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Math Help - Complex numbers help

  1. #1
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    Complex numbers help

    In my exercise book I couldn't find a way to solve what I've typed below (I know for final solutions anyway):

    PROBLEM 1

    Find Re(w) and Im(w) [where z=x+yi]:

    w=z+\frac{1}{z}

    So we have to write a complex number in a standard form and then extract a real and an imaginary part from there. We work here with the unknown variables ( x and y ).

    My way would be:

    w=z+\frac{1}{z}=x+yi+\frac{1}{x+yi}=\frac{(x+yi)^{  2}+1}{x+yi}=\frac{((x+yi)^{2}+1)(x-yi)}{(x+yi)(x-yi)}=
    \frac{((x+yi)^{2}+1)(x-yi)}{x^{2}+y^{2}}=...

    If I expand the expression further (i.e. numerator) I don't find a way to write it in a standard form.

    However, the final solution is:

    Re(w)=x+\frac{x}{x^{2}+y^{2}} and Im(w)=y-\frac{y}{x^{2}+y^{2}}

    Does anybody know the steps leading to the final solution?

    PROBLEM 2

    Find all solutions to the equation [where z=x+yi]:

    2z^{2}-3\overline{z}^{2}=10i

    It is a quadratic equation and has a conjugate of complex number in it.

    Thanks for help in advance.
    Last edited by CaptainBlack; February 10th 2007 at 09:11 AM. Reason: trying to make thread accessible despite LaTeX problems
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  2. #2
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    Quote Originally Posted by summerBoy View Post
    In my exercise book I couldn't find a way to solve what I've typed below (I know for final solutions anyway):

    PROBLEM 1

    Find Re(w) and Im(w) [where z=x+yi]:

    w=z+\frac{1}{z}

    So we have to write a complex number in a standard form and then extract a real and an imaginary part from there. We work here with the unknown variables ( x and y ).

    My way would be:

    w=z+\frac{1}{z}=x+yi+\frac{1}{x+yi}=\frac{(x+yi)^{  2}+1}{x+yi}=\frac{((x+yi)^{2}+1)(x-yi)}{(x+yi)(x-yi)}=
    \frac{((x+yi)^{2}+1)(x-yi)}{x^{2}+y^{2}}=...

    If I expand the expression further (i.e. numerator) I don't find a way to write it in a standard form.
    .
    You did everything correct except expand which is important.

    For example if you have,
    (1+i)(2+i)/(25)
    Then if you expand,
    (1+3i)/25=(1/25)+(3/35)i
    Thus,
    Re(z)=1/25 and Im(z)=3/35

    Note, this is the the solution to your problem but this is the general way to do it.
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  3. #3
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    Quote Originally Posted by summerBoy View Post
    In my exercise book I couldn't find a way to solve what I've typed below (I know for final solutions anyway):

    PROBLEM 1

    Find Re(w) and Im(w) [where z=x+yi]:

    w=z+\frac{1}{z}

    So we have to write a complex number in a standard form and then extract a real and an imaginary part from there. We work here with the unknown variables ( x and y ).

    My way would be:

    w=z+\frac{1}{z}=x+yi+\frac{1}{x+yi}=\frac{(x+yi)^{  2}+1}{x+yi}=\frac{((x+yi)^{2}+1)(x-yi)}{(x+yi)(x-yi)}=
    \frac{((x+yi)^{2}+1)(x-yi)}{x^{2}+y^{2}}=...

    If I expand the expression further (i.e. numerator) I don't find a way to write it in a standard form.

    However, the final solution is:

    Re(w)=x+\frac{x}{x^{2}+y^{2}} and Im(w)=y-\frac{y}{x^{2}+y^{2}}

    Does anybody know the steps leading to the final solution?

    PROBLEM 2

    Find all solutions to the equation [where z=x+yi]:

    2z^{2}-3\overline{z}^{2}=10i

    It is a quadratic equation and has a conjugate of complex number in it.

    Thanks for help in advance.
    Oh, it can be opened now? I tried to answer this since many hours ago but it wouldn't open.

    Anyway, I tried the 1st Problem only. I did expand it paiently and I came up to the shown Final solution.

    Here goes:

    W = [((x +yi)^2 +1)(x -yi)] / (x^2 +y^2)

    W = [(x^2 +2xyi -y^2 +1)(x -yi)] / (x^2 +y^2)

    W = [x^3 -(x^2)yi +2(x^2)yi -2x(y^2)(i^2) -xy^2 +(y^3)i +x -yi] / (x^2 +y^2)

    W = [x^3 -(x^2)yi +2(x^2)yi +2xy^2 -xy^2 +(y^3)i +x -yi] / (x^2 +y^2)

    W = [x^3 +(x^2)yi +xy^2 +(y^3)i +x -yi] / (x^2 +y^2)

    Separate the imaginaries, or collect the reals and the imaginaries separately
    W = [(x^3 +xy^2 +x) +i(yx^2 +y^3 -y)] / (x^2 +y^2)

    So,
    Re(W) = (x^3 +xy^2 +x) /(x^2 +y^2)
    Im(W) = [(y^3 +yx^2 -y) /(y^2 +x^2)](i).

    Do long divisions on both and you'd arrive to the shown solution.
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  4. #4
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    Thanks for help.

    What about the 2nd problem?
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  5. #5
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    Quote Originally Posted by summerBoy View Post
    Find all solutions to the equation [where z=x+yi]:

    2z^{2}-3\overline{z}^{2}=10i

    It is a quadratic equation and has a conjugate of complex number in it.

    Thanks for help in advance.
    Important note: x and y are assumed to be real numbers!!!

    We have:
    2(x + iy)^2 - 3(x - iy)^2 = 10i

    2x^2 + 4ixy - 2y^2 - 3x^2 + 6ixy + 3y^2 = 10i

    (-x^2 + y^2 ) + i(10xy - 10) = 0

    Now, both the real and imaginary parts of this expression must be 0 so we have the system of equations:
    -x^2 + y^2 = 0
    10xy - 10 = 0

    I would solve the bottom equation for y:
    10xy = 10
    xy = 1
    y = 1/x

    and insert this into the top equation:
    -x^2 + (1/x)^2 = 0
    -x^4 + 1 = 0
    x^4 = 1

    Now, remember that x is a REAL number. There are only two real solutions to this equation:
    x = -1, 1

    So y = 1/x means that y = -1, 1 respectively.

    Thus your solution set is (-1, -1), (1, 1). Thus z = -1 - i and z = 1 + i. You can check these yourself.

    -Dan
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  6. #6
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    Thank you very much for your help.
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