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Thread: 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 $\displaystyle z$ that satisfies the equation:

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

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

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

    This is what I've done so far:

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

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

    $\displaystyle a^2 + b^2 = 20$

    $\displaystyle a^2 = 20 - b^2$

    $\displaystyle a = \sqrt {20 - b^2}$

    ---

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


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

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

    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 $\displaystyle z$ that satisfies the equation:

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

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

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

    This is what I've done so far:

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

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

    $\displaystyle a^2 + b^2 = 20$

    $\displaystyle a^2 = 20 - b^2$

    $\displaystyle a = \sqrt {20 - b^2}$

    ---

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


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

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

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


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

    We know that $\displaystyle z\overline{z}=|z|^2$. 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 :

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

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


    Thus, the equation is now :

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

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

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

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


    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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