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  1. #1
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    complex number

    If z is a complex number such that modulus of z is 1 , find the real part of $\displaystyle \frac{1}{1-z}$
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    Hello,
    Quote Originally Posted by thereddevils View Post
    If z is a complex number such that modulus of z is 1 , find the real part of $\displaystyle \frac{1}{1-z}$
    If the modulus of z is 1, this means that $\displaystyle z\bar z=1$, where $\displaystyle \bar z$ is the complex conjugate of z ($\displaystyle \overline{x+iy}=x-iy$)

    Property :
    $\displaystyle \forall a\in\mathbb{R} ~,~ \overline{z+a}=\bar z+a$
    (that's because the conjugate of a sum is the sum of the conjugates, and because the conjugate of a real number is itself)


    Thus the conjugate of $\displaystyle 1-z$ is $\displaystyle 1-\bar{z}$

    Multiply by the conjugate of the denominator :

    $\displaystyle \frac{1-\bar{z}}{(1-z)(1-\bar z)}=\frac{1-\bar z}{1+z\bar z-(z+\bar z)}=\frac{1-\bar z}{2-2 \Re (z)}$

    ($\displaystyle \Re$ denotes the real part)

    so finally, the real part of $\displaystyle \frac{1}{1-z}$ is $\displaystyle \frac{1-\Re(z)}{2(1-\Re(z))}=\frac 12$


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  3. #3
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    Quote Originally Posted by thereddevils View Post
    If z is a complex number such that modulus of z is 1 , find the real part of $\displaystyle \frac{1}{1-z}$
    If $\displaystyle z = x + iy$ and $\displaystyle |z| = 1$, then

    $\displaystyle \sqrt{x^2 + y^2} = 1$

    $\displaystyle x^2 + y^2 = 1$

    Let's keep that in mind...


    Now let's work on the $\displaystyle \frac{1}{1 - z}$.

    $\displaystyle \frac{1}{1 - z} = \frac{1}{1 - x - iy}$

    $\displaystyle = \frac{1}{1-x-iy}\times\frac{1-x + iy}{1 - x + iy}$

    $\displaystyle = \frac{1 - x + iy}{(1-x)^2 + y^2}$

    $\displaystyle = \frac{1 - x}{(1 - x)^2 + y^2} + i\frac{y}{(1 - x)^2 + y^2}$.


    So the real part of $\displaystyle \frac{1}{1 - z}$ is $\displaystyle \frac{1 - x}{(1 - x)^2 + y^2}$, provided that $\displaystyle x^2 + y^2 = 1$


    $\displaystyle = \frac{1 - x}{1 - 2x + x^2 + y^2}$

    $\displaystyle = \frac{1 - x}{1 - 2x + 1}$ since $\displaystyle x^2 + y^2 = 1$

    $\displaystyle = \frac{1 - x}{2 - 2x}$

    $\displaystyle = \frac{1 - x}{2(1 - x)}$

    $\displaystyle = \frac{1}{2}$.



    So the Real part of $\displaystyle \frac{1}{1 - z}$ is $\displaystyle \frac{1}{2}$.
    Last edited by Prove It; Jun 3rd 2009 at 03:20 AM.
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