anonymous
  • anonymous
if |z|>= 3 the least value of |z + 1/z| is?
Mathematics
schrodinger
  • schrodinger
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hartnn
  • hartnn
i think we need to use |x|+|y|>|x+y| but not sure.
anonymous
  • anonymous
But wat Does that Least Value For?
hartnn
  • hartnn
i m just trying.... if |z|>=3 what can u say about |1/z| ??

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anonymous
  • anonymous
|z -(- 1/z)| > = |z| - |-1/z|
anonymous
  • anonymous
I Did nt Understand....wat is here Least..""
hartnn
  • hartnn
means |z+1/z| is > = 'some number' u need to find that number.
hartnn
  • hartnn
but |z+1/z| <= |z|+|1/z| how did u get > sign there ?
anonymous
  • anonymous
u see..i have inserted a -ve there....Since it asks for least value."
Callisto
  • Callisto
Hmm.. Am I right for this: |z| ≥ 3 i.e. z ≥ 3 or z ≤ -3 ?
hartnn
  • hartnn
i think z is complex here, isn't it @Yahoo! ??
anonymous
  • anonymous
Yes!
hartnn
  • hartnn
so |z|>= 3 represents region outside the circle with radius 3.....if i have not mis-interpreted
Callisto
  • Callisto
Wow! Something I don't know! I'm sorry!!
anonymous
  • anonymous
"OUTSIDE" hw
hartnn
  • hartnn
because of > sign, if < then inside.
anonymous
  • anonymous
Ok...)
anonymous
  • anonymous
@hartnn Any Idea....
hartnn
  • hartnn
i got something from yahoo answers , trying to understand it, see if that makes some sense to u..... http://in.answers.yahoo.com/question/index?qid=20100423221137AAAgV9K
anonymous
  • anonymous
i think we have to find the least dist b/w Circles
anonymous
  • anonymous
@hartnn i Did nt Understand.....Can u Explain...
anonymous
  • anonymous
\[|z| \geq 3\implies z\leq-3\;or\;z\geq3\] \(\left|z + \frac1z\right|\) is just a symmetrical hyperbola about Y-axis. If you're well versed with hyperbola you will get the idea of the graph with constraint \(|z|\geq3\).
anonymous
  • anonymous
lol...i Dont knw Hyperbola....
anonymous
  • anonymous
Okay. \[x + \frac1x\] \(x\) is ever increasing, so to find the minima of \(x + \frac1x\) you should find value for which \(x\) is minimum i.e 3 as \(1/x\) is decreasing for \(x>1\).
anonymous
  • anonymous
Important to note \(1/x\) is decreasing not negative.
anonymous
  • anonymous
Negative with respect to \(x\).
anonymous
  • anonymous
express Z= x+ iy.......and its mod value is suar root of x^2 +y^2.....now mod of Z+1 can be expressed as (x+1 )+ iy......and then solve it.....i know its very lengthy but it works when u have no other option....
anonymous
  • anonymous
If r is the modulus of z, then |z+ 1/z| >= r - 1/r. Now the minimum of r - 1/r for r>=3 is clearly 3 - 1/3 = 8/3. The minimum for |z+1/z| is attained if |z| = 3 and if z and 1/z have opposite directions. Take z = 3i so that z = 1/3i = - i / 3, and the minimum 8/3 is attained for that value. Can u Explain this!
anonymous
  • anonymous
I was afk. Complex Numbers?
anonymous
  • anonymous
afk. ===?
anonymous
  • anonymous
away from the keyboard.
anonymous
  • anonymous
yup. complex nos.
anonymous
  • anonymous
Yes.. it is Complex Numbers
anonymous
  • anonymous
let Z= x + iy and try to get value of mod 1/z
anonymous
  • anonymous
can u do it?
anonymous
  • anonymous
if mod z is r then ull get mod i/z as 1/r.....
anonymous
  • anonymous
I will be back, my trackpad isn't working somehow.
anonymous
  • anonymous
shouldn't it be this way? or is something wrong with this? |dw:1347108277591:dw|
hartnn
  • hartnn
if its <= then its max value , not least value.
anonymous
  • anonymous
The answer should be 8/3
hartnn
  • hartnn
and yes,thats the correct way to find max value.
anonymous
  • anonymous
lol.....Can we Do like this |z -(- 1/z)| > = |z| - |-1/z|
anonymous
  • anonymous
= 3 - 1/3 = 8/3
hartnn
  • hartnn
is this true ? |x-y| >= |x|-|y| if yes, then u can do it like that.
hartnn
  • hartnn
and i guess its true.
anonymous
  • anonymous
Yes.....

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