At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
i think we need to use |x|+|y|>|x+y| but not sure.
But wat Does that Least Value For?
i m just trying.... if |z|>=3 what can u say about |1/z| ??
|z -(- 1/z)| > = |z| - |-1/z|
I Did nt Understand....wat is here Least..""
means |z+1/z| is > = 'some number' u need to find that number.
but |z+1/z| <= |z|+|1/z| how did u get > sign there ?
u see..i have inserted a -ve there....Since it asks for least value."
Hmm.. Am I right for this: |z| ≥ 3 i.e. z ≥ 3 or z ≤ -3 ?
i think z is complex here, isn't it @Yahoo! ??
so |z|>= 3 represents region outside the circle with radius 3.....if i have not mis-interpreted
Wow! Something I don't know! I'm sorry!!
because of > sign, if < then inside.
@hartnn Any Idea....
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
i think we have to find the least dist b/w Circles
@hartnn i Did nt Understand.....Can u Explain...
\[|z| \geq 3\implies z\leq-3\;or\;z\geq3\] \(\left|z + \frac1z\right|\) 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 \(|z|\geq3\).
lol...i Dont knw Hyperbola....
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 \(x>1\).
Important to note \(1/x\) is decreasing not negative.
Negative with respect to \(x\).
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....
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!
I was afk. Complex Numbers?
away from the keyboard.
yup. complex nos.
Yes.. it is Complex Numbers
let Z= x + iy and try to get value of mod 1/z
can u do it?
if mod z is r then ull get mod i/z as 1/r.....
I will be back, my trackpad isn't working somehow.
shouldn't it be this way? or is something wrong with this? |dw:1347108277591:dw|
if its <= then its max value , not least value.
The answer should be 8/3
and yes,thats the correct way to find max value.
lol.....Can we Do like this |z -(- 1/z)| > = |z| - |-1/z|
= 3 - 1/3 = 8/3
is this true ? |x-y| >= |x|-|y| if yes, then u can do it like that.
and i guess its true.