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Homework Help: (Complex analysis). Show that the inequality holds

  1. Jan 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Show that the inequality[tex]\left|\frac{z^2-2z+4}{3x+10}\right|\leq3[/tex]holds for all [itex] z\in\mathbb{C}[/itex] such that [itex]|z|=2[/itex]

    2. Relevant equations

    Triangle inequality

    3. The attempt at a solution

    I'm not really sure how to go about this. the x is throwing me off. Should I write it out with [itex]z=x+iy[/itex]?
  2. jcsd
  3. Jan 19, 2014 #2


    Staff: Mentor

    Could it be a typo and they really intended x to be z? That x really looks out of place to me.
  4. Jan 19, 2014 #3
    I don't know if it is a typo. If it is not then I could say that

    [tex]3x+10=Re(3z+10)[/tex][tex]|Re(3z+10)|\leq | 3z+10 | = 3|z|+10=16[/tex]
  5. Jan 19, 2014 #4
    on the top I say that [tex]|z^2-2z+4|\leq |z|^2+2|z|+4 = 16[/tex]
  6. Jan 19, 2014 #5
    not sure how that would help me...
  7. Jan 20, 2014 #6
    Must be a type: suppose x => -10/3 and y => sqrt(4-(10/3)^2)* i with z=x+yi then the denominator heads towards zero and the fraction goes to inf.
  8. Jan 20, 2014 #7
    An outline of a solution, assuming the d should be a z.

    Think about what the numerator and the denominator do to the circle mod z=2.

    The denominator converts it to a circle of radius 6, 10 units shifted to the right. Thus the distance from the origin ranges from 4 to 12.

    The numerator is a bit more tricky. Thinking of it as (z-1)^2 +3 you we that it first shifts the circle to the left one unit (making the distance to origin range from 1 to 3). Then you square it, which meAns the distance ranges from 1 to 9, with these extrema being on the real axis. Then you add 3 to get a distance range of 4 to 12.

    Thus both numerator and denominator have modulus between 4 and 12, so their ratio cannot be greater than 3
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