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  1. Calculator
    • 2 years ago
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    check |dw:1348800828515:dw| how do you draw the inequalities?

  2. Calculator
    • 2 years ago
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    @UnkleRhaukus

  3. vf321
    • 2 years ago
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    2+2i will be at angle pi/2, but has a distance from the origin of \(2\sqrt 2\), not 1.

  4. Calculator
    • 2 years ago
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    |dw:1348803360924:dw|

  5. vf321
    • 2 years ago
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    correct, but not to scale

  6. vf321
    • 2 years ago
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    im working on the ineqs btw, just wait plz

  7. Calculator
    • 2 years ago
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    k

  8. vf321
    • 2 years ago
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    Okay. got it.

  9. Calculator
    • 2 years ago
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    can you draw it out?

  10. vf321
    • 2 years ago
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    Yes. But ill just give u the eq., and you'll draw it.

  11. vf321
    • 2 years ago
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    Denote imaginary number z as a vector (a, b) where a, b are reals. Note that this means that on Argand Diagram x = a and y = b. Note also that |z| = |(a,b)| = sqrt(a^2+b^2). Then since \[|z-1|\le|z-i|\]\[|z-1|^2\le|z-i|^2\]\[|(a-1, b)|^2\le|(a, b-1)|^2\]\[(a-1)^2+b^2\le a^2+(b-1)^2\]

  12. vf321
    • 2 years ago
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    Simplify that and you'll get a nice cartesian inequality

  13. vf321
    • 2 years ago
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    Follow similar steps for the other one.

  14. UnkleRhaukus
    • 2 years ago
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    |dw:1348805078839:dw|

  15. Calculator
    • 2 years ago
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    why is it pi/2? \[\arg ,~z=\tan^{-1}(\frac{b}{a})=\tan^{-1}\frac{2}{2}=\frac{\pi}{4}\]

  16. UnkleRhaukus
    • 2 years ago
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    oh, i ment that

  17. UnkleRhaukus
    • 2 years ago
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    |dw:1348805469963:dw|

  18. Calculator
    • 2 years ago
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    so how do you sketch the inequalities? i have to go now , can you leave some tips or answer later ill go through it. thanks!

  19. UnkleRhaukus
    • 2 years ago
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    hmm im not sure about the inequalities yet , ill think about them

  20. Calculator
    • 2 years ago
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    @Directrix

  21. perl
    • 2 years ago
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    how did you derive |z - 1 | < | z - i | ?

  22. Directrix
    • 2 years ago
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    We can't let this question pass without paying homage of Jean Robert Argand who was an amateur mathematician. His day job was that of an accountant. It is remarkable and inspirational that although this diagram is Argand's only contribution to mathematics, it is, nonetheless, a significant one. http://www-groups.dcs.st-and.ac.uk/history/Biographies/Argand.html

  23. perl
    • 2 years ago
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    see if you can solve the first question

  24. perl
    • 2 years ago
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    | x + 3a | > 2 | x - 2a | , a > 0 , solve

  25. perl
    • 2 years ago
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    i think you can do that in cases, case 1 : x > -3a

  26. Calculator
    • 2 years ago
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    for that you just square both sides and get answer

  27. Calculator
    • 2 years ago
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    \[x^2+6ax+9a^2 > 4(x^2-4ax+4a^2)\]

  28. perl
    • 2 years ago
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    ok

  29. Calculator
    • 2 years ago
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    \[-3x^2+22ax-7>0\]

  30. perl
    • 2 years ago
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    yes

  31. perl
    • 2 years ago
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    then use the quadratic formula?

  32. Calculator
    • 2 years ago
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    \[x=\frac{11}{3}a \pm\frac{\sqrt{547}}{3}\]

  33. perl
    • 2 years ago
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    what about the other ones, can you solve them

  34. myko
    • 2 years ago
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    |dw:1348823976443:dw| hint for 7 ii

  35. Directrix
    • 2 years ago
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    Spoilers/Hints/Maybe from Mark Scheme for Exam http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s10_ms_31.pdf

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  36. Calculator
    • 2 years ago
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    |dw:1348824437459:dw| u could be there

  37. Calculator
    • 2 years ago
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    |dw:1348824571707:dw|

  38. Calculator
    • 2 years ago
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    @amistre64

  39. amistre64
    • 2 years ago
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    im not to confident in what args are. Ive never really delved into the language for them to be sure that what i know is what i know :)

  40. phi
    • 2 years ago
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    |z-1| ≤ |z -i| if this were the equality |z-1| = |z -i| it can be interpreted as the points z equidistant from the points (1,0) and (0,1) on the complex plane. The locus of points equidistant from 2 points is the perpendicular bisector of the line joining the points. For the inequality |z-1| ≤ |z -i|, the locus is all points on and below the line defined by the points (0,0) and (1/2, 1/2) |z-u| ≤ 1 if an equality, the locus is the circle with radius 1 located at u for the inequality, it is the filled in circle. the intersection of the two constraints is the half circle with center 2+2i and radius 1, (the half region below the line going from (0,0) through (2,2)

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