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    l z+6 l > l z+2+8i l, what is the method to sketch these?
    if you equal them, the cartesian would be y = 0.5x - 2
    so obviously you'd draw this line, then do you sub z=0 or something? i dont get that.
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    (Original post by cooldudeman)
    l z+6 l > l z+2+8i l, what is the method to sketch these?
    if you equal them, the cartesian would be y = 0.5x - 2
    so obviously you'd draw this line, then do you sub z=0 or something? i dont get that.
    I think it's fairly clear that the point (-2,-8) will be in the desired reqion as that makes the RHS of the inequality zero, whereas the LHS will be >0.
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    (Original post by ghostwalker)
    I think it's fairly clear that the point (-2,-8) will be in the desired reqion as that makes the RHS of the inequality zero, whereas the LHS will be >0.
    so if it satisfies the inequality, i shade the region? or the opposite. book is making it confusing.
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    (Original post by cooldudeman)
    so if it satisfies the inequality, i shade the region? or the opposite. book is making it confusing.
    Shade the region.

    Note that since this is a strict inequality, the dividing line is not included, so you should put it in as a dashed line.
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    (Original post by ghostwalker)
    Shade the region.

    Note that since this is a strict inequality, the dividing line is not included, so you should put it in as a dashed line.
    ok but i still dont get the part when you said (-2,-8), so do i just say if z= -2 -8i, and sub it in the inequality?
    so l z+6 l > l z+2+8i l => l -2 -8i +6 l > l -2 -8i +2+8i l so l 4 - 8i l > l 0 + 0i l then 4^2 + 8^2 > 0^2? is this the method or am i way off.
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    (Original post by cooldudeman)
    ok but i still dont get the part when you said (-2,-8), so do i just say if z= -2 -8i, and sub it in the inequality?
    so l z+6 l > l z+2+8i l => l -2 -8i +6 l > l -2 -8i +2+8i l so l 4 - 8i l > l 0 + 0i l then 4^2 + 8^2 > 0^2? is this the method or am i way off.
    I wouldn't actually do any working out. I'd say the point -2-8i clearly satisfy the inequality, hence it is in the desired region.

    I choose -2-8i since |z+2+8i| will be zero, and the other side of the inequality must be >0.
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    (Original post by ghostwalker)
    I wouldn't actually do any working out. I'd say the point -2-8i clearly satisfy the inequality, hence it is in the desired region.

    I choose -2-8i since |z+2+8i| will be zero, and the other side of the inequality must be >0.
    yeah i get ya, thanks
 
 
 
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