TargetIIT

Find the number of roots of the equation [im]http://codecogs.izyba.com/gif.latex?z%5E7+4z%5E2+11%3D0[/im] satisfying [im]http://codecogs.izyba.com/gif.latex?1%3C%7Cz%7C%3C2[/im]

All the roots satisfy the given condition.

is that from Mathematica, or any reasoning?

sir are there 4 roots ?
since the sum of roots is 0
is one root is z then -z is another root
                                           _                           _
also if z is one root then   z is also a root and -  z
   so four roots are there                                                                  
                                   
                                   

I dont know how many distinct roots there are. There is atleast one real root (any odd degree polynomial has this propery).

But how on earth did u get that if z is a root, so is -z?

Divide the given eqn on both sides by [im]http://codecogs.izyba.com/gif.latex?z%5E2[/im] to have

[im]http://codecogs.izyba.com/gif.latex?z%5E5+4+%5Cfrac%7B11%7D%7Bz%5E2%7D%3D0[/im]

Taking modulo on both sides we have [im]http://codecogs.izyba.com/gif.latex?%5Cleft%7Cz%5E5+4%20%5Cright%7C%3D%5Cfrac%7B11%7D%7B%7Cz%5E2%7C%7D%5Cle%20%7Cz%7C%5E5+4[/im] from triangle's inequality, from which we get the obvious reasoning that [im]http://codecogs.izyba.com/gif.latex?%7Cz%7C%3E1[/im].

Similarly we have [im]http://codecogs.izyba.com/gif.latex?%7Cz%7C%5E2%3D%5Cleft%7C%5Cfrac%7B11%7D%7Bz%5E5+4%7D%20%5Cright%7C%5Cle%20%5Cfrac%7B11%7D%7B%7Cz%7C%5E5+4%7D[/im]

From this we have that [im]http://codecogs.izyba.com/gif.latex?%7Cz%7C%3C2[/im]....thus any z satisfying the given eqn must have its modulo between 1 and 2.

@theprophet
now that you mentioned it, this is what Mathematica gives

[image]38562338.jpg[/image]

Not much useful.

As far as I am concerned, I used the Rouche's theorem (which should be familiar if any one does a course on complex analysis) which gives easily the required number of zeroes in annulus 1<|z| <2.

Rouche's theorem doesnt (directly) give the number of zeroes in an annulus.

You will first need soumik's result that we need |z|>1.

Soumik, your second inequality is incorrect. Would you like to try some more?

More or less its direct by Rouche's.
For |z|=2,
|4z[p]2[/p]+11| ≤ 4|z|[p]2[/p] + 11 =27 < 128 =|z[p]7[/p]|
So in the disc |z|<2, the polynomial z[p]7[/p] + 4z[p]2[/p] +11 has the same number of roots as z[p]7[/p], namely, 7 (counting multiplicities).

On the other hand, on the unit circle |z|=1, the least value of |4z[p]2[/p]+11| is 7 which is greater than 1=|z[p]7[/p]|. As such on |z|=1,
|z[p]7[/p]|<|4z[p]2[/p]+11|

Hence,  in the disc |z|<1, the polynomial z[p]7[/p] + 4z[p]2[/p] +11 has the same number of roots as 4z[p]2[/p]+11, namely, 0.

And, obviously there are no roots with |z|=1.

Hence, all the roots of z[p]7[/p] + 4z[p]2[/p] +11 lie in the annulus 1 < |z| < 2.

ok but only location of roots is proved ...but we have to find no.of roots

That's a 7th degree polynomial equation, so there are seven roots (counting multiplicities).

@soumik. Nice work, that was the approach I was looking for