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Math Help - Complex numbers help please?

  1. #1
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    Complex numbers help please?

    Can somebody please help me with this problem?

    ==================================

    Find the complex number z that satisfies the equation:

    \frac {25}{z} - \frac {15}{\bar z} = 1 - 8i

    Given \bar z is the conjugate of z and |z| = 2\sqrt{5}

    ==================================

    This is what I've done so far:

    Let z = a + bi
    Therefore: \bar z = a - bi

    \sqrt {a^2 + b^2} = 2\sqrt{5}

    a^2 + b^2 = 20

    a^2 = 20 - b^2

    a = \sqrt {20 - b^2}

    ---

    \frac {25}{a + bi} - \frac {15}{a - bi} = 1 - 8i


    \frac {25}{\sqrt {20 - b^2} + bi} - \frac {15}{\sqrt {20 - b^2} - bi} = 1 - 8i

    -------------------------------------------------

    Should I continue or should I stop before I start stuffing everything up? Is this correct? If not, could you please show me how to continue on?

    Thank you - all help is of course, appreciated.
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by sqleung View Post
    Can somebody please help me with this problem?

    ==================================

    Find the complex number z that satisfies the equation:

    \frac {25}{z} - \frac {15}{\bar z} = 1 - 8i

    Given \bar z is the conjugate of z and |z| = 2\sqrt{5}

    ==================================

    This is what I've done so far:

    Let z = a + bi
    Therefore: \bar z = a - bi

    \sqrt {a^2 + b^2} = 2\sqrt{5}

    a^2 + b^2 = 20

    a^2 = 20 - b^2

    a = \sqrt {20 - b^2}

    ---

    \frac {25}{a + bi} - \frac {15}{a - bi} = 1 - 8i


    \frac {25}{\sqrt {20 - b^2} + bi} - \frac {15}{\sqrt {20 - b^2} - bi} = 1 - 8i

    -------------------------------------------------

    Should I continue or should I stop before I start stuffing everything up? Is this correct? If not, could you please show me how to continue on?

    Thank you - all help is of course, appreciated.
    Correct method, but I think you can shorten it a lot

    Let z=a+ib and \overline{z}=a-ib.


    \frac{25}{z}-\frac{15}{\overline{z}}=1-8i

    We know that z\overline{z}=|z|^2. This is the key of the trick. You can check it, it's not difficult

    So the reflex will be to convert to a common denominator the LHS :

    \frac{25 \overline{z}-15 z}{z \overline{z}}=1-8i

    z\overline{z}=|z|^2=(2 \sqrt{5})^2=20


    Thus, the equation is now :

    \frac{25 \overline{z}-15z}{20}=1-8i

    Substituting z and \overline{z}, and simplifying :

    \frac{10a-40ib}{20}=1-8i

    \frac 12 a-2ib=1-8i


    Equate the real parts together, and the imaginary parts together and you will be done
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  3. #3
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    Thank you very much for your answer!

    It was just what I needed!
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