Trigonometric Form of a Complex Number...1

Discussion in 'Geometry and Trigonometry' started by nycmathguy, Feb 19, 2022.

  1. nycmathguy

    nycmathguy

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    Section 6.6

    Screenshot_20220218-170321_Samsung Notes.jpg

    IMG_20220219_164038.jpg

    IMG_20220219_164047.jpg

    IMG_20220219_164102.jpg
     
    nycmathguy, Feb 19, 2022
    #1
  2. nycmathguy

    MathLover1

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    you got r=2sqrt(5) and θ= -π/4, that was enough to do

    z=r(cos(θ)+i*sin(θ))

    answer is: z=2sqrt(5)(cos(-π/4)+i*sin(-π/4))
     
    MathLover1, Feb 20, 2022
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Ok. I tried. Thanks for the correction.
     
    nycmathguy, Feb 20, 2022
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  4. nycmathguy

    nycmathguy

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    Are you sure that r = 2•sqrt{5}? I think r = 5•sqrt{2}.
    You say?
     
    nycmathguy, Feb 20, 2022
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