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Math Help - [SOLVED] Is this answer and steps correct?

  1. #1
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    [SOLVED] Is this answer and steps correct?

    If not, please do correct me.

    The question is,

    The complex number z is given by z = 1 - i. Express z^2 - \frac{1}{z} in the form p+qi where p and q are real and find the modulus and argument of z^2 - \frac{1}{z}.

    I don't know, I just feel that something is not right here even though I've checked many times. Here's my answer:

    -1/2 - (5/2)i

    Ok. I've figured out how to use LaTex.

    The complex number z is given by z = 1 - i. Express z^2 - \frac{1}{z} in the form p+qi where p and q are real and find the modulus and argument of z^2 - \frac{1}{z}

    Here is how I expressed it in the form p + qi:

    = z^2 - \frac{1}{z}

    = (1 - i)^2 - \frac{1}{1 - i}

    = (1 - 2i - 1) - \frac{1}{1 - i}

    = \frac{(1 - 2i - 1)(1 - i) - 1}{1 - i}

    = \frac{(-2i - 2 - 1)}{1 - i}

    = \frac{(-2i - 2 - 1)}{1 - i}\;\text{x}\,\frac{1 + i}{1 + i}

    = \frac{(-3 - 2i)(1 + i)}{2}

    = \frac{-3 - 5i + 2}{2}

    = \frac{-1 - 5i}{2}

    = \text{-}\,\frac{1}{2}\;\text{-}\,\frac{5}{2}\;\text{i}

    =Correct or Wrong? Please help to evaluate! Thanks. If this is correct then that means my modulus and argument are also correct.
    Last edited by mark1950; June 3rd 2009 at 07:17 PM.
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  2. #2
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    try again
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  3. #3
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    Quote Originally Posted by mark1950 View Post
    If not, please do correct me.
    The question is,
    The complex number z is given by z = 1 - i. Express z^2 - 1/z in the form p+qi where p and q are real and find the modulus and argument of z^2 - 1/z.
    First, notice that your exponents did not show up in your OP.
    I added the correct LaTeX codes.
    Now it is not clear what you are asking.
    Is it  z^2  - \frac{1}<br />
{z}\;\text{ or } \,\frac{{z^2  - 1}}<br />
{z}
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  4. #4
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    Ok. I've figured out how to use LaTex.

    The complex number z is given by z = 1 - i. Express z^2 - \frac{1}{z} in the form p+qi where p and q are real and find the modulus and argument of z^2 - \frac{1}{z}

    Here is how I expressed it in the form p + qi:

    = z^2 - \frac{1}{z}

    = (1 - i)^2 - \frac{1}{1 - i}

    = (1 - 2i - 1) - \frac{1}{1 - i}

    = \frac{(1 - 2i - 1)(1 - i) - 1}{1 - i}

    = \frac{(-2i - 2 - 1)}{1 - i}

    = \frac{(-2i - 2 - 1)}{1 - i}\;\text{x}\,\frac{1 + i}{1 + i}

    = \frac{(-3 - 2i)(1 + i)}{2}

    = \frac{-3 - 5i + 2}{2}

    = \frac{-1 - 5i}{2}

    = \text{-}\,\frac{1}{2}\;\text{-}\,\frac{5}{2}\;\text{i}

    =Correct or Wrong? Please help to evaluate! Thanks. If this is correct then that means my modulus and argument are also correct.
    Last edited by mark1950; June 3rd 2009 at 07:11 PM.
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  5. #5
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    Quote Originally Posted by mark1950 View Post
    Ok. I've figured out how to use LaTex.

    The complex number z is given by z = 1 - i. Express z^2 - \frac{1}{z} in the form p+qi where p and q are real and find the modulus and argument of z^2 - \frac{1}{z}

    Here is how I expressed it in the form p + qi:

    = z^2 - \frac{1}{z}

    = (1 - i)^2 - \frac{1}{1 - i}

    = (1 - 2i - 1) - \frac{1}{1 - i}
    At this point I would combine like terms to simplify things: (1 - 2i - 1) = -2i:

    = -2i - \frac{1}{1 - i}

    Now combine:

    = \frac{-2i(1 - i) - 1}{1 - i}

    = \frac{-2i - 2 - 1}{1 - i}

    = \frac{-2i - 3}{1 - i}

    Now multiply the numerator & denominator by the denominator's conjugate:

    = \frac{-2i - 3}{1 - i}

    = \frac{-2i - 3}{1 - i} \times \frac{1 + i}{1 + i}

    = \frac{-2i + 2 - 3 -3i}{1 + 1}

    = \frac{-1 -5i}{2}

    = -\frac{1}{2} -\frac{5}{2}i

    We get the same answer nevertheless.

    The modulus of this complex number would be

    |z| = \sqrt{\left( -\frac{1}{2} \right)^2 + \left( -\frac{5}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{25}{4}} = \frac{\sqrt{26}}{2}

    The argument would be found by solving

    tan(\theta) = \frac{-5/2}{-1/2} = 5

    \theta = tan^{-1}(5) + 180\ deg (because the point would be in Quadrant III)

    \theta = 258.69\ deg


    01
    Last edited by yeongil; June 3rd 2009 at 10:06 PM.
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  6. #6
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    Oh, ok thanks!
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