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Fp2 chapter 3 Loci of complex numbers question Watch

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    Hi, I'm having some problems with the questions in fp2, an example of one would be:
    sketch the locus of z when:arg (z-3i) - arg(z-5) = pi/4
    in the example in the book, it says to do:

    let arg (z-3i) = theta (i cant find the symbol)
    arg (z-5) = alpha

    theta - alpha = pi/4

    then draw both individually on an argand diagram, and the loci is the circle round the points where they intersect.

    In mostly okay with that I think, but what I don't understand is sometimes the line they draw for arg (z- whatever) has a positive gradient, and sometimes it has a negative gradient - i.e. sometimes they take theta to be obtuse and sometimes they take it to be accute, and the same for alpha.
    How do you know whether theta/alpha is accute or obtuse, or when to use negative theta/alpha on your diagram?

    Sorry I know I havent explained it very well but can anyone help?
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    (Original post by Emma09)
    Hi, I'm having some problems with the questions in fp2, an example of one would be:
    sketch the locus of z when:arg (z-3i) - arg(z-5) = pi/4
    in the example in the book, it says to do:

    let arg (z-3i) = theta (i cant find the symbol)
    arg (z-5) = alpha

    theta - alpha = pi/4

    then draw both individually on an argand diagram, and the loci is the circle round the points where they intersect.

    In mostly okay with that I think, but what I don't understand is sometimes the line they draw for arg (z- whatever) has a positive gradient, and sometimes it has a negative gradient - i.e. sometimes they take theta to be obtuse and sometimes they take it to be accute, and the same for alpha.
    How do you know whether theta/alpha is accute or obtuse, or when to use negative theta/alpha on your diagram?

    Sorry I know I havent explained it very well but can anyone help?
    What the book is basically saying is:

    let  arg(z - 3i) = \theta and  arg(z - 5) = \phi

    now if you try to draw it such that  \theta - \phi = \frac{\pi}{4} you will see it wont be possible, as  \phi > \theta (so the angle made by the two half lines will never be pi/4) hence let  arg(z - 3i) = -\theta and  arg(z - 5) = -\phi (then  -\theta - (-\phi) = \frac{\pi}{4} which works as  \phi > \theta ) then you will see that it works, just draw it so that the angle made by the half lines is pi/4 and you will see how it works
 
 
 
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