Find the range of this expression

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


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]


Homework Equations



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
[itex]6-2(z+\overline{z})-2(ω+\overline{ω})+4|ω|^{2}[/itex]
[itex]6-2(z+\overline{z})+2(ω+\overline{ω})[/itex]
Since [itex](ω+\overline{ω})=-1[/itex]
[itex]4-2(z+\overline{z})[/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:
 

Answers and Replies

  • #3
utkarshakash
Gold Member
855
13
How did you deduce that

?
Since ω is a cube root of unity therefore [itex]\overline{ω}[/itex] will be [itex]
ω^{2}[/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]
 
  • #4
631
0
Since ω is a cube root of unity therefore [itex]\overline{ω}[/itex] will be [itex]
ω^{2}[/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]
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.
 
  • #5
utkarshakash
Gold Member
855
13
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.
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
6+4Re(ω)+4Re(z)

Now what to do??
 
  • #6
631
0
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).
 
  • #7
utkarshakash
Gold Member
855
13
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).
Hey thanks. At last I got my answer. The complication arised only because of ambiguity in the question.
 
  • #8
631
0
Very good. :smile:
 

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