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Complex functions - loci in the complex plane Watch

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    A transformation, T, maps the point P from the z plane to Q in the w plane, defined below.

     \displaystyle w = \frac{2-iz}{z}, z \in \mathbb{C}, z\neq 0

    What is the locus of the point P, if P is mapped to Q which has locus  arg(w)=\frac{ \pi }{3} .

    I got  \displaystyle arg \left(\frac{z+2i}{z} \right )=\frac{5 \pi}{6} .
    Can someone just quickly check this and tell me what answer they got. Would be appreciated.
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    (Original post by Ano123)
    A transformation, T, maps the point P from the z plane to Q in the w plane, defined below.

     \displaystyle w = \frac{2-iz}{z}, z \in \mathbb{C}, z\neq 0

    What is the locus of the point P, if P is mapped to Q which has locus  arg(w)=\frac{ \pi }{3} .

    I got  \displaystyle arg \left(\frac{z+2i}{z} \right )=\frac{5 \pi}{6} .
    Can someone just quickly check this and tell me what answer they got. Would be appreciated.
    well it is correct but is there a reason why you want it like this?
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    (Original post by TeeEm)
    well it is correct but is there a reason why you want it like this?
    What do you mean? The form that it is in?
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    (Original post by Ano123)
    What do you mean? The form that it is in?
    yes
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    (Original post by TeeEm)
    yes
    I think it probably the easiest to sketch in this form.
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    (Original post by Ano123)
    I think it probably the easiest to sketch in this form.
    it is
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    (Original post by TeeEm)
    it is
    (Original post by Ano123)
    I think it probably the easiest to sketch in this form.
    Hi, sorry to hijack your thread but i have a question for you both

    So i was trying to sketch  \displaystyle \displaystyle arg \left(\frac{z+2i}{z} \right )=\frac{5 \pi}{6} but give up and went to wolfham alpha to see it, only to find it was a line

    usually my method is to pick some values for the argument and find where the lines meet. An example below;

    (note: i will quote angles in degrees rather than radians, just because i cba latexing it, i know its wrong and im sorry :spank:)

    I would do arg(z) = 0 therefore arg(z-2i) = -45, draw the lines and "X" where they cross
    arg(z) = 60 therefore arg(z-2i) = 15 and do the same etc until I see the part circle which i can effectively dot to dot



    however unless i am being really dumb this method cant be applied to OPs question, so what other (and probably better) method is there for these questions?

    thanks, and once again sorry for hijacking OP
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    (Original post by DylanJ42)
    Hi, sorry to hijack your thread but i have a question for you both

    So i was trying to sketch  \displaystyle \displaystyle arg \left(\frac{z+2i}{z} \right )=\frac{5 \pi}{6} but give up and went to wolfham alpha to see it, only to find it was a line

    usually my method is to pick some values for the argument and find where the lines meet. An example below;

    (note: i will quote angles in degrees rather than radians, just because i cba latexing it, i know its wrong and im sorry :spank:)

    I would do arg(z) = 0 therefore arg(z-2i) = -45, draw the lines and "X" where they cross
    arg(z) = 60 therefore arg(z-2i) = 15 and do the same etc until I see the part circle which i can effectively dot to dot



    however unless i am being really dumb this method cant be applied to OPs question, so what other (and probably better) method is there for these questions?

    thanks, and once again sorry for hijacking OP
    these are arcs of circles
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    (Original post by TeeEm)
    these are arcs of circles
    yep.. is this it?


    but is there a better method, my current method seems very unreliable
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    (Original post by DylanJ42)
    yep.. is this it?


    but is there a better method, my current method seems very unreliable
    Look up the "arc of circles" video On examsolutions under complex loci
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    (Original post by DylanJ42)
    Hi, sorry to hijack your thread but i have a question for you both

    So i was trying to sketch  \displaystyle \displaystyle arg \left(\frac{z+2i}{z} \right )=\frac{5 \pi}{6} but give up and went to wolfham alpha to see it, only to find it was a line

    usually my method is to pick some values for the argument and find where the lines meet. An example below;

    (note: i will quote angles in degrees rather than radians, just because i cba latexing it, i know its wrong and im sorry :spank:)

    I would do arg(z) = 0 therefore arg(z-2i) = -45, draw the lines and "X" where they cross
    arg(z) = 60 therefore arg(z-2i) = 15 and do the same etc until I see the part circle which i can effectively dot to dot



    however unless i am being really dumb this method cant be applied to OPs question, so what other (and probably better) method is there for these questions?

    thanks, and once again sorry for hijacking OP
    You will start at -2i, then draw an arc of a circle (minor arc as angle is greater than  \pi/2 ) anticlockwise to the origin, o+oi. As Zacken has done.
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    (Original post by DylanJ42)
    yep.. is this it?


    but is there a better method, my current method seems very unreliable
    the geometric approach is pants, but in exams will be adequate
    I was educated to use algebra to obtain the equation of the circle, and then get the arc from algebraic constraints
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    (Original post by Zacken)
    Look up the "arc of circles" video On examsolutions under complex loci
    i will do once I start FP2 revision again, it has been too long since I done it

    (Original post by B_9710)
    You will start at -2i, then draw an arc of a circle (minor arc as angle is greater than  \pi/2 ) anticlockwise to the origin, o+oi.
    and that's all you do?

    so in my example above;  \displaystyle \arg(z) - \arg(z-2i) = \frac{\pi}{4} you start at (0,0) and draw an arc to 2i, go anticlockwise and its major because angle is less than  \displaystyle \frac{\pi}{2} ?

    and im guessing if the angle is minus you do a clockwise arc? with minor if |angle| > pi/2 and major if |angle| < pi/2?
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    (Original post by TeeEm)
    the geometric approach is pants, but in exams will be adequate
    I was educated to use algebra to obtain the equation of the circle, and then get the arc from algebraic constraints
    damn right, it's very tedious

    I will educate myself to use algebra, it seems much better

    thank you
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    (Original post by DylanJ42)
    ...
    Sounds like you got it, yep.
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    (Original post by Zacken)
    Sounds like you got it, yep.
    wtf :laugh: that's so much easier than my method, thanks a lot
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    (Original post by DylanJ42)
    wtf :laugh: that's so much easier than my method, thanks a lot
    You understand why we start at (0,0) in your case though, right? If it was arg(z -(a+ib)) - arg(blah) you'd start from (a,b).
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    The equation of the circle that the arc lies on is
     \displaystyle (x+\sqrt3)^2+(y+1)^2=4 (please correct me if I have made a mistake).

    Let's let the arc, C be defined by the parametric equations
     x=2\sin t-\sqrt3 and  y=2\cos t-1 , for  \pi/3 \leqslant t \leqslant 2\pi/3 .
    That's the arc in the x-y plane.
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    (Original post by Zacken)
    You understand why we start at (0,0) in your case though, right? If it was arg(z -(a+ib)) - arg(blah) you'd start from (a,b).
    if this is right then im all good

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    (Original post by DylanJ42)
    if this is right then im all good

    Correct.
 
 
 
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