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    Question:
    "z=x+iy
    |z-(3+2i)|<|z-6|"

    Can someone please explain how I would go about solving this step by step? I can't find a worked example online to try it with this question.

    Cheers

    Edit: I've got 6x-4y-23<0 after inputting the value of z=x+iy into the equation and replacing the modulus with squaring and rooting. Firstly I don't know whether this is correct and secondly, do I need to anything else to solve it or do I leave it like this?
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    Substitute z=x+iy (x, y \in \mathbb R), ie.

    |x+iy-(3+2i)|&lt;|x+iy-6|

    It may help you to write this as,

    |(x-3)+(y-2)i| &lt; | (x-6)+iy|
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    (Original post by MathsMeme)
    Question:
    "z=x+iy
    |z-(3+2i)|<|z-6|"

    Can someone please explain how I would go about solving this step by step? I can't find a worked example online to try it with this question.

    Cheers
    I recommend this video and then watch example 2 in this video.

    An explanation of this topic on a forum is hard to do and may not be very useful.
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    I think you are correct ( 6x-4y-23<0 ), You can double check by drawing the graph and I don’t think there’s more thing you can do.
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    (Original post by MathsMeme)
    Edit: I've got 6x-4y-23<0 after inputting the value of z=x+iy into the equation and replacing the modulus with squaring and rooting. Firstly I don't know whether this is correct and secondly, do I need to anything else to solve it or do I leave it like this?
    Can you please post the whole question?
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    (Original post by Notnek)
    Can you please post the whole question?
    Name:  Screenshot_20171009-114018.jpg
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    (Original post by MathsMeme)
    Question:
    "z=x+iy
    |z-(3+2i)|<|z-6|"

    Can someone please explain how I would go about solving this step by step? I can't find a worked example online to try it with this question.
    |z-a| is the distance from fixed complex number a to arbitrary complex number z in the complex plane. So geometrically, the question wants you to find the region of the plane where all the complex numbers (i.e. all the points in the region) are nearer to 3+2i than to 6.
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    (Original post by MathsMeme)
    Name:  Screenshot_20171009-114018.jpg
Views: 58
Size:  234.0 KB
    I'm not actually sure what your teacher means by "solve" in this question. Do you know if you are meant to sketch the locus?
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    May I quickly ask which FP unit this is from?
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    (Original post by W. A. Mozart)
    May I quickly ask which FP unit this is from?
    For Edexcel old (modular) spec this is FP2.
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    (Original post by Notnek)
    For Edexcel old (modular) spec this is FP2.
    Ah thank you! I was getting a little worried as I hadn’t seeing inequalities involving complex numbers yet in FP1 (AQA).
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    (Original post by atsruser)
    |z-a| is the distance from fixed complex number a to arbitrary complex number z in the complex plane. So geometrically, the question wants you to find the region of the plane where all the complex numbers (i.e. all the points in the region) are nearer to 3+2i than to 6.
    Agreed. I think it's important for students to realise that many (most) complex number loci questions are essentially geometry exercises in disguise.
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    If your teacher is asking for the (x,y) that satisfy the inequality, your answer is correct. I doubt that they're looking for a sketch from the steps that they have provided, and the use of the word "solve".
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    (Original post by _gcx)
    If your teacher is asking for the (x,y) that satisfy the inequality, your answer is correct. I doubt that they're looking for a sketch from the steps that they have provided, and the use of the word "solve".
    Yes looking again that's probably what they want. I'm sure a textbook/exam question would be much clearer.
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    To everyone who has made an effort to help me, thank you very much. I went and saw my teacher and she said what I did was correct, just rearrange for y and then I'm done.

    Cheers again!
 
 
 
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