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Homework Help: Find the range of this expression

  1. Oct 11, 2012 #1


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    1. The problem statement, all variables and given/known data
    If |z|=1 and |ω-1|=1, where z, ω [itex]\in[/itex] C, then find the range of [itex] |2z-1|^{2}+|2ω-1|^{2}.[/itex]

    2. Relevant equations

    3. The attempt at a solution

    Since |ω-1|=1
    Squaring both sides and simplifying
    [itex] |ω|^{2}=ω+\overline{ω}[/itex]

    Also simplifying the expression given in the question
    Since [itex](ω+\overline{ω})=-1[/itex]
    Since [itex](z+\overline{z}) = 2Re(z)[/itex]
    Now the expression reduces to

    [itex] \large 4 \left\{ 1-Re(z) \right\} [/itex]

    Since |z|=1
    ∴Locus of z will be a circle with centre at origin and unit radius. So the max Re(z) can be 1 and min Re(z) can be -1. Substituting these in my expression for max and min I get [0,8] but the answer is [2,18]. :frown:
  2. jcsd
  3. Oct 11, 2012 #2
    How did you deduce that
  4. Oct 12, 2012 #3


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    Since ω is a cube root of unity therefore [itex]\overline{ω}[/itex] will be [itex]

    [itex] ω = \frac{-1}{2} + \frac{\sqrt{3}}{2}i and \overline{ω}=\frac{-1}{2} - \frac{\sqrt{3}}{2}i[/itex]

    Adding two I get [itex]ω+\overline{ω}=-1[/itex]
  5. Oct 12, 2012 #4
    Cube root of unity?? But the definition of ω you have in your original question doesn't comply with this.

    I think you're confusing the variable ω used in this question with the standard notation used for cube root of unity. For present question, ω has nothing to do with cube root of unity.
  6. Oct 12, 2012 #5


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    OK, I understand you. So correcting my mistake I am left with this

    [itex] 6+2(ω+\overline{ω})-2(z+\overline{z}) [/itex]

    which reduces to

    Now what to do??
  7. Oct 12, 2012 #6
    Now you have to find the maximum and minimum of Re(ω) and Re(z). In your first post, you have done so for Re(z), you can do the same for Re(ω).
  8. Oct 12, 2012 #7


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    Hey thanks. At last I got my answer. The complication arised only because of ambiguity in the question.
  9. Oct 12, 2012 #8
    Very good. :smile:
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