Polarrepresentation of 1/(1/2 - 1/2i)?

Discussion in 'Linear and Abstract Algebra' started by Humboldt, Mar 28, 2021.

  1. Humboldt

    Humboldt

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    Hello,
    What's the Polarrepresentation of 1/(1/2 - 1/2i)?

    My result was: √2e^(i(π/4)) , but the solution on my worksheet says 2e^(i(π/4)). Can someone explain me if I'm wrong or if they forget the root on the solution.

    I'm sorry if I posted in the wrong category.
    Thx in advance.
     
    Humboldt, Mar 28, 2021
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  2. Humboldt

    MathLover1

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    you are not wrong

    1/(1/2 - 1/2i)

    1(1/2 +1/2i)/((1/2 - 1/2i)(1/2 + 1/2i))

    (1/2( 1+i))/((1/2)^2 - (1/2i)^2)

    (1/2( 1+i))/(1/4 - 1/4*(i)^2)

    (1/2( 1+i))/(1/4 - 1/4*(-1))

    (1/2( 1+i))/(1/4 + 1/4)

    (1/2( 1+i))/(2/4)

    (1/2( 1+i))/(1/2)

    1+i =>a+bi

    in polar form:

    r=sqrt(a^2+b^2)

    r=sqrt(1^2+1^2)
    r=sqrt(2)

    r*cos(θ)=a =>sqrt(2)*cos(θ)=1

    =>cos(θ)=1/sqrt(2)

    θ=cos^-1(1/sqrt(2))

    θ=π/4 (result in radians)

    r*sin(θ)=b =>sqrt(2)*sin(θ)=1

    =>sin(θ)=1/sqrt(2)

    θ=sin^-1(1/sqrt(2))
    θ=π/4

    sqrt(2) (cos(π/4)+i* sin(π/4)) or sqrt(2) e^((i π)/4)
     
    MathLover1, Jun 27, 2021
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    nycmathguy likes this.
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  3. Humboldt

    nycmathguy

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    I can't wait to be at this level of math. I hope Humboldt will at least thank you for this reply. This site needs a major renovation. Where is the owner? Where are the tutors?
     
    nycmathguy, Jul 5, 2021
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