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Thread: complex locus

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    complex locus

    given Arg(z-1) = 2Arg(z), deduce locus of z is a circle
    part ii. find z in mod arg form if z also satisfies |z| = |z-1|

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    Re: complex locus

    Since , the equation can be rearranged to



    Note that an argument of zero means that could be a positive real number , as and any real that is less than 1 produces a negative number with argument .

    This means there are restrictions on the real and negative parts of .

    Also, the identity might be useful.

    The locus is not a circle, because as stated before any satisfies the equation.
    Last edited by fan96; 3 Mar 2018 at 2:11 PM.

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    Re: complex locus

    for i) z-1/z^2 = a+0i, because arg is 0, therefore a number real. equate the imaginary part of the z-1/z^2 to 0 and get it into the form of a circle.
    for ii) i think it creates a straight line; so simultaneously solve the line and the circle to get the intended values.
    im too lazy to do it thou but thats how to solve it.
    Last edited by mrbunton; 25 Feb 2018 at 8:38 AM.

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    Re: complex locus

    Quote Originally Posted by fan96 View Post
    Since , the equation can be rearranged to



    Note that an argument of zero means that could be a positive real number , as and any real that is less than 1 produces a negative number with argument .

    This means there are restrictions on the real and negative parts of .

    Also, the identity might be useful.

    The locus is not a circle, because as stated before any satisfies the equation.
    Can you pleaes explain how you rearranged it to get
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    Re: complex locus



    Using ,



    Using ,

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    Re: complex locus

    Quote Originally Posted by fan96 View Post


    Using ,



    Using ,

    Ah thanks
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