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  1. Calculator Group Title
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    check |dw:1348800828515:dw| how do you draw the inequalities?

    • one year ago
  2. Calculator Group Title
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    @UnkleRhaukus

    • one year ago
  3. vf321 Group Title
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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.

    • one year ago
  4. Calculator Group Title
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    |dw:1348803360924:dw|

    • one year ago
  5. vf321 Group Title
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    correct, but not to scale

    • one year ago
  6. vf321 Group Title
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    im working on the ineqs btw, just wait plz

    • one year ago
  7. Calculator Group Title
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    k

    • one year ago
  8. vf321 Group Title
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    Okay. got it.

    • one year ago
  9. Calculator Group Title
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    can you draw it out?

    • one year ago
  10. vf321 Group Title
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    Yes. But ill just give u the eq., and you'll draw it.

    • one year ago
  11. vf321 Group Title
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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 \[|z-1|\le|z-i|\]\[|z-1|^2\le|z-i|^2\]\[|(a-1, b)|^2\le|(a, b-1)|^2\]\[(a-1)^2+b^2\le a^2+(b-1)^2\]

    • one year ago
  12. vf321 Group Title
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    Simplify that and you'll get a nice cartesian inequality

    • one year ago
  13. vf321 Group Title
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    Follow similar steps for the other one.

    • one year ago
  14. UnkleRhaukus Group Title
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    |dw:1348805078839:dw|

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

    • one year ago
  16. UnkleRhaukus Group Title
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    oh, i ment that

    • one year ago
  17. UnkleRhaukus Group Title
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    |dw:1348805469963:dw|

    • one year ago
  18. Calculator Group Title
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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!

    • one year ago
  19. UnkleRhaukus Group Title
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    hmm im not sure about the inequalities yet , ill think about them

    • one year ago
  20. Calculator Group Title
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    @Directrix

    • one year ago
  21. perl Group Title
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    how did you derive |z - 1 | < | z - i | ?

    • one year ago
  22. Directrix Group Title
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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

    • one year ago
  23. perl Group Title
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    see if you can solve the first question

    • one year ago
  24. perl Group Title
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    | x + 3a | > 2 | x - 2a | , a > 0 , solve

    • one year ago
  25. perl Group Title
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    i think you can do that in cases, case 1 : x > -3a

    • one year ago
  26. Calculator Group Title
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    for that you just square both sides and get answer

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

    • one year ago
  28. perl Group Title
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    ok

    • one year ago
  29. Calculator Group Title
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    \[-3x^2+22ax-7>0\]

    • one year ago
  30. perl Group Title
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    yes

    • one year ago
  31. perl Group Title
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    then use the quadratic formula?

    • one year ago
  32. Calculator Group Title
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    \[x=\frac{11}{3}a \pm\frac{\sqrt{547}}{3}\]

    • one year ago
  33. perl Group Title
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    what about the other ones, can you solve them

    • one year ago
  34. myko Group Title
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    |dw:1348823976443:dw| hint for 7 ii

    • one year ago
  35. Directrix Group Title
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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

    • one year ago
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  36. Calculator Group Title
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    |dw:1348824437459:dw| u could be there

    • one year ago
  37. Calculator Group Title
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    |dw:1348824571707:dw|

    • one year ago
  38. Calculator Group Title
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    @amistre64

    • one year ago
  39. amistre64 Group Title
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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 :)

    • one year ago
  40. phi Group Title
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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)

    • one year ago
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