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    Heya! Still on my maths
    If i've been told arg (z/z-2)=pi/4
    i'm asked to draw the locus of this
    so i said z=x+iy
    then did arg z-arg z-2=pi/4
    then would the locus look like the attatched file?
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    (Original post by Goldenratio)
    Heya! Still on my maths
    If i've been told arg (z/z-2)=pi/4
    i'm asked to draw the locus of this
    so i said z=x+iy
    then did arg z-arg z-2=pi/4
    then would the locus look like the attatched file?
    I don't think it does. arg(z+2) would look like that diagram.

    Thing is, I never figured out how those worked. Apparently the locus is a circle which passes through certain points. If you have a P6 Heinemann textbook, have a look at the end of the Complex Numbers chapter; otherwise I'm sure someone else will be able to tell you.
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    Cheers. Yeh it's supposed to be a circle with point z the outer pount, with z1 and Z2 the points in the opposite segment. Then with a pizza slice out of the circle between all three points. But if z=x+iy then what would arg z be? 0 rads?
    arg z+2 = arg (x-2)+iy? but then how do i find out x?
    From what i can see arctan(y/x)-arctan(y/x-2)=pi/4
    so tanpi/4=(y/x)-(y/x-2)
    so 1=(y/x)-(y/x-2)
    Is this so?
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    fraid not as tan(A-B) doesn't equal tanA - tanB

    and why does pi/4 become pi/2 - and tan(pi/2) is infinite? :confused:
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    sorry my mistake
    pi/4 all the way.
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    so what do i do with it?
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    (Original post by Goldenratio)
    so what do i do with it?
    Apply the tan(A-B) formula correctly to

    arg z-arg z-2=pi/4

    noting tan(argz) = Imz/Rez
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    okay
    then I take tan(A-B)=1
    so
    (z-(z-2))/(1+z(z-2))=1
    2/(z^2-2z+1)=1
    (z-1)^2-1-1=0
    z=1+-sqrt2
    but then i'm left with an equation with two variables: namely
    x+iy=1+-sqrt2
    correct?
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    (Original post by Goldenratio)
    okay
    then I take tan(A-B)=1
    so
    (z-(z-2))/(1+z(z-2))=1
    2/(z^2-2z+1)=1
    (z-1)^2-1-1=0
    z=1+-sqrt2
    but then i'm left with an equation with two variables: namely
    x+iy=1+-sqrt2
    correct?
    No as tan(argz) isn't z - it's y/x
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    even so, it still leaves me with 1 equation with both x and y.
    [math]\frac {\frac{y}{x}-\frac {y}{(x-2)}}{1+\frac{y}{x}\frac{y}{x-2}}[/math]=[math]1[/math]
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    ah no LaTeX.
    okay. y/x-y/(x-2)
    -----------=1
    1+x/y(x/x-2)
    so
    y(x-2)-xy 1+(x/x-2)
    ---------=
    x(x-2)

    => (xy-2y-yx)(x-2)=(2x-2)x(x-2)
    therefore -2y=2x-2
    so y=1-x
    How does this help me?
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    \frac {\frac{y}{x}-\frac {y}{(x-2)}}{1+\frac{y}{x}\frac{y}{x-2}}=1
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    thanks for that Gaz
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    I think the point you're missing is that this isn't going to be a single point.

    It will actually define the arc of a circle
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    Yes Riche, i do know it will difine the acr of a circle, but to find this circle, i first need another point on its circumference which is neither z1 nor z2.
    I'm only asking you to help me with the method, as i'm completely new to complex numbers (started last week). After which i should be able to answer the rest of the questions in the excercise.

    Thank you for your very kind help so far.
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    (Original post by Goldenratio)
    Yes Riche, i do know it will difine the acr of a circle, but to find this circle, i first need another point on its circumference which is neither z1 nor z2.
    I'm only asking you to help me with the method, as i'm completely new to complex numbers (started last week). After which i should be able to answer the rest of the questions in the excercise.

    Thank you for your very kind help so far.
    You've managed to understand everything you need to know on complex numbers in a week!? Wow! It took me ages to get all the individual bits (indeed I wasn't sure on the nth roots of unity until an hour before my P6 exam! Such a shame that it didn't appear on the paper)
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    (Original post by Goldenratio)
    Yes Riche, i do know it will difine the acr of a circle, but to find this circle, i first need another point on its circumference which is neither z1 nor z2.
    I'm only asking you to help me with the method, as i'm completely new to complex numbers (started last week). After which i should be able to answer the rest of the questions in the excercise.

    Thank you for your very kind help so far.
    Well I didn't want to give you the answer straight up. But if you take Gaz's equation

    [(y/x) - (y/(x-2))]/[1+(y/x)(y/(x-2))] = 1

    you can rearrange it into something of the form

    (x-a)^2 + (y-b)^2 = r^2
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    No, no i only have some basic grasp of them. Indeed only the first chapter, of which there are two pages left :P yipee. But i do see that there's another one, called further complex numbers, coming up in a but though. hehe
    Pitty that the nth root didn't come up for you. hate it when that happens lol.
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    (Original post by Goldenratio)
    No, no i only have some basic grasp of them. Indeed only the first chapter, of which there are two pages left :P yipee. But i do see that there's another one, called further complex numbers, coming up in a but though. hehe
    Pitty that the nth root didn't come up for you. hate it when that happens lol.
    I'm not complaining. I didn't understand the differential equations bit in my P6 until after I understood the nth roots, and that came up . I was thinking "Ahh yes... Let's whip this big boy out onto the page".

    If I were you, I'd go back and get more of the basics licked. Don't do this further complex numbers until you know what you've learnt inside out. From my experience, there are lots of little fiddly bits which are really easy to forget.
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    Ahhhh! i see i made a silly mistake...
    so now i have
    \frac{-2y}{x^2-2x+y^2}=1
    which as you have pointed out will give me the eq of a circle.
    (1,-1) radius 2
    THANK YOU!
 
 
 
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