if |z|>= 3 the least value of |z + 1/z| is?

- anonymous

if |z|>= 3 the least value of |z + 1/z| is?

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- hartnn

i think we need to use
|x|+|y|>|x+y|
but not sure.

- anonymous

But wat Does that Least Value For?

- hartnn

i m just trying....
if |z|>=3 what can u say about |1/z| ??

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## More answers

- anonymous

|z -(- 1/z)| > = |z| - |-1/z|

- anonymous

I Did nt Understand....wat is here Least..""

- hartnn

means |z+1/z| is > = 'some number'
u need to find that number.

- hartnn

but |z+1/z| <= |z|+|1/z|
how did u get > sign there ?

- anonymous

u see..i have inserted a -ve there....Since it asks for least value."

- Callisto

Hmm.. Am I right for this:
|z| ≥ 3
i.e. z ≥ 3
or z ≤ -3
?

- hartnn

i think z is complex here, isn't it @Yahoo! ??

- anonymous

Yes!

- hartnn

so |z|>= 3 represents region outside the circle with radius 3.....if i have not mis-interpreted

- Callisto

Wow! Something I don't know! I'm sorry!!

- anonymous

"OUTSIDE" hw

- hartnn

because of > sign, if < then inside.

- anonymous

Ok...)

- anonymous

@hartnn Any Idea....

- 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

i think we have to find the least dist b/w Circles

- anonymous

@hartnn i Did nt Understand.....Can u Explain...

- 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

lol...i Dont knw Hyperbola....

- 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

Important to note \(1/x\) is decreasing not negative.

- anonymous

Negative with respect to \(x\).

- 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

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

I was afk.
Complex Numbers?

- anonymous

afk. ===?

- anonymous

away from the keyboard.

- anonymous

yup. complex nos.

- anonymous

Yes.. it is Complex Numbers

- anonymous

let Z= x + iy and try to get value of mod 1/z

- anonymous

can u do it?

- anonymous

if mod z is r then ull get mod i/z as 1/r.....

- anonymous

I will be back, my trackpad isn't working somehow.

- anonymous

shouldn't it be this way? or is something wrong with this?
|dw:1347108277591:dw|

- hartnn

if its <= then its max value , not least value.

- anonymous

The answer should be 8/3

- hartnn

and yes,thats the correct way to find max value.

- anonymous

lol.....Can we Do like this
|z -(- 1/z)| > = |z| - |-1/z|

- anonymous

= 3 - 1/3 = 8/3

- hartnn

is this true ?
|x-y| >= |x|-|y|
if yes, then u can do it like that.

- hartnn

and i guess its true.

- anonymous

Yes.....

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