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Mathematics 7 Online
OpenStudy (anonymous):

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

hartnn (hartnn):

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

OpenStudy (anonymous):

But wat Does that Least Value For?

hartnn (hartnn):

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

OpenStudy (anonymous):

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

OpenStudy (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 ?

OpenStudy (anonymous):

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

OpenStudy (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! ??

OpenStudy (anonymous):

Yes!

hartnn (hartnn):

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

OpenStudy (callisto):

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

OpenStudy (anonymous):

"OUTSIDE" hw

hartnn (hartnn):

because of > sign, if < then inside.

OpenStudy (anonymous):

Ok...)

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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 \(|z|\geq3\).

OpenStudy (anonymous):

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

OpenStudy (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 \(x>1\).

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Negative with respect to \(x\).

OpenStudy (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....

OpenStudy (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!

OpenStudy (anonymous):

I was afk. Complex Numbers?

OpenStudy (anonymous):

afk. ===?

OpenStudy (anonymous):

away from the keyboard.

OpenStudy (anonymous):

yup. complex nos.

OpenStudy (anonymous):

Yes.. it is Complex Numbers

OpenStudy (anonymous):

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

OpenStudy (anonymous):

can u do it?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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