nth Root of Complex Number...2

Discussion in 'Geometry and Trigonometry' started by nycmathguy, Feb 27, 2022.

  1. nycmathguy

    nycmathguy

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    Section 6.6

    Screenshot_20220225-174744_Samsung Notes.jpg

    IMG_20220227_172240.jpg

    IMG_20220227_172250.jpg

    IMG_20220227_172304.jpg

    IMG_20220227_172315.jpg IMG_20220227_172326.jpg

    IMG_20220227_172334.jpg
     
    nycmathguy, Feb 27, 2022
    #1
  2. nycmathguy

    MathLover1

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    upload_2022-2-27_20-30-39.jpeg

    upload_2022-2-27_20-31-0.jpeg
    upload_2022-2-27_20-31-16.jpeg
     
    MathLover1, Feb 28, 2022
    #2
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  3. nycmathguy

    nycmathguy

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    nycmathguy, Feb 28, 2022
    #3
  4. nycmathguy

    MathLover1

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    The polar form of -4 is 4 (cos(pi ) + i *sin(pi))

    ->should be r=4, and θ = π (above was mistake, I just put value of sin(π) which gives you θ =0)

    so, answer will change and should be:

    upload_2022-2-28_11-54-47.jpeg

    answer:

    upload_2022-2-28_11-56-55.gif
    upload_2022-2-28_11-57-13.gif
    upload_2022-2-28_11-57-37.gif
    upload_2022-2-28_11-57-53.gif
     

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    MathLover1, Feb 28, 2022
    #4
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  5. nycmathguy

    nycmathguy

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    nycmathguy, Feb 28, 2022
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  6. nycmathguy

    MathLover1

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    yes, it is correct
     
    MathLover1, Feb 28, 2022
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  7. nycmathguy

    nycmathguy

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    Can you please go back and check my answers again to the other nth roots of complex numbers threads? I am almost sure I got the others right, too. Going to bed now. Back to work tonight.
     
    nycmathguy, Feb 28, 2022
    #7
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