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Math Help - Complex numbers

  1. #1
    Junior Member Cyberman86's Avatar
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    Question Complex numbers

    Given Z1= sqr.root of 3+i and Z2= 6(cos(pie/6) + i sin(pie/6)) respond to the following:

    *Convert Z1 to polar form

    *Find Z1 dot Z2 (there is a dot in the middle) Leave your answer in polar form.
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  2. #2
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    For those who understand complex numbers \sqrt {3 + i} does not exists.
    Now there are two square roots of {3 + i}!
    But the radical sign is applied only to real numbers.
    You seem to have some real confusion in mathematical notation and symbols.
    What are you trying to ask?
    BTW: Pie is a somthing one eats, like Amercian Pie. It is not a number.
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  3. #3
    Junior Member Cyberman86's Avatar
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    Post

    i just wrote everything just like my teacher gave it to me, and i wrote pie because i don't know how to put the symbol.
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  4. #4
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    Quote Originally Posted by Cyberman86 View Post
    i just wrote everything just like my teacher gave it to me, and i wrote pie because i don't know how to put the symbol.
    The Greek letter is pi, \pi.
    It is realy sad but your ‘teacher’ may not be up to speed.
    Those easily misunderstand complex numbers or Complex Analysis who are minimally prepared.
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  5. #5
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    Help

    Z1= sqr.root of 3+i and Z2= 6(cos(pie/6) + i sin(pie/6))

    Okay, I don't know if you can take it from here, but Euler's Formula gives:

    cos(x)+i sin(x)=e^(ix)

    And:

    sqrt.(3+i)=(3+i)^(1/2)=e^ln((3+i)^(1/2))=e^(.5(ln(3+i)))

    And another formula gives:

    ln(x+yi)=.5(ln(x^2+y^2))+i(arctan(y/x))

    So Z(1)=sqrt.(3+i)=e^(.5(.5(ln(9+1))+i arctan(1/3))) (I may be missing a paranthesis, but you get the idea)

    So now you have:

    Z(1)=[e^(.25(ln(10)))][e^(i[.5(arctan(1/3)))]

    Thus:

    Z(1)=[e^(.25(ln(10)))][cos(.5(arctan(1/3)))+i sin(.5(artan(1/3)))]

    And from there you can find the real and imaginary part of Z(1)

    I may have made a typo somewhere in there, but the formulas are correct.
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