_lol Group Title
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    well i know that i means the y on a plane and it is the imaginary part?

    • 2 years ago
  16. frx Group Title
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    That's also, kind of, correct when dealing with imaginary numbers the Y-X plane is called Im-Re

    • 2 years ago
  17. frx Group Title
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    So based on the definition of \[i=\sqrt{-1}\] what's, for example \[\sqrt{-7}\]

    • 2 years ago
  18. frx Group Title
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    ?

    • 2 years ago
  19. frx Group Title
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    \[i \sqrt{10}(11+i \sqrt{5})\]=\[11i \sqrt{10}+i ^{2}\sqrt{10}\sqrt{5}= 11i \sqrt{10}-\sqrt{50}\]

    • 2 years ago
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