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Math Help - Complex number questions

  1. #1
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    Complex number questions

    It seems complex numbers has just gotten a new standard of difficulty. I'am unable to post my working at the moment due to not having any start with the Maths Syntax used here. The question as a whole is connected in a manner that I can't solve the later parts getting done with the first part so please help me out.

    The variable complex number z is given b:

    z = 2cos@ + i(1-2sin@).

    where @ takes all values in the interval pi < @ <= pi.

    a, Show that lz-il = 2 for all values of @. Hence sketch, in an Argand Diagram, the locus of the point representing z.

    b, Prove that the real part of 1/z+2-i is constant for -pi < @ < pi
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  2. #2
    MHF Contributor

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    For part a, note that z - i = 2\cos \left( \Theta \right) - i2\sin \left( \Theta \right).

    Part b needs a bit more.
    \Re e\left( {\frac{1}{w}} \right) = \frac{{\Re e(w)}}<br />
{{\left| w \right|^2 }}
    So \Re e\left( {\frac{1}{{z + 2 - i}}} \right) = \frac{{2\cos \left( \Theta \right) + 2}}{{8 + 8\cos \left( \Theta \right)}}
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  3. #3
    Junior Member Ziaris's Avatar
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    a. z=2(\cos\theta-i\sin\theta)+i, and given that |z|=\sqrt{Re^2(z) + Im^2(z)} the modulus of z in our case is

    |z-i|=\sqrt{4}\sqrt{\cos^2\theta+\sin^2\theta}=\sqrt{  4}\sqrt{1}=2. Therefore |z-i| is constant at 2.
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  4. #4
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    Thank you
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