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Math Help - can someone please help with this problem involving complex numbers

  1. #1
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    can someone please help with this problem involving complex numbers

    Y=1/Z1+1/Z2+1/Z3

    Z1= 2+j2,
    Z2=1+j5 ,
    Z3=j6

    i cant figure out how to find Y, could someone please explain how to,or just point me in the right direction. I would be very grateful for any help on this??Thanks!!
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  2. #2
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    Re: can someone please help with this problem involving complex numbers

    Y= (1-j)/2
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    Re: can someone please help with this problem involving complex numbers

    could you please explain how you came to the answer?
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  4. #4
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    Re: can someone please help with this problem involving complex numbers

    Hello, rich1979!

    \text{Given: }\:\begin{Bmatrix}z_1 &=& 2+2i \\ z_2 &=& 1+5i \\ z_3 &=& 6i \end{Bmatrix}

    \text{Find: }\:y \:=\:\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3}

    We must rationalize those denominators.

    \frac{1}{z_1} \;=\;\frac{1}{2+2i} \;=\;\frac{1}{2+2i}\cdot\frac{2-2i}{2-2i} \;=\;\frac{2-2i}{8} \:=\:\frac{1}{4} - \frac{1}{4}i

    \frac{1}{z_2} \;=\;\frac{1}{1+5i} \;=\;\frac{1}{1+5i}\cdot\frac{1-5i}{1-5i} \;=\;\frac{1-5i}{26} \;=\;\frac{1}{26} - \frac{5}{26}i

    \frac{1}{z_3} \;=\;\frac{1}{6i} \;=\;\frac{1}{6i}\cdot\frac{-6i}{-6i} \;=\;\frac{-6i}{36} \;=\;-\frac{1}{6}i


    \text{Hence: }\:y \;=\;\left(\frac{1}{4}-\frac{1}{4}i\right) + \left(\frac{1}{26} - \frac{5}{26}y\right) + \left(-\frac{1}{6}i\right)

    n . . . . y \;=\;\left(\frac{1}{4} + \frac{1}{26}\right) - \left(\frac{1}{4} + \frac{5}{26} + \frac{1}{6}\right)i

    n . . . . y \;=\;\frac{15}{52} - \frac{95}{156}i
    Thanks from rich1979
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  5. #5
    Senior Member MaxJasper's Avatar
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    Re: can someone please help with this problem involving complex numbers

    I assumed j2, j5, j6 mean j^2, j^5, j^6
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    Re: can someone please help with this problem involving complex numbers

    Thank for your help soroban.
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    Re: can someone please help with this problem involving complex numbers

    maxjasper no j2 is a complex number.
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  8. #8
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    Re: can someone please help with this problem involving complex numbers

    Quote Originally Posted by rich1979 View Post
    maxjasper no j2 is a complex number.
    Ok, that means j2=a+b*i where i=sqrt(-1). We can write j5, j6 in similar manner and then simplify y:
    \text{j2}\text{:=}a+b I;\text{j5}\text{:=}c+d I;\text{j6}\text{:=}e+f I;\text{z1}\text{:=}2+\text{j2};\text{z2}\text{:=}  1+\text{j5};\text{z3}\text{:=}\text{j6};y\text{:=}  \frac{1}{\text{z1}}+\frac{1}{\text{z2}}+\frac{1}{\  text{z3}}

    Then we will have:

    y=\frac{2}{(2+a)^2+b^2}+\frac{a}{(2+a)^2+b^2}+\fra  c{1}{(1+c)^2+d^2}+\frac{c}{(1+c)^2+d^2}+\frac{e}{e  ^2+f^2}+i \left(-\frac{b}{(2+a)^2+b^2}-\frac{d}{(1+c)^2+d^2}-\frac{f}{e^2+f^2}\right)
    Last edited by MaxJasper; August 16th 2012 at 03:52 PM.
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