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    complex number

    If z is a complex number such that modulus of z is 1 , find the real part of \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 \frac{1}{1-z}
    If the modulus of z is 1, this means that z\bar z=1, where \bar z is the complex conjugate of z ( \overline{x+iy}=x-iy)

    Property :
    \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 1-z is 1-\bar{z}

    Multiply by the conjugate of the denominator :

    \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)}

    ( \Re denotes the real part)

    so finally, the real part of \frac{1}{1-z} is \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 \frac{1}{1-z}
    If z = x + iy and |z| = 1, then

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

    x^2 + y^2 = 1

    Let's keep that in mind...


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

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

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

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

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


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


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

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

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

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

     = \frac{1}{2}.



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