Hard complex number problem :?

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2 Answers

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  • 7 months ago
    Favorite Answer

    This is the required solution -

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  • atsuo
    Lv 6
    7 months ago

    One method with only calculations : 

     

    x+iy = √[(a+ib)/(c+id)] 

     

    Square both sides, 

     

    (x+iy)^2 = (a+ib)/(c+ib) 

     

    We know 

     

    (x+iy)^2 = x^2-y^2 + 2ixy and 

    (a+ib)/(c+id) 

    = (a+ib)(c-id)/((c+ib)(c-id)) 

    = (ac+bd+i(bc-ad))/(c^2+d^2) 

    = (ac+bd)/(c^2+d^2) + i(bc-ad)/(c^2+d^2) 

     

    Therefore 

     

    x^2-y^2 + 2ixy = (ac+bd)/(c^2+d^2) + i(bc-ad)/(c^2+d^2) 

     

    That is, 

     

    x^2-y^2 = (ac+bd)/(c^2+d^2) ---(#1) 

    2xy = (bc-ad)/(c^2+d^2) ---(#2) 

     

    We must find the value of (x^2+y^2)^2. 

     

    (x^2+y^2)^2 

    = x^4 + 2x^2y^2 + y^4 

    = x^4 - 2x^2y^2 + y^4 + 4x^2y^2 

    = (x^2-y^2)^2 + (2xy)^2 ---(#3) 

     

    Substitute (#1) and (#2) into (#3). 

     

    (x^2+y^2)^2 

    = ((ac+bd)/(c^2+d^2))^2 + ((bc-ad)/(c^2+d^2))^2 

    = [(a^2c^2 + 2abcd + b^2d^2) + (b^2c^2 - 2abcd + a^2d^2)]/(c^2+d^2)^2 

    = (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2)/(c^2+d^2)^2 

    = (a^2+b^2)(c^2+d^2)/(c^2+d^2)^2 

    = (a^2+b^2)/(c^2+d^2) <--- the answer

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