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    (Original post by EricPiphany)
    Show that the locus of \displaystyle z=x+yi, \arg\left(\frac{z-3i}{z-5}\right)=\frac{\pi}{4}, has the cartesian equation \displaystyle (x-1)^2+(y+1)^2=17, subject to \displaystyle 5y+3x<15.
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    I think this is a little bit above the syllabus, so don't worry if you don't know how to do it. Also I may have made an error.
    \displaystyle

\begin{equation*} \frac{z-3i}{z-5} = \frac{x + (y-3)i}{(x-5) + iy} = \frac{(x +(y-3)i)((x-5) -iy) }{(x-5)^2 + y^2}

    Doing this all on LaTeX, forgive the excessive working:

    \displaystyle 

\begin{equation*}\frac{x(x-5) + y(y+3) - iyx + (y-3)(x-5)i}{(x-5)^2 + y^2} \end{equation*}

    Imaginary and real parts need to be identical:

    \displaystyle

\begin{equation*}x^2 - 5x + y^2 + 3y = -5y - 3x + 15\end{equation*}

    Completing the square:

    \displaystyle

\begin{equation*}(x-1)^2 + (y+1)^2 = 15 + 2 = 17\end{equation*}

    As required.

    Blah blah required conditions, pish posh. We require both the real and imaginary part to be positive, which gives us

    \displaystyle x(x-5) + y(y+3) > 0 and -5y - 3x  + 15> 0 \iff 5y + 3x < 15.

    I don't really see how this is above the syllabus. :dontknow:
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    (Original post by Zacken)
    \displaystyle

\begin{equation*} \frac{z-3i}{z-5} = \frac{x + (y-3)i}{(x-5) + iy} = \frac{(x +(y-3)i)((x-5) -iy) }{(x-5)^2 + y^2}

    Doing this all on LaTeX, forgive the excessive working:

    \displaystyle 

\begin{equation*}\frac{x(x-5) + y(y+3) - iyx + (y-3)(x-5)i}{(x-5)^2 + y^2} \end{equation*}

    Imaginary and real parts need to be identical:

    \displaystyle

\begin{equation*}x^2 - 5x + y^2 + 3y = -5y - 3x + 15\end{equation*}

    Completing the square:

    \displaystyle

\begin{equation*}(x-1)^2 + (y+1)^2 = 15 + 2 = 17\end{equation*}

    As required.

    Blah blah required conditions, pish posh. We require both the real and imaginary part to be positive, which gives us

    \displaystyle x(x-5) + y(y+3) > 0 and -5y - 3x  + 15> 0 \iff 5y + 3x < 15.

    I don't really see how this is above the syllabus. :dontknow:
    Good. I feel bad you put so much work into the latex
    They just don't ask these types of problem in the textbook.
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    (Original post by EricPiphany)
    Good. I feel bad you put so much work into the latex
    They just don't ask these types of problem in the textbook.
    You're nicer than I am! Had I seen this problem, I'd have asked: Show that \arg \frac{z-ai}{z-b} = \frac{\pi}{4} has cartesian equation blah subject to by + ax < ab :rofl:
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    (Original post by Zacken)
    You're nicer than I am! Had I seen this problem, I'd have asked: Show that \arg \frac{z-ai}{z-b} = \frac{\pi}{4} has cartesian equation blah subject to by + ax < ab :rofl:
    Go for the full \displaystyle \arg\left(\frac{z-(a+bi)}{z-(c+di)}\right)=\theta. lol
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    (Original post by EricPiphany)
    Go for the full \displaystyle \arg\left(\frac{z-(a+bi)}{z-(c+di)}\right)=\theta. lol
    I wouldn't mind doing that one, but there's no way I'm doing it algebraically. :lol:
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    (Original post by Zacken)
    I wouldn't mind doing that one, but there's no way I'm doing it algebraically. :lol:
    At least it's one for all. You might have to make cases for which quadrant θ is in.
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    DONT DO IT LOL
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    Hi, could anyone explain how to do question 3b on this paper please? https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf I'm confused how the modulus changes things? Thanks
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    (Original post by economicss)
    Hi, could anyone explain how to do question 3b on this paper please? https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf I'm confused how the modulus changes things? Thanks
    Sketch the graph and you'll see
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    Flopped my FP2 mock today :lol:
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    (Original post by Student403)
    Flopped my FP2 mock today :lol:
    :console:

    Sorry about your 74
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    (Original post by Kvothe the arcane)
    :console:

    Sorry about your 74
    No seriously :rofl: I ain't Zain lol

    Reason being we do a joint FP2-FP3 (I think the maths department wanted to kill us) and it's impossible to revise for both of them the night before :cry2:
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    (Original post by Student403)
    Flopped my FP2 mock today :lol:
    It'll be alright on the night
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    (Original post by SeanFM)
    It'll be alright on the night
    Hope so mate - thanks

    How was your test today?
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    (Original post by Student403)
    Hope so mate - thanks

    How was your test today?
    :lol: good memory, PRSOM (I see we are both really close to 11 gems...)

    It was alright, not sure if I've met my GYG goal but we shall see :borat:
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    (Original post by SeanFM)
    :lol: good memory, PRSOM (I see we are both really close to 11 gems...)

    It was alright, not sure if I've met my GYG goal but we shall see :borat:
    I'm sure it was alright!

    And holy cow.. Yeah :eek:
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    Please can anyone explain why the limits aren't between 0 and pi/6 on question 4b of this paper https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf and how you know to subtract a shape from the area? Thanks
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    (Original post by economicss)
    Please can anyone explain why the limits aren't between 0 and pi/6 on question 4b of this paper https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf and how you know to subtract a shape from the area? Thanks
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    (Original post by Zacken)
    \displaystyle

\begin{equation*} \frac{z-3i}{z-5} = \frac{x + (y-3)i}{(x-5) + iy} = \frac{(x +(y-3)i)((x-5) -iy) }{(x-5)^2 + y^2}

    Doing this all on LaTeX, forgive the excessive working:

    \displaystyle 

\begin{equation*}\frac{x(x-5) + y(y+3) - iyx + (y-3)(x-5)i}{(x-5)^2 + y^2} \end{equation*}

    Imaginary and real parts need to be identical:

    \displaystyle

\begin{equation*}x^2 - 5x + y^2 + 3y = -5y - 3x + 15\end{equation*}

    Completing the square:

    \displaystyle

\begin{equation*}(x-1)^2 + (y+1)^2 = 15 + 2 = 17\end{equation*}

    As required.

    Blah blah required conditions, pish posh. We require both the real and imaginary part to be positive, which gives us

    \displaystyle x(x-5) + y(y+3) > 0 and -5y - 3x  + 15> 0 \iff 5y + 3x < 15.

    I don't really see how this is above the syllabus. :dontknow:
    Whaaat is this in FP2? tbh we have done all the chapters except end of 3 this looks horrible

    edit I just went through it again and its not that bad
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    (Original post by Cpj16)
    Whaaat is this in FP2? tbh we have done all the chapters except end of 3 this looks horrible

    edit I just went through it again and its not that bad
    Glad to hear that, it seems fairly standard to me but mileage varies; don't be worried if you weren't able to do it.
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    Hi can someone pls explain this integration, I don't understand it

    thanks
 
 
 
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