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    you can say ((1 + i)/(1 - i)) = (ai + b)2 , where ai + b is what you are trying to find.

    now simplify, cross multiply & compare coefficients to find a & b
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    (Original post by jlf98)
    How would I put Root((1+i)/(1-i)) in the form x+iy.
    I multiplied the fraction by the conjugate of the denominator (1+i) to get root(i) then i dont know where to go from there?
    do you know what de moivres theorum is?
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    (Original post by bruh2132)
    do you know what de moivres theorum is?
    I wouldn't jump to use De Moivre's here considering that the OP hasn't mentioned their study level. I'd say an approach by comparing coefficients would likely be more accessible to the OP.
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    (Original post by _gcx)
    I wouldn't jump to use De Moivre's here considering that the OP hasn't mentioned their study level. I'd say an approach by comparing coefficients would likely be more accessible to the OP.
    yh I was just asking, if he did know about it it might have been easier
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    yes would you say to approach the question using de moivres theorem ?
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    (Original post by jlf98)
    yes would you say to approach the question using de moivres theorem ?
    I would do it that way if you know how, since i can be written in the form cos(theta)+isin(theta), you can raise that to a power and put it in whatever form you want
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    (Original post by jlf98)
    How would I put Root((1+i)/(1-i)) in the form x+iy.
    I multiplied the fraction by the conjugate of the denominator (1+i) to get root(i) then i dont know where to go from there?
    When you multiply top and bottom by 1+i you might want to leave the numerator as (1+i)^2 - you should see the answer straight away (N.B. you might think there could be 2 possible answers, but remember that the root sign indicates the principal square root)
 
 
 
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