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    The z-plane is mapped onto the w-plane by the transformation w=z+1/z . Given that z lies on the circle |z|=1, show that w lies on an interval of the real axis. Identify this interval precisely.

    I've done:
    w=(z^2 +1 )/z
    |w|=|z^2 +1|/|z|
    |w|=|z^2 +1|

    Not sure what to do next
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    (Original post by bobbricks)
    The z-plane is mapped onto the w-plane by the transformation w=z+1/z . Given that z lies on the circle |z|=1, show that w lies on an interval of the real axis. Identify this interval precisely.

    I've done:
    w=(z^2 +1 )/z
    |w|=|z^2 +1|/|z|
    |w|=|z^2 +1|

    Not sure what to do next
    I'd take another approach.

    \dfrac{1}{z}=\dfrac{1}{x+iy}= \dfrac{x-iy}{x^2+y^2}=?
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    (Original post by BuryMathsTutor)
    I'd take another approach.

    \dfrac{1}{z}=\dfrac{1}{x+iy}= \dfrac{x-iy}{x^2+y^2}=?
    So 1/z =x-iy as |z|=x^2 + y^2 =1 ?

    So |w|=x-iy+1 ...?
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    (Original post by bobbricks)
    So 1/z =x-iy as |z|=x^2 + y^2 =1 ?

    So |w|=x-iy+1 ...?

    Yes, 1/z=x-iy.

    So what is z +\dfrac{1}{z}?
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    (Original post by BuryMathsTutor)
    Yes, 1/z=x-iy.

    So what is z +\dfrac{1}{z}?
    Z+1/Z = 2x

    so |w|=2|x| ...?
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    (Original post by bobbricks)
    Z+1/Z = 2x

    so |w|=2|x| ...?

    and -1 \le x \le 1.
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    (Original post by BuryMathsTutor)
    and -1 \le x \le 1.
    How do you know that?
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    (Original post by bobbricks)
    How do you know that?
    Your question said: "Given that z lies on the circle |z|=1".
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    (Original post by bobbricks)
    The z-plane is mapped onto the w-plane by the transformation w=z+1/z . Given that z lies on the circle |z|=1, show that w lies on an interval of the real axis. Identify this interval precisely.

    I've done:
    w=(z^2 +1 )/z
    |w|=|z^2 +1|/|z|
    |w|=|z^2 +1|

    Not sure what to do next

    although this particular question can be done in Cartesian
    this type of problem is solved by the substitution z = e, which represent a unit circle
 
 
 
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