1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

(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

    Mark44

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: (Complex analysis). Show that the inequality holds
Loading...